A key theme in real analysis is that of studying general functions or by first approximating them by “simpler” or “nicer” functions. But the precise class of “simple” or “nice” functions may vary from context to context. In measure theory, for instance, it is common to approximate measurable functions by indicator functions or simple functions. But in other parts of analysis, it is often more convenient to approximate rough functions by continuous or smooth functions (perhaps with compact support, or some other decay condition), or by functions in some algebraic class, such as the class of polynomials or trigonometric polynomials.
In order to approximate rough functions by more continuous ones, one of course needs tools that can generate continuous functions with some specified behaviour. The two basic tools for this are Urysohn’s lemma, which approximates indicator functions by continuous functions, and the Tietze extension theorem, which extends continuous functions on a subdomain to continuous functions on a larger domain. An important consequence of these theorems is the Riesz representation theorem for linear functionals on the space of compactly supported continuous functions, which describes such functionals in terms of Radon measures.
Sometimes, approximation by continuous functions is not enough; one must approximate continuous functions in turn by an even smoother class of functions. A useful tool in this regard is the Stone-Weierstrass theorem, that generalises the classical Weierstrass approximation theorem to more general algebras of functions.
As an application of this theory (and of many of the results accumulated in previous lecture notes), we will present (in an optional section) the commutative Gelfand-Neimark theorem classifying all commutative unital -algebras.
— 1. Urysohn’s lemma —
Let be a topological space. An indicator function in this space will not typically be a continuous function (indeed, if is connected, this only happens when is the empty set or the whole set). Nevertheless, for certain topological spaces, it is possible to approximate an indicator function by a continuous function, as follows.
Lemma 1 (Urysohn’s lemma) Let be a topological space. Then the following are equivalent:
- (i) Every pair of disjoint closed sets in can be separated by disjoint open neighbourhoods , .
- (ii) For every closed set in and every open neighbourhood of , there exists an open set and a closed set such that .
- (iii) For every pair of disjoint closed sets in , there exists a continuous function which equals on and on .
- (iv) For every closed set in and every open neighbourhood of , there exists a continuous function such that for all .
A topological space which obeys any (and hence all) of (i-iv) is known as a normal space; definition (i) is traditionally taken to be the standard definition of normality. We will give some examples of normal spaces shortly.
Proof: The equivalence of (iii) and (iv) is clear, as the complement of a closed set is an open set and vice versa. The equivalence of (i) and (ii) follows similarly.
To deduce (i) from (iii), let be disjoint closed sets, let be as in (iii), and let be the open sets and .
The only remaining task is to deduce (iv) from (ii). Suppose we have a closed set and an open set with . Applying (ii), we can find an open set and a closed set such that
Applying (ii) two more times, we can find more open sets and closed sets such that
Iterating this process, we can construct open sets and closed sets for every dyadic rational in such that for all , and for any .
If we now define , where ranges over dyadic rationals between and , and with the convention that the empty set has sup and inf , one easily verifies that the sets and are open for every real number , and so is continuous as required.
The definition of normality is very similar to the Hausdorff property, which separates pairs of points instead of closed sets. Indeed, if every point in is closed (a property known as the property), then normality clearly implies the Hausdorff property. The converse is not always true, but (as the term suggests) in practice most topological spaces one works with in real analysis are normal. For instance:
Exercise 1 Show that every metric space is normal.
Exercise 2 Let be a Hausdorff space.
- Show that a compact subset of and a point disjoint from that set can always be separated by open neighbourhoods.
- Show that a pair of disjoint compact subsets of can always be separated by open neighbourhoods.
- Show that every compact Hausdorff space is normal.
Exercise 3 Let be the real line with the usual topology , and let be the topology on generated by and the set consisting only of the rationals ; in other words, is the coarsest refinement of the usual topology that makes the set of rationals an open set. Show that is Hausdorff, with every point closed, but is not normal.
The above example was a simple but somewhat artificial example of a non-normal space. One can create more “natural” examples of non-normal Hausdorff spaces (with every point closed), but establishing non-normality becomes more difficult. The following example is due to Stone.
- Show that is Hausdorff, and every point is closed.
- For , let be the set of all tuples such that for all outside of a countable set, and such that is injective on this finite set (i.e. there do not exist distinct such that ). Show that are disjoint and closed.
- Show that given any open neighbourhood of , there exists disjoint finite subsets of and an injective function such that for any , any such that for all and is identically on , lies in .
- Show that any open neighbourhood of and any open neighbourhood of necessarily intersect, and so is not normal.
- Conclude that with the product topology is not normal.
The property of being normal is a topological one, thus if one topological space is normal, then any other topological space homeomorphic to it is also normal. However, (unlike, say, the Hausdorff property), the property of being normal is not preserved under passage to subspaces:
Exercise 5 Given an example of a subspace of a normal space which is not normal. (Hint: use Exercise 4, possibly after replacing with a homeomorphic equivalent.)
Let be the space of real continuous compactly supported functions on . Urysohn’s lemma generates a large number of useful elements of , in the case when is locally compact Hausdorff:
Exercise 6 Let be a locally compact Hausdorff space, let be a compact set, and let be an open neighbourhood of . Show that there exists such that for all . (Hint: First use the local compactness of to find a neighbourhood of with compact closure; then restrict to this neighbourhood. The closure of is now a compact set; restrict everything to this set, at which point the space becomes normal.)
One consequence of this exercise is that tends to be dense in many other function spaces. We give an important example here:
Definition 2 (Radon measure) Let be a locally compact Hausdorff space that is also -compact, and let be the Borel -algebra. An (unsigned) Radon measure is a unsigned measure with the following properties:
Example 1 Lebesgue measure on is a Radon measure, as is any absolutely continuous unsigned measure , where . More generally, if is Radon and is a finite unsigned measure which is absolutely continuous with respect to , then is Radon. On the other hand, counting measure on is not Radon (it is not locally finite). It is possible to define Radon measures on Hausdorff spaces that are not -compact or locally compact, but the theory is more subtle and will not be considered here. We will study Radon measures more thoroughly in the next section.
Proposition 3 Let be a locally compact Hausdorff space that is also -compact, and let be a Radon measure on . Then for any , is a dense subset in (real-valued) . In other words, every element of can be expressed as a limit (in ) of continuous functions of compact support.
Proof: Since continuous functions of compact support are bounded, and compact sets have finite measure, we see that is a subspace of . We need to show that the closure of this space contains all of .
Let be a Borel set of finite measure. Applying inner and outer regularity, we can find a sequence of compact sets and open sets such that . Applying Exercise 6, we can then find such that . In particular, this implies (by the squeeze theorem) that converges in to (here we use the finiteness of ); thus lies in for any measurable set . By linearity, all simple functions lie in ; taking closures, we see that any function lies in , as desired.
Of course, the real-valued version of the above proposition immediately implies a complex-valued analogue. On the other hand, the claim fails when :
Exercise 7 Let be a locally compact Hausdorff space that is -compact, and let be a Radon measure. Show that the closure of in is , the space of continuous real-valued functions which vanish at infinity (i.e. for every there exists a compact set such that for all ). Thus, in general, is not dense in .
Thus we see that the norm is strong enough to preserve continuity in the limit, whereas the norms are (locally) weaker and permit discontinuous functions to be approximated by continuous ones.
Another important consequence of Urysohn’s lemma is the Tietze extension theorem:
Theorem 4 (Tietze extension theorem) Let be a normal topological space, let be a bounded interval, let be a closed subset of , and let be a continuous function. Then there exists a continuous function which extends , i.e. for all .
Proof: It suffices to find an continuous extension taking values in the real line rather than in , since one can then replace by (note that min and max are continuous operations).
Let be the restriction map . This is clearly a continuous linear map; our task is to show that it is surjective, i.e. to find a solution to the equation for each . We do this by the standard analysis trick of getting an approximate solution to first, and then using iteration to boost the approximate solution to an exact solution.
Let have sup norm , thus takes values in . To solve the problem , we approximate by . By Urysohn’s lemma, we can find a continuous function such that on the closed set and on the closed set . Now, is not quite equal to ; but observe from construction that has sup norm .
Scaling this fact, we conclude that, given any , we can find a decomposition , where and .
Starting with any , we can now iterate this construction to express for all , where and . As is a Banach space, we see that converges absolutely to some limit , and that , as desired.
Remark 1 Observe that Urysohn’s lemma can be viewed the special case of the Tietze extension theorem when is the union of two disjoint closed sets, and is equal to on one of these sets and equal to on the other.
Remark 2 One can extend the Tietze extension theorem to finite-dimensional vector spaces: if is a closed subset of a normal vector space and is bounded and continuous, then one has a bounded continuous extension . Indeed, one simply applies the Tietze extension theorem to each component of separately. However, if the range space is replaced by a space with a non-trivial topology, then there can be topological obstructions to continuous extension. For instance, a map from a two-point set into a topological space is always continuous, but can be extended to a continuous map if and only if and lie in the same path-connected component of . Similarly, if is a map from the unit circle into a topological space , then a continuous extension from to exists if and only if the closed curve is contractible to a point in . These sorts of questions require the machinery of algebraic topology to answer them properly, and are beyond the scope of this course.
There are analogues for the Tietze extension theorem in some other categories of functions. For instance, in the Lipschitz category, we have
Exercise 8 Let be a metric space, let be a subset of , and let be a Lipschitz continuous map with some Lipschitz constant (thus for all ). Show that there exists an extension of which is Lipschitz continuous with the same Lipschitz constant . (Hint: A “greedy” algorithm will work here: pick to be as large as one can get away with (or as small as one can get away with.))
One can also remove the requirement that the function be bounded in the Tietze extension theorem:
Exercise 9 Let be a normal topological space, let be a closed subset of , and let be a continuous map (not necessarily bounded). Then there exists an extension of which is still continuous. (Hint: first “compress” to be bounded by working with, say, (other choices are possible), and apply the usual Tietze extension theorem. There will be some sets in which one cannot invert the compression function, but one can deal with this by a further appeal to Urysohn’s lemma to damp the extension out on such sets.)
There is also a locally compact Hausdorff version of the Tietze extension theorem:
Proposition 3 shows that measurable functions in can be approximated by continuous functions of compact support (cf. Littlewood’s second principle). Another approximation result in a similar spirit is Lusin’s theorem:
Theorem 5 (Lusin’s theorem) Let be a locally compact Hausdorff space that is -compact, and let be a Radon measure. Let be a measurable function supported on a set of finite measure, and let . Then there exists which agrees with outside of a set of measure at most .
Proof: Observe that as is finite everywhere, it is bounded outside of a set of arbitrarily small measure. Thus we may assume without loss of generality that is bounded. Similarly, as is -compact (or by inner regularity), the support of differs from a compact set by a set of arbitrarily small measure; so we may assume that is also supported on a compact set . By Exercise 10, it then suffices to show that is continuous on the complement of an open set of arbitrarily small measure; by outer regularity, we may delete the adjective “open” from the preceding sentence.
As is bounded and compactly supported, lies in for every , and using Proposition 3 and Chebyshev’s inequality, it is not hard to find, for each , a function which differs from by at most outside of a set of measure at most (say). In particular, converges uniformly to outside of a set of measure at most , and is therefore continuous outside this set. The claim follows.
Another very useful application of Urysohn’s lemma is to create partitions of unity.
Lemma 6 (Partitions of unity) Let be a normal topological space, and let be a collection of closed sets that cover . For each , let be an open neighbourhood of , which are finitely overlapping in the sense that each has a neighbourhood that intersects at most finitely many of the . Then there exists a continuous function supported on for each such that for all .
If is locally compact Hausdorff instead of normal, and the are compact, then one can take the to be compactly supported.
Proof: Suppose first that is normal. By Urysohn’s lemma, one can find a continuous function for each which is supported on and equals on the closed set . Observe that the function is well-defined, continuous and bounded below by . The claim then follows by setting .
The final claim follows by using Exercise 6 instead of Urysohn’s lemma.
- Show that an indicator function is upper semi-continuous if and only if is closed, and lower semi-continuous if and only if is open.
- If is normal, show that a function is upper semi-continuous if and only if for all , and lower semi-continuous if and only if for all , where we write if for all .
— 2. The Riesz representation theorem —
Let be a locally compact Hausdorff space which is also -compact. In Definition 2 we defined the notion of a Radon measure. Such measures are quite common in real analysis. For instance, we have the following result.
Proof: As is finite, it is locally finite, so it suffices to show inner and outer regularity. Let be the collection of all Borel subsets of such that
It will then suffice to show that every Borel set lies in (note that as is compact, a subset of is closed if and only if it is compact).
Clearly contains the empty set and the whole set , and is closed under complements. It is also closed under finite unions and intersections. Indeed, given two sets , we can find a sequences , of closed sets and open sets such that and . Since
we have (by monotonicity of ) that
and so .
One can also show that is closed under countable disjoint unions and is thus a -algebra. Indeed, given disjoint sets and , pick a closed and open such that ; then
for any , and the claim follows from the squeeze test.
To finish the claim it suffices to show that every open set lies in . For this it will suffice to show that is a countable union of closed sets. But as is a compact metric space, it is separable (Lemma 4 from Notes 10), and so has a countable dense subset . One then easily verifies that every point in the open set is contained in a closed ball of rational radius centred at one of the that is in turn contained in ; thus is the countable union of closed sets as desired.
This result can be extended to more general spaces than compact metric spaces, for instance to Polish spaces (provided that the measure remains finite). For instance:
Exercise 12 Let be a locally compact metric space which is -compact, and let be an unsigned Borel measure which is finite on every compact set. Show that is a Radon measure.
When the assumptions of are weakened, then it is possible to find locally finite Borel measures that are not Radon measures, but they are somewhat pathological in nature.
Exercise 13 Let be a locally compact Hausdorff space which is -compact, and let be a Radon measure. Define a set to be a countable union of closed sets, and a set to be a countable intersection of open sets. Show that every Borel set can be expressed as the union of an set and a null set, and as a set with a null subset removed.
If is a Radon measure on , then we can define the integral for every , since assigns every compact set a finite measure. Furthermore, is a linear functional on which is positive in the sense that whenever is non-negative. If we place the uniform norm on , then is continuous if and only if is finite; but we will not use continuity for now, relying instead on positivity.
The fundamentally important Riesz representation theorem for such spaces asserts that this is the only way to generate such linear functionals:
Theorem 8 (Riesz representation theorem for , unsigned version) Let be a locally compact Hausdorff space which is also -compact. Let be a positive linear functional. Then there exists a unique Radon measure on such that .
Remark 3 The -compactness hypothesis can be dropped (after relaxing the inner regularity condition to only apply to open sets, rather than to all sets); but I will restrict attention here to the -compact case (which already covers a large fraction of the applications of this theorem) as the argument simplifies slightly.
Proof: We first prove the uniqueness, which is quite easy due to all the properties that Radon measures enjoy. Suppose we had two Radon measures such that ; in particular, we have
for all . Now let be a compact set, and let be an open neighbourhood of . By Exercise 6, we can find with ; applying this to (1), we conclude that
for all . Now let be a compact set, and let be an open neighbourhood of . By Exercise 6, we can find with ; applying this to (1), we conclude that
Taking suprema in and using inner regularity, we conclude that ; exchanging and we conclude that and agree on open sets; by outer regularity we then conclude that and agree on all Borel sets.
Now we prove existence, which is significantly trickier. We will initially make the simplifying assumption that is compact (so in particular ), and remove this assumption at the end of the proof.
Observe that is monotone on , thus whenever .
We would like to define the measure on Borel sets by defining . This does not work directly, because is not continuous. To get around this problem we shall begin by extending the functional to the class of bounded lower semi-continuous non-negative functions. We define for such functions by the formula
(cf. Exercise 11). This definition agrees with the existing definition of in the case when is continuous. Since is finite and is monotone, one sees that is finite (and non-negative) for all . One also easily sees that is monotone on : whenever and , and homogeneous in the sense that for all and . It is also easy to verify the super-additivity property for ; this simply reflects the linearity of on , together with the fact that if and , then .
We now complement the super-additivity property with a countably sub-additive one: if is a sequence, and is such that for all , then .
Pick a small . It will suffice to show that (say) whenever is such that , and denotes a quantity bounded in magnitude by , where is a quantity that is independent of .
Fix . For every , we can find a neighbourhood of such that for all ; we can also find such that . By shrinking if necessary, we see from the lower semicontinuity of the and that we can also ensure that for all and .
By normality, we can find open neighbourhoods of whose closure lies in . The form an open cover of . Since we are assuming to be compact, we can thus find a finite subcover of . Applying Lemma 6, we can thus find a partition of unity , where each is supported on .
Let be such that . Then we can write . If is in this sum, then , and thus (for small enough) , and hence . We can then write
(here we use the fact that and that the continuous compactly supported function is bounded). Observe that only finitely many summands are non-zero. We conclude that
(here we use that and so is finite). On the other hand, for any and any , the expression
is bounded from above by
since and , this is bounded above in turn by
We conclude that
and the sub-additivity claim follows.
Combining sub-additivity and super-additivity we see that is additive: for .
Now that we are able to integrate lower semi-continuous functions, we can start defining the Radon measure . When is open, we define by
which is well-defined and non-negative since is bounded, non-negative and lower semi-continuous. When is closed we define by complementation:
this is compatible with the definition of on open sets by additivity of , and is also non-negative. The monotonicity of implies monotonicity of : in particular, if a closed set lies in an open set , then .
Given any set , define the outer measure
and the inner measure
thus . We call a set measurable if . By arguing as in the proof of Theorem 7, we see that the class of measurable sets is a Boolean algebra. Next, we claim that every open set is measurable. Indeed, unwrapping all the definitions we see that
Each in this supremum is supported in some closed subset of , and from this one easily verifies that . Similarly, every closed set is measurable. We can now extend to measurable sets by declaring when is measurable; this is compatible with the previous definitions of .
Next, let be a countable sequence of disjoint measurable sets. Then for any , we can find open neighbourhoods of and closed sets in such that and . Using the sub-additivity of on , we have . Similarly, from the additivity of we have . Letting , we conclude that is measurable with . Thus the Boolean algebra of measurable sets is in fact a -algebra, and is a countably additive measure on it. From construction we also see that it is finite, outer regular, and inner regular, and therefore is a Radon measure. The only remaining thing to check is that for all . If is a finite non-negative linear combination of indicator functions of open sets, the claim is clear from the construction of and the additivity of on ; taking uniform limits, we obtain the claim for non-negative continuous functions, and then by linearity we obtain it for all functions.
This concludes the proof in the case when is compact. Now suppose that is -compact. Then we can find a partition of unity into continuous compactly supported functions , with each being contained in the support of finitely many . (Indeed, from -compactness and the locally compact Hausdorff property one can find a nested sequence of compact sets, with each in the interior of , such that . Using Exercise 6, one can find functions that equal on and are supported on ; now take and .) Observe that for all . From the compact case we see that there exists a finite Radon measure such that for all ; setting one can verify (using the monotone convergence theorem) that obeys the required properties.
Remark 4 One can also construct the Radon measure using the Carátheodory extension theorem; this proof of the Riesz representation theorem can be found in many real analysis texts. A third method is to first create the space by taking the completion of with respect to the norm , and then define . It seems to me that all three proofs are about equally lengthy, and ultimately rely on the same ingredients; they all seem to have their strengths and weaknesses, and involve at least one tricky computation somewhere (in the above argument, the most tricky thing is the countable subadditivity of on lower semicontinuous functions). I have yet to find a proof of this theorem which is both clean and conceptual, and would be happy to learn of other proofs of this theorem.
Remark 5 One can use the Riesz representation theorem to provide an alternate construction of Lebesgue measure, say on . Indeed, the Riemann integral already provides a positive linear functional on , which by the Riesz representation theorem must come from a Radon measure, which can be easily verified to assign the value to every interval and thus must agree with Lebesgue measure. The same approach lets one define volume measures on manifolds with a volume form.
Exercise 14 Let be a locally compact Hausdorff space which is -compact, and let be a Radon measure. For any non-negative Borel measurable function , show that
Similarly, for any non-negative lower semi-continuous function , show that
Now we consider signed functionals on , which we now turn into a normed vector space using the uniform norm. The key lemma here is the following variant of the Jordan decomposition theorem.
Proof: For , we define
Clearly for ; one also easily verifies the homogeneity property and super-additivity property for and . On the other hand, if are such that , then we can decompose for some with and ; for instance we can take and . From this we can complement super-additivity with sub-additivity and conclude that .
Every function in can be expressed as the difference of two functions in . From the additivity and homogeneity of on we may thus extend uniquely to be a linear functional on . Since is bounded on , we see that is also. If we then define , one quickly verifies all the required properties.
Exercise 15 Show that among all possible choices for the functionals appearing in the above lemma, there is a unique choice which is minimal in the sense that for any other functionals obeying the conclusions of the lemma, one has and for all .
Define a signed Radon measure on a -compact, locally compact Hausdorff space to be a signed Borel measure whose positive and negative variations are both Radon. It is easy to see that a signed Radon measure generates a linear functional on as before, and is continuous if is finite. We have a converse:
Exercise 16 (Riesz representation theorem, signed version) Let be a locally compact Hausdorff space which is also -compact, and let be a continuous linear functional. Then there exists a unique signed finite Radon measure such that . (Hint: combine Theorem 8 with Lemma 9.)
The space of signed finite Radon measures on is denoted , or for short.
Exercise 17 Show that the space , with the total variation norm , is a real Banach space, which is isomorphic to the dual of both and its completion , thus
Remark 6 Note that the previous exercise generalises the identifications from previous notes. For compact Hausdorff spaces , we have , and thus . For locally compact Hausdorff spaces that are -compact but not compact, we instead have , where is the Stone-Cech compactification of , which we will discuss in later notes.
Remark 7 One can of course also define complex Radon measures to be those complex finite Borel measures whose real and imaginary parts are signed Radon measures, and define to be the space of all such measures; then one has analogues of the above identifications. We omit the details.
Exercise 18 Let be two locally compact Hausdorff spaces that are also -compact, and let be a continuous map. If is an unsigned finite Radon measure on , show that the pushforward measure on , defined by , is a Radon measure on . What happens for infinite measures? Establish the same fact for signed finite Radon measures.
Let be locally compact Hausdorff and -compact. As is equivalent to the dual of the Banach space , it acquires a weak* topology (see Notes 11), known as the vague topology. A sequence of Radon measures then converges vaguely to a limit if and only if for all .
Exercise 19 Let be Lebesgue measure on the real line (with the usual topology).
Exercise 20 Let be locally compact Hausdorff and -compact. Show that for every unsigned Radon measure , the map defined by sending to the measure is an isometry, thus can be identified with a subspace of . Show that this subspace is closed in the norm topology, but give an example to show that it need not be closed in the vague topology. Show that , where ranges over all unsigned Radon measures on ; thus one can think of as many ‘s “glued together”.
Exercise 21 Let be a locally compact Hausdorff space which is -compact. Let be a sequence of functions, and let be another function. Show that converges weakly to in if and only if the are uniformly bounded and converge pointwise to .
Exercise 22 Let be a locally compact metric space which is -compact.
- Show that the space of finitely supported measures in is a dense subset of in the vague topology.
- Show that a Radon probability measure in can be expressed as the vague limit of a sequence of discrete (i.e. finitely supported) probability measures.
— 3. The Stone-Weierstrass theorem —
We have already seen how rough functions (e.g. functions) can be approximated by continuous functions. Now we study in turn how continuous functions can be approximated by even more special functions, such as polynomials. The natural topology to work with here is the uniform topology (since uniform limits of continuous functions are continuous).
For non-compact spaces, such as , it is usually not possible to approximate continuous functions uniformly by a smaller class of functions. For instance, the function cannot be approximated uniformly by polynomials on , since is bounded, the only bounded polynomials are the constants, and constants cannot converge to anything other than another constant. On the other hand, on a compact domain such as , one can easily approximate uniformly by polynomials, for instance by using Taylor series. So we will focus instead on compact Hausdorff spaces such as , in which continuous functions are automatically bounded.
The space of (real-valued) polynomials is a subspace of the Banach space . But it is also closed under pointwise multiplication , making an algebra, and not merely a vector space. We can then rephrase the classical Weierstrass approximation theorem as the assertion that is dense in .
One can then ask the more general question of when a sub-algebra of – i.e. a subspace closed under pointwise multiplication – is dense. Not every sub-algebra is dense: the algebra of constants, for instance, will not be dense in when has at least two points. Another example in a similar spirit: given two distinct points in , the space is a sub-algebra of , but it is not dense, because it is already closed, and cannot separate and in the sense that it cannot produce a function that assigns different values to and .
The remarkable Stone-Weierstrass theorem shows that this inability to separate points is the only obstruction to density, at least for algebras with the identity.
Theorem 10 (Stone-Weierstrass theorem, real version) Let be a compact Hausdorff space, and let be a sub-algebra of which contains the constant function and separates points (i.e. for every distinct , there exists at least one in such that . Then is dense in .
Remark 8 Observe that this theorem contains the Weierstrass approximation theorem as a special case, since the algebra of polynomials clearly separates points. Indeed, we will use (a very special case) of the Weierstrass approximation theorem in the proof.
Proof: It suffices to verify the claim for algebras which are closed in the topology, since the claim follows in the general case by replacing with its closure (note that the closure of an algebra is still an algebra).
Observe from the Weierstrass approximation theorem that on any bounded interval , the function can be expressed as the uniform limit of polynomials ; one can even write down explicit formulae for such a , though we will not need such formulae here. Since continuous functions on the compact space are bounded, this implies that for any , the function is the uniform limit of polynomial combinations of . As is an algebra, the lie in ; as is closed; we see that lies in .
Using the identities , , we conclude that is a lattice in the sense that one has whenever .
Now let and . We would like to find such that for all .
Given any two points , we can at least find a function such that and ; this follows since the vector space separates points and also contains the identity function (the case needs to be treated separately). We now use these functions to build the approximant . First, observe from continuity that for every there exists an open neighbourhood of such that for all . By compactness, for any fixed we can cover by a finite number of these . Taking the max of all the associated to this finite subcover, we create another function such that and for all . By continuity, we can find an open neighbourhood of such that for all . Again applying compactness, we can cover by a finite number of the ; taking the min of all the associated to this finite subcover we obtain with for all , and the claim follows.
There is an analogue of the Stone-Weierstrass theorem for algebras that do not contain the identity:
The Stone-Weierstrass theorem is not true as stated in the complex case. For instance, the space of complex-valued functions on the closed unit disk has a closed proper sub-algebra that separates points, namely the algebra of functions in that are holomorphic on the interior of this disk. Indeed, by Cauchy’s theorem and its converse (Morera’s theorem), a function lies in if and only if for every closed contour in , and one easily verifies that this implies that is closed; meanwhile, the holomorphic function separates all points. However, the Stone-Weierstrass theorem can be recovered in the complex case by adding one further axiom, namely that the algebra be closed under conjugation:
Exercise 24 (Stone-Weierstrass theorem, complex version) Let be a compact Hausdorff space, and let be a complex sub-algebra of which contains the constant function , separates points, and is closed under the conjugation operation . Then is dense in .
Exercise 25 Let be the space of trigonometric polynomials on the unit circle , where and the are complex numbers. Show that is dense in (with the uniform topology), and that is dense in (with the topology) for all .
Exercise 26 Let be a locally compact Hausdorff space that is -compact, and let be a sub-algebra of which separates points and contains the identity function. Show that for every function there exists a sequence which converges to uniformly on compact subsets of .
Exercise 27 Let be compact Hausdorff spaces. Show that every function can be expressed as the uniform limit of functions of the form , where and .
Exercise 28 Let be a family of compact Hausdorff spaces, and let be the product space (with the product topology). Let . Show that can be expressed as the uniform limit of continuous functions , each of which only depend on finitely many of the coordinates in , thus there exists a finite subset of and a continuous function such that for all .
One useful application of the Stone-Weierstrass theorem is to demonstrate separability of spaces such as .
Proposition 11 Let be a compact metric space. Then and are separable.
Proof: It suffices to show that is separable. By Lemma 4 of Notes 10, has a countable dense subset . By Urysohn’s lemma, for each we can find a function which equals on and is supported on . The can then easily be verified to separate points, and so by the Stone-Weierstrass theorem, the algebra of polynomial combinations of the in are dense; this implies that the algebra of rational polynomial combinations of the are dense, and the claim follows.
Combining this with the Riesz representation theorem and the sequential Banach-Alaoglu theorem, we obtain
Corollary 12 If is a compact metric space, then the closed unit ball of is sequentially compact in the vague topology.
Corollary 13 (Prokhorov’s theorem, compact case) Let be a compact metric space, and let be a sequence of Borel (hence Radon) probability measures on . Then there exists a subsequence of which converge vaguely to another Borel probability measure .
Exercise 29 (Prokhorov’s theorem, non-compact case) Let be a locally compact metric space which is -compact, and let be a sequence of Borel probability measures. We assume that the sequence is tight, which means that for every there exists a compact set such that for all . Show that there is a subsequence of which converges vaguely to another Borel probability measure . If tightness is not assumed, show that there is a subsequence which converges vaguely to a non-negative Borel measure , but give an example to show that this measure need not be a probability measure.
This theorem can be used to establish Helly’s selection theorem:
Exercise 30 (Helly’s selection theorem) Let be a sequence of functions whose total variation is uniformly bounded in , and which is bounded at one point (i.e. is bounded). Show that there exists a subsequence of which converges pointwise almost everywhere on compact subsets of . (Hint: one can deduce this from Prokhorov’s theorem using the fundamental theorem of calculus for functions of bounded variation.)
— 4. The commutative Gelfand-Naimark theorem (optional) —
One particularly beautiful application of the machinery developed in the last few notes is the commutative Gelfand-Naimark theorem, that classifies commutative -algebras, and is of importance in spectral theory, operator algebras, and quantum mechanics.
Definition 14 A complex Banach algebra is a complex Banach space which is also a complex algebra, such that for all . An algebra is unital if it contains a multiplicative identity , and commutative if for all . A -algebra is a complex Banach algebra with an anti-homomorphism map from to (thus , , and for and ) which is an isometry (thus for all ), an involution (thus for all ), and obeys the identity for all .
A homomorphism between two -algebras is a continuous algebra homomorphism such that for all . An isomorphism is an homomorphism whose inverse exists and is also a homomorphism; two -algebras are isomorphic if there exists an isomorphism between them.
Exercise 31 If is a Hilbert space, and is the algebra of bounded linear operators on this space, with the adjoint map and the operator norm, show that is a unital -algebra (not necessarily commutative). Indeed, one can think of -algebras as an abstraction of a space of bounded linear operators on a Hilbert space (this is basically the content of the non-commutative Gelfand-Naimark theorem, which we will not discuss here).
The remarkable (unital commutative) Gelfand-Naimark theorem asserts the converse statement to Exercise 32:
There are analogues of this theorem for non-unital or non-commutative -algebras, but for simplicity we shall restrict attention to the unital commutative case. We first need some spectral theory.
Exercise 33 Let be a unital Banach algebra. Show that if is such that , then is invertible. (Hint: use Neumann series.) Conclude that the space of invertible elements of is open.
Define the spectrum of an element to be the set of all such that is not invertible.
Exercise 35 (Beurling-Gelfand spectral radius formula) If is a unital Banach algebra and , show that is non-empty with . (Hint: To get the upper bound, observe that if is invertible for some , then so is , then use Exercise 34. To get the lower bound, first observe that for any , the function is holomorphic on the complement of , which is already enough (with Liouville’s theorem) to show that is non-empty. Let be arbitrary, then use Laurent series to show that for all and some independent of . Then divide by and use the uniform boundedness principle to conclude.)
for all and . Conclude that any homomorphism between -algebras has operator norm at most . Also conclude that
when is self-adjoint.
The next important concept is that of a character.
Definition 16 Let be a unital commutative -algebra. A character of is be an element in the dual Banach space such that , , and for all ; equivalently, a character is a homomorphism from to (viewed as a (unital) algebra). We let be the space of all characters; this space is known as the spectrum of .
Exercise 37 If is a unital commutative -algebra, show that is a compact Hausdorff subset of in the weak-* topology. (Hint: first use the spectral radius formula to show that all characters have operator norm , then use the Banach-Alaoglu theorem.)
Exercise 38 Define an ideal of a unital commutative -algebra to be a proper subspace of such that for all and . Show that if , then the kernel is a maximal ideal in ; conversely, if is a maximal ideal in , show that is closed, and there is exactly one such that . Thus the spectrum of can be canonically identified with the space of maximal ideals in .
Exercise 39 Let be a compact Hausdorff space, and let be the -algebra . Show that for each , the operation is a character of . Show that the map is a homeomorphism from to ; thus the spectrum of can be canonically identified with . (Hint: use Exercise 23 to show the surjectivity of , Urysohn’s lemma to show injectivity, and Corollary 2 of Notes 10 to show the homeomorphism property.)
Inspired by the above exercise, we define the Gelfand representation , by the formula .
Exercise 41 Let be a non-invertible element of a unital commutative -algebra . Show that vanishes at some . (Hint: the set is a proper ideal of , and thus by Zorn’s lemma is contained in a maximal ideal.)