Notes on Chapter 5 (Examples of Banach Space Techniques) of Walter Rudin's Real and Complex Analysis.

Chapter 5

The concept of a Banach space, i.e. a complete normed linear space, is introduced as a direct generalization of a Hilbert space (as the norm does not necessarily have to arise from an inner product). This does mean that we cannot necessarily define the notion of 'orthogonal' in this context, but gives us a little more freedom.

The norm of of a linear transform \(\Lambda\) between normed spaces is introduced, defined as $$\|\Lambda\| = \sup\{\|\Lambda x\|: x\in X, \|x\|\leq 1\}$$ which corresponds to the maximum amount a linear transformation can scale a vector. Equivalently this can be stated as the smallest number such that the inequality \(\|\Lambda x\|\leq\|\Lambda\|\|x\|\) holds for all \(x\). If the norm is finite, the linear operator is called bounded.

In particular, one can show that a linear transform is a bounded iff it is continuous at any point.

The chapter goes on to discuss Baire's Theorem and several of its consequences, which are an important way in which the completeness of Banach spaces is used.

Baire Category Theorem: If \(X\) is a complete metric space, the intersection of every countable collection of dense open subsets of \(X\) is dense in \(X\). □

The proof goes by inductively selecting a point in the intersection of the first $n$ subsets and an arbitrary open set using the fact that they are dense. The points are selected to form a Cauchy sequence, and by completeness a limit exists, which can be shown to be in the originally selected arbitrary open set.

Corollary: In a complete metric space, the intersection of any countable collection of dense \(G_\delta\)'s is again a dense \(G_\delta\). □

The BCT gives rise to a few other useful theorems:

Banach-Steinhaus Theorem (Uniform Boundedness Principle): Suppose \(X\) is a Banach space, \(Y\) is a normed linear space, and \(\{\Lambda_\alpha\}\) is a collection of bounded linear transformations of \(X\) into \(Y\), where \(\alpha\) ranges over some index set \(A\). The one of the following holds:

• There is an \(M<\infty\) such that \(\|\Lambda_\alpha\|\leq M\) for all \(\alpha\)
• \(\sup_\alpha \|\Lambda_\alpha x\| = \infty\) for all \(x\) belonging to some dense \(G_\delta\) in \(X\). □

Open Mapping Theorem: Let \(U\) and \(V\) be the open unit balls of the Banach spaces \(X\) and \(Y\). To every bounded surjective linear transformation of \(X\) onto \(Y\) there is a \(\delta > 0\) s.t. \(\Lambda(U)\supset\delta V\). □

In other words, the image of every open set is open. Alternatively, to every \(\|y\| < \delta\) there corresponds an \(x\) with \(\|x\| < 1\) so that \(\Lambda x = y\).

Corollary: If \(X\) and \(Y\) are Banach spaces and if \(\Lambda\) is a bounded surjective linear transformation of \(X\) onto \(Y\) which is also one-to-one, then there is a \(\delta > 0\) s.t. \(\|\Lambda x\| \geq \delta\|x\|\), i.e. \(\Lambda^{-1}\) is a bounded linear transformation of \(Y\) onto \(X\). □

Rudin then proceeds to discuss the convergence of Fourier series for \(C(T)\):

Is it true for every \(f\in C(T)\) that the Fourier series of \(f\) converges to \(f(x)\) at every point \(x\)?

As noted in Chapter 4, the partial sums do converge to \(f\), but in the \(L^2\) norm. Then each \(f\in L^2(T) \supset C(T)\) is the pointwise limit a.e. of some subsequence of the full sequence of partial sums (using that for every convergent Cauchy sequence in \(L^p\) there is a subsequence that converges pointwise a.e. to the limit of the Cauchy sequence). However, this does not answer the question (of whether the partials converge to f everywhere).

Using the Banach-Steinhaus theorem (and BCT), it can be shown that the answer is negative:

Failure of point-wise convergence: There is a set \(E\subset C(T)\) which is a dense \(G_\delta\) in \(C(T)\) and which has the following property: For each \(f\in E\), the set $$Q_f=\{x:s^*(f;x)=\infty\}$$ is a dense \(G_\delta\) in \(\mathbb{R}\). □

Here, \(s_n(f;x)\) denotes the \(n\)th partial of the fourier series (summing from \(-n\) to \(n\)) evaluated at \(x\), and \(s^*(f;x)=\sup_n |s_n(f;x)|\).

This is proven by showing that the collection of linear functionals \(f\mapsto s_n(f;x)\) (for a fixed \(x\)) is a collection of bounded linear functionals, and their norm goes to infinity as \(n\) goes to infinity. The Banach-Steinhaus theorem then asserts that for every real \(x\), we have \(s^*(f;x)=\infty\) for every \(f\in E_x\) where \(E_x\) is a dense \(G_\delta\) in \(C(T)\). The BCT is then used to derive the result considering \(E_x\) at a countable dense set of reals and taking their intersection.

Furthermore, it is actually the case that \(E\) and \(Q_f\) are both uncountable:

Theorem (5.13): In a complete metric space \(X\) with no isolated points, no countable dense set is a \(G_\delta\). □

Other questions arise when considering the Fourier coefficients of \(L^1\) functions. We have the following lemma:

Riemann-Lebesgue Lemma: \(\hat{f}(n)\rightarrow 0\) as \(n\rightarrow\pm\infty\). □

The question is if the converse is true, i.e. if \(a_n\) is a sequence of complex numbers going to 0 as \(n\rightarrow\pm\infty\), are these the Fourier coefficients of some \(f\in L^1(T)\)? This would be a Riesz-Fischer theorem-like statement.

However, this is not the case. Denote by \(c_0\) the space of complex functions \(\varphi\) on \(\mathbb{Z}\) such that \(\varphi(n)\rightarrow 0\) as \(n\rightarrow\pm\infty\), with the supremum norm. This is a Banach space. Then, we have the following:

Failure of converse: The mapping \(f\mapsto\hat{f}\) is a one-to-one bounded linear transformation of \(L^1(T)\) into but not onto \(c_0\). □

This follows from the (corollary to) the open mapping theorem, by showing that the inequality cannot hold and thereform the mapping is not onto.

The text moves to the important Hahn-Banach theorem:

Hahn-Banach: If \(M\) is a (not necessarily closed) subspace of a normed linear space \(X\) and if \(f\) is a bounded linear functional on \(M\), then \(f\) can be extended to a bounded linear functional \(F\) on \(X\) s.t. \(\|F\| = \|f\|\). □

Note that the norms are computed wrt. their respective domains.

The proof of the Hahn-Banach theorem consists of two steps. First, it is shown more or less directly that for any single point \(x_0\in X\setminus M\), if we add it to \(M\) and take the span \(M_1\), we can construct a norm-preserving extension of \(f\) to \(M_1\).

Second, consider the poset of all norm-preserving extensions of \(f\) whose domain contains \(M\), ordered by inclusion on the domain. Zorn/Hausdorff maximality theorem gives a mixmial totally ordered subcollection of this poset, with which the desired extension can be constructed. The domain of this construction must be all of \(X\), otherwise by the first part it could be further extended (contradicting maximality).

The following are two important consequences:

Well-behavedness of closure: Let \(M\) be a linear subspace of a normed linear space \(X\), and let \(x_0\in X\). Then \(x_0\) is in the closure \(\bar{M}\) of \(M\) iff there is no bounded linear functional \(f\) on \(X\) s.t. \(f(x)=0\) for all \(x\in M\) but \(f(x_0)\neq 0\). □

Existence of special functionals: If \(X\) is a normed linear space and if \(x_0\in X\), \(x_0\neq =0\), there is a bounded linear functional \(f\) on \(X\) of norm 1, with \(f(x_0)=\|x_0\|\). □

Finally, the text applies the Hahn-Banach theorem to derive the Poisson integral.

Let \(K\) be a compact Hausdorff space, \(H\) a compact subset of \(K\), and \(A\) a subspace of \(C(K)\) such that \(1\in A\) and $$\|f\|_K = \|f\|_H\qquad(f\in A),$$ where $$\|f\|_E = \sup\{|f(x)|: x\in E\}.$$ \(H\) is sometimes called a boundary of \(K\), corresponding to the space \(A\).

Let \(M\) be the set of all functions on \(H\) that are restrictions to \(H\) of members of \(A\). This is a bijection between \(M\) and \(A\) because of the boundary property. By considering the linear functional \(\Lambda f = f(x)\) for fixed \(x\), and applying Hahn-Banach and the Riesz representation theorem, we get the representation formula $$f(x) = \int_H f\ d\mu_x\qquad(f\in A).$$

Hence for each \(x\in K\) there corresponds a positive measure \(\mu_x\) on the boundary \(H\) which represents \(x\) in the above sense.

Note that this representation is not unique, because although the Riesz representation theorem determines the measure uniquely, the Hahn-Banach extension is not necessarily at all unique. NB: The Riesz representation theorem does not guarantee the uniqueness of the \(\sigma\)-algebra (hence the measure) in general. As stated in the book, uniqueness of the measure is only guaranteed given the \(\sigma\)-algebra. However (according to the MathSE link), the \(\sigma\)-algebra is unique when the space is \(\sigma\)-compact or more generally when \(\mu\) is \(\sigma\)-finite (TODO: actually read this). This is the case in our scenario (as \(K\) is comapact).

In particular, if we let \(U\) be the open unit disc in the complex plane, \(K\) the closed unit disc, and \(T\) the boundary of the disc, then it can be shown that every polynomial \(f\) satisfies the relation \(\|f\|_U = \|f\|_T\). This is a special case of the maximum modulus principle.

Now the following can be shown:

Poisson Integral Representation: Suppose \(A\) is a vector space of continuous complex functions on the closed unit disc \(\bar{U}\). If \(A\) contains all polynomials, and if $$\|f\|_U = \|f\|_T$$ for every \(f\in A\), then the Poisson integral representation $$f(z) = \frac{1}{2\pi}\int_{-\pi}^\pi \frac{1-r^2}{1-2r\cos(\theta - t) + r^2} f(e^{it})\ dt\qquad (z = re^{i\theta})$$ is valid for every \(f\in A\) and every \(z\in U\). □

From the previous results, we can see that to each \(z\in U\) there corresponds a positive Borel measure \(\mu_z\) on \(T\) s.t. $$f(x) = \int_T f\ d\mu_z\qquad(f\in A).$$

Considering the Poisson kernel $$P_r(\theta - t) = \sum_{n=-\infty}^\infty r^{|n|}e^{in(\theta - t)}$$ where \(t\) is real, one can show that $$\int_T f\ d\mu_z = \frac{1}{2\pi}\int_{-\pi}^\pi f(e^{it})P_r(\theta - t)\ dt$$ holds for every trigonometric polynomial \(f\) and hence for every \(f\in C(T)\). This also shows that \(\mu_z\) is uniquely determined in this case, as this fixes the value of the (extended) linear operator \(\Lambda f = \int_T f\ d\mu_z\) on the whole of \(C(T)\).

In particular, for \(f\in A\), this value will also be equal to \(f(z)\). Summing the series \(P_r(\theta - t)\) explicitly gives the Poission integral representation.