polynomial approximation in Hardy space $H^\infty$

Question

$H^\infty$ is the Hardy space of bounded analytic functions on the open unit disk $|z| < 1$ with the norm $$\|f\| = \sup_{ |z| \,< \,1}|f(z)|$$ It has two important subspaces :

• $H^\infty_C$ the (closed) subspace of analytic functions that stay continuous on the closed unit disk $|z|\le 1$.

• $H^\infty_K $ the subspace of functions whose Taylor series stay convergent on $|z| = 1$

Question : Can you show that $H^\infty_C \subseteq H^\infty_K$, or find a counter-example $f \in H^\infty_C, f \not\in H^\infty_K$ ?

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Attempt :

• $f \in H^\infty_C, f \not\in H^\infty_K$ means its Fourier series diverges, so it can't be Holder continuous on the boundary, and the Dini's criterion should diverge. That's why I thought to $f(z) = \frac{z}{\log(1-z)}$, but it seems its Taylor coefficients are all of the same sign, so Abel's lemma guarantees it converges. Next try : things like $\frac{1}{\log \log (1-z)}$ or $\frac{1}{\log \log \log \log (1-z)}$ (the branch $\log(1) = 2i\pi$)

• there is a proof of existence (but no example) of a $2\pi$-periodic continuous function whose Fourier series diverges at one point (but it doesn't have to be analytic, so it doesn't apply here)

• $f_N(z) = \sum_{n=0}^N \frac{f^{(n)}(0)}{n!} z^n$ is the truncated Taylor series. With $r \to 1^-$ , and $N \to \infty$ : $$\begin{eqnarray}\|f(z)-f_N(rz)\| & \ =\ & \|(f(z)-f(rz))+(f(rz)-f_N(rz))\| \\ & \le & \|f(z)-f(rz)\|+\|f(rz)-f_N(rz)\| \end{eqnarray} $$ $f \in H^\infty_C$ means it is uniformly continuous on $|z| \le 1$ so that $ \|f(z)-f(rz)\| = o(r)$, and $f(rz)$ is analytic on $1/r > 1$ hence it is approximated uniformly on $|z| \le 1$ by its truncated Taylor series $f_N(rz)$, so that $\|f(rz)-f_N(rz)\| = o_r(N)$ and $$\|f(z)-f_N(rz)\| = o(r)+o_r(N)$$ showing that the polynomials are dense in $H^\infty_C$.

• But doing the same for $\|f(z) - f_N(z)\|$ is more complicated : $$\begin{eqnarray}\|f(z)-f_N(z)\| & \ =\ & \|(f(z)-f_N(rz))+(f_N(rz)-f_N(z))\| \\ & \le & \|f(z)-f_N(rz)\|+\|f_N(rz)-f_N(z)\| \\ & = & o(r)+o_r(N) + \mathcal{O}((r-1)N) \end{eqnarray} $$ Ideally, if $f \in H^\infty_K$ then we would have $\|f_N(rz)-f_N(z)\|< \alpha \|f(rz)-f(z)\| = o(r)$ so that $\|f(z)-f_N(z)\| = o(r) + o_r(N)$.

  Hence, for showing $f \in H^\infty_K$, one needs a good bound for $\|f_N(rz)-f_N(z)\|$, probably using that $$f_N(rz)-f_N(z) = \frac{1}{2i\pi} \int_{|s| = R} \frac{f(s)}{s} \left(\frac{1-(rz/s)^{N+1}}{1-rz/s}-\frac{1-(z/s)^{N+1}}{1-z/s}\right)ds$$

Answer 1

Question : Can you show that $H^\infty_C \subseteq H^\infty_K$, or find a counter-example $f \in H^\infty_C, f \not\in H^\infty_K$ ?

Answer : Neither. The two spaces are certainly different, but one cannot write down a simple example showing this.

The second assertion is not something I'm going to try to prove mathematically. It's true in the same sort of sense that "there is no simple proof of Fermat's last theorem" is true. A simple example of a function in $H^\infty_C\setminus H^\infty_K$ would in particular be a simple example of a continuous function on the circle with a divergent Fourier series, and there is no such example. At least nobody has ever given such an example; this "shows" that simple constructions as in the OP and the comments "cannot" succeed, because if it were possible someone would have done it by now.

The space $H^\infty_C$ is commmonly known as the "disk algebra", denoted $A$; I will use this notation below.

The existence of $f\in A$ such that the power series for $f$ does not converge on the boundary follows fairly easily from the fact mentioned above, that there exists a function in $C(\Bbb T)$ with a divergent Fourier series. The implication would be utterly trivial if the Hilbert transform, i.e. the conjugate-function operator, were bounded on $C(\Bbb T)$. But the Hilbert transform is not bounded on $C(\Bbb T)$; the one result nonetheless follows from the other by a little trick.

For $n=1,2\dots$ define $$s_n:C(\Bbb T)\to C(\Bbb T)$$by saying $s_n f$ is the $n$-th partial sum of the Fourier series for $f$. Let $||s_n||$ denote the norm of $s_n$ as a operator on $C(\Bbb T)$. Note that $s_n:A\to A$; let $||s_n||_H$ be the norm of the restriction of $s_n$ to $A$. It is enough to show that $$||s_n||_H\to\infty;$$then the Uniform Boundedness Principle shows that there exists $f\in A$ with $||s_nf||$ unbounded.

Recall that the standard proof of the existence of a continuous function with a divergent Fourier series proceeds by showing that $$||s_n||\to\infty;$$hence it's enough to show that $$||s_n||\le2\sup_k||s_k||_H:=2c.$$

Suppose that $p(t)=\sum_{n=-N}^Na_ne^{int}$is a trigonometric polynomial, and let $$q(t)=e^{iNt}p(t),$$so that $q\in A$ (or rather, so that there exists $f\in A$ with $q(t)=f(e^{it})$). If $n\ge N$ then $s_np=p$. On the other hand, if $n<N$ then $$s_np(t)=e^{-iNt}(s_{N+n}q-s_{N-n}q),$$so that $$||s_np||\le2c||q||= 2c||p||.$$Since the trigonometric polynomials are dense in the continuous functions (and certainly $c\ge1$) this shows that $$||s_n||\le 2c.$$