# Completeness of 'Hardy Space' $H^2(D)$

Define Hardy Space $H^2(D)$ as a space of holomorphic functions $f$ on unit open disc $D=\{z\in\mathbb{C}:|z|<1\}$ endowed with the norm

$$||f||^2=\sup_{0<r<1} \int_0^{2\pi} |f(re^{i\theta})|^2\,d\theta$$

By Taylor expanding $f$, write the norm in terms of its Taylor coefficient and deduce that $H^2(D)$ is complete

My Attempts:


Given $f\in H^{2}\left( \D \right)$, by Taylor's Theorem, there exists $a_n\in \C$ such that

$$f(z)=\sum_{n=0}^{\infty} a_n z^{n} \;\;\; \forall z\in \D$$

Now for each $N$, we have $\sum_{n=0}^{N} \cl{a_nz^{n}} = \cl{\sum_{n=0}^{N}a_nz^n} \to \cl{f(z)}$. Hence

$$\cl{f(z)}=\sum_{n=0}^{\infty} \cl{a_nz^n} \;\;\; \forall z\in \D$$

and both series expansions $f,\cl{f}$ have radius of convergence $\ge 1$.

Therefore, since we can multiply the sums term by term within radius of convergence, for every $z=re^{i\theta}\in \D$ we have

\begin{align*} \left| f(z)\right|^2&=f(z)\cl{f(z)}=\left( \sum_{k=0}^{\infty} a_kz^k \right) \left( \sum_{l=0}^{\infty} \cl{a_lz^l} \right)\\ &=\left( \sum_{k=0}^{\infty} a_kr^ke^{ik\theta} \right) \left( \sum_{l=0}^{\infty} \cl{a_l}\cl{r^le^{il\theta}} \right)\\ &=\left( \sum_{k=0}^{\infty} a_kr^ke^{ik\theta} \right) \left( \sum_{l=0}^{\infty} \cl{a_l}r^le^{-il\theta} \right)\\ &= \sum_{k=0}^{\infty}\sum_{l=0}^{\infty} a_k\cl{a_l}r^{k+l}e^{i(k-l)\theta}\\ &=\sum_{n=-\infty}^{\infty} e^{in\theta} \sum_{k-l=n} a_k\cl{a_l}r^{k+l} \end{align*}

Where $k,l\ge 0$. But we also have

$$\left|e^{in\theta} \sum_{k-l=n} a_k\cl{a_l}r^{k+l}\right| \le \left|\sum_{k-l=n} a_k\cl{a_l}r^{k+l}\right| \in L^{1}(0,2\pi)$$

Therefore, by DCT

$$I(r):=\int_{0}^{2\pi} \left|f(re^{i\theta})\right|^{2}\,d\theta = \sum_{n=-\infty}^{\infty}\int_{0}^{2\pi} \left[e^{in\theta} \sum_{k-l=n} a_k\cl{a_l}r^{k+l}\right]\,d\theta$$

Now, note that since $e^{in\theta}=\cos(n\theta) +i\sin(n\theta)$, by periodicity of $\sin$ and $\cos$, we have $\int_{0}^{2\pi} e^{in\theta}\,d\theta = 2\pi \delta_{0,n}$ where $\delta_{0,n}$ denotes Kronecker delta function. Therefore,

$$I(r)=2\pi \sum_{0\le k=l} a_k\cl{a_l}r^{k+l} = 2\pi\sum_{n=0}^{\infty} |a_n|^2 r^{2n}$$

Hence finally, it follows that

$$\n{f}_{H}^{2} = \sup_{0< r < 1} I(r)=2\pi \sup_{0< r < 1}\sum_{n=0}^{\infty} |a_n|^2 r^{2n} = 2\pi \sum_{n=0}^{\infty} |a_n|^2$$

To show that $H^2(\D)$ is a Hilbert Space, we need to show two things: that it is complete with respect to the given norm, and that this norm induces a corresponding inner product.

• Completeness

Suppose that $f_m$ is a Cauchy sequence in $H^2(\D)$. By Taylor's theorem, there exist $a_{n,m}$ such that

$$f_m(z)=\sum_{n=0}^{\infty} a_{n,m} z^n\;\;\; \forall z\in \D$$

Then $f_k(z)-f_l(z)=2\pi \sum_{n=0}^{\infty} (a_{n,k}-a_{n,l})z^n$, so that $\n{f_k-f_l}_H^2=\sum_{n=0}^{\infty} |a_{n,k}-a_{n,l}|^2$. Then

$$\n{(a_{n,k})_{k}-(a_{n,l})_{l}}_2^2 = \sum_{n=0}^{\infty} |a_{n,k}-a_{n,l}|^2=\frac{1}{2\pi} \n{f_k-f_l}_H^2 \to 0$$

as $k,l\to\infty$. By completeness of $l^{2}$, it follows that $(a_{n,k})_{k} \to (b_n)$ in $l^2$. No consider

$$f(z)=\sum_{n=0}^{\infty} b_nz^n$$

for $z\in \D$. Then, for instance, by Young's inequality we have $|b_nz^n| \le \frac{|b_n|^2+|z^n|^2}{2}$. Since $|b_n|$ is square summable (in $l^2$) and $|z|\in [0,1)$, $|z_n|$ is also square summable, and therefore we conclude that $\sum_{n=0}^{\infty} |b_nz^n|$ converges for all $z\in \D$. Therefore, by completeness of $\C$ it follows that $f(z)$ exists for all $z\in\D$. Moreover, by definition, $f$ is clearly differentiable within its radius of convergence, so is holomorphic on $\D$. Hence $f\in H^2(\D)$.\

Finally, note that since convergence in $l^2$ implies pointwise convergence, so $a_{n,l} \to b_n$ as $l\to \infty$. Therefore, letting $l\to \infty$

\begin{align*} \n{f_m-f}_H^2&=\lim_{l\to\infty} \n{f_m-f_l}_H^2= 2\pi \n{(a_{n,m})_{m} - (b_n)}^2_2 \to 0 \end{align*}

by continuity of norms. So $f_n\to f$ in $H^2(\D)$ and we are done.

However, I'm particularly skeptical about the part where I've manipluated doubly infinite sum, and applied DCT to it to find $\n{f}^2$. Generally, I feel like I've forced out conclusion through either wrong or unncessarily complicated argument. It'd be really nice if you could check whether the proof is valid, and suggest better way to do it.

• You have shown that $H^2(D)$ is isometric to $\ell^2$, so it is complete. – user99914 Jun 13 '16 at 8:25
• @JohnMa Yes, I guess I was doing something unncessary for completeness part, but do you think parts before it where lots of multiplication of two infinite series stuff were valid? – user340297 Jun 13 '16 at 9:47
• O.K, I see. "Your" $H^2(D)$ is the space of holomorphic functions $f$ on $D$ s.t. $||f||<+\infty$ (not all, for example $\frac{1}{1-z}$ is not there). Then, as @ArcticChar says with reason, it is a recoding of ${\it l}^2$ and hence complete. – Duchamp Gérard H. E. Jul 21 '16 at 16:00