Ch4. Hilbert Space: An Introduction (Ex.01 ~ Ex.09)

(i) Show that \(|(f,g)| \le \|f\|\|g\|\) for all \(f,g \in \mathcal{H}\)

proof
① \(g=0\) (wlog)

\((\alpha f,0)=\alpha(f,0)=(f,\overline{\alpha}\,0)=(f,0)\) and this holds for all \(\alpha\in\mathbb{C}\). Letting \(\alpha=0,\ (f,0)=(0,0)=0\). Note that \(\|f\|\|g\|=0\), so the inequality holds.

② \(f,g\ne 0\)

Let \(h=f-\dfrac{(f,g)}{\|g\|^2}g\).

\[ \begin{aligned} \|h\|^2 &=\left(f-\frac{(f,g)}{\|g\|^2}g,\ f-\frac{(f,g)}{\|g\|^2}g\right) \\ &=\|f\|^2+\frac{|(f,g)|^2}{\|g\|^2}+\left(f,-\frac{(f,g)}{\|g\|^2}g\right)+\left(-\frac{(f,g)}{\|g\|^2}g,f\right) \\ &=\|f\|^2+\frac{|(f,g)|^2}{\|g\|^2}-\frac{(f,g)}{\|g\|^2}(f,g)-\frac{\overline{(f,g)}}{\|g\|^2}(g,f) \\ &=\|f\|^2-\frac{|(f,g)|^2}{\|g\|^2}\ge 0. \end{aligned} \]

Thus, \(|(f,g)|\le \|f\|\|g\|\).

(ii) Show that \(\|f+g\|\le \|f\|+\|g\|\)

proof
\[ \begin{aligned} \|f+g\|^2 &=(f+g,f+g)=\|f\|^2+(f,g)+(g,f)+\|g\|^2 \\ &=\|f\|^2+2\operatorname{Re}(f,g)+\|g\|^2 \\ &\le \|f\|^2+2|(f,g)|+\|g\|^2 \\ &\le \|f\|^2+2\|f\|\|g\|+\|g\|^2=(\|f\|+\|g\|)^2 \quad (\because \text{Cauchy}) \end{aligned} \]

(a) neither \(L^2(\mathbb{R}^d)\subset L^1(\mathbb{R}^d)\) nor \(L^1(\mathbb{R}^d)\subset L^2(\mathbb{R}^d)\) is valid

proof
① \(f(x)=\chi_{\{|x|<1\}}(x)\cdot\frac{1}{|x|^{d/2}}\), \(L^1\nsubseteq L^2\)

\(\int_{\mathbb{R}^d}|f|=\int_{|x|<1}\frac{1}{|x|^{d/2}}\,dx<\infty\ \ (\because \ d/2<d),\) while \(\int_{\mathbb{R}^d}|f|^2=\int_{|x|<1}\frac{1}{|x|^{d}}\,dx=\infty\).

② \(g(x)=\chi_{\{|x|>1\}}(x)\cdot\frac{1}{|x|^{d}}\), \(L^2\nsubseteq L^1\)

\(\int_{\mathbb{R}^d}|g|^2=\int_{|x|>1}\frac{1}{|x|^{2d}}\,dx<\infty\ \ (\because \ 2d>d),\) while \(\int_{\mathbb{R}^d}|g|=\int_{|x|>1}\frac{1}{|x|^{d}}\,dx=\infty\).

(b) if \(f\in L^2\) is supported on a set \(E\) of finite measure, then \(f\in L^1\).

proof
\(\|f\|_1\) can be written as \[ \|f\|_1=\int_E|f|=\int_{\mathbb{R}^d}|f|\chi_E=\int_{\mathbb{R}^d}|f|\overline{\chi_E}.\quad(\because \chi_E\ \text{is real}) \]

Since \(|f|,\chi_E\in L^2\), \[ \begin{aligned} \int_{\mathbb{R}^d}|f|\overline{\chi_E} &\le \|f\|_2\|\chi_E\|_2 \ \ (\because \text{Cauchy-Schwarz ineq.})\\ &=\|f\|_2\left\{\int_{\mathbb{R}^d}\chi_E^2\right\}^{1/2} \\ &=\|f\|_2\left\{\int_{\mathbb{R}^d}\chi_E\right\}^{1/2} \\ &=\|f\|_2\{m(E)\}^{1/2}<\infty \quad (f\in L^2) \end{aligned} \]

\(\|f\|_1<\infty\), so \(f\in L^1\).

(c) if \(|f|\le M\) and \(f\in L^1\), then \(f\in L^2\) \((M>0)\)

proof
\[ \|f\|_2^2=\int_{\mathbb{R}^d}|f|^2=\int_{\mathbb{R}^d}|f||f|\le M\int_{\mathbb{R}^d}|f|=M\|f\|_1<\infty\ \ (\because f\in L^1), \] so \(\|f\|_2<\infty\) and \(f\in L^2\).

(a) Simple functions

proof
① \(f\ge 0\)
Choose \(f\in L^2\) such that \(f\ge 0\) and let \(\varepsilon>0\) be fixed.

\(f\) is measurable, so there is a sequence of non-negative simple functions \(\{\varphi_n\}\) such that \(\varphi_n\le \varphi_{n+1}\) and \(\lim_{n\to\infty}\varphi_n=f\) (thm 1.4.1).

\(0\le f-\varphi_n\le f\) so \(|f-\varphi_n|^2\le |f|^2\). Then \(\lim_{n\to\infty}\int_{\mathbb{R}^d}|f-\varphi_n|^2=0\) (DCT), and there is \(N_\varepsilon\in\mathbb{N}\) such that \(\|f-\varphi_{N_\varepsilon}\|_{L^2}<\varepsilon\). This holds for all \(\varepsilon>0\) and all non-negative \(f\in L^2\). So, simple functions are dense in nonnegative \(L^2(\mathbb{R}^d)\).

② \(f\in\mathbb{R}\)
Consider \(f\in\mathbb{R}\) s.t. \(f=f^+-f^-\). Each \(f^+\) and \(f^-\) is non-negative, so there exist non-negative simple functions \(\{\psi_n^+\}\), \(\{\psi_n^-\}\) and \(N_\varepsilon\in\mathbb{N}\) such that \(\|f^+-\psi_{N_\varepsilon}^+\|_{L^2}<\varepsilon/2\) and \(\|f^--\psi_{N_\varepsilon}^-\|_{L^2}<\varepsilon/2\) for given \(\varepsilon>0\) (\(\because ①\)).

Note that \(\varphi_n=\psi_n^+-\psi_n^-\) is simple function. Then, \[ \begin{aligned} \|f-\varphi_{N_\varepsilon}\|_{L^2} &=\|(f^+-f^-)-(\psi_{N_\varepsilon}^+-\psi_{N_\varepsilon}^-)\|_{L^2} \\ &=\|(f^+-\psi_{N_\varepsilon}^+)-(f^--\psi_{N_\varepsilon}^-)\|_{L^2} \\ &\le \|f^+-\psi_{N_\varepsilon}^+\|_{L^2}+\|f^--\psi_{N_\varepsilon}^-\|_{L^2} \\ &<\varepsilon \qquad(\because \Delta\ \text{ineq of }\|\cdot\|_2). \end{aligned} \]

This holds for all \(\varepsilon>0\) and all real \(f\), so simple functions are dense in real \(L^2(\mathbb{R}^d)\).

③ general \(f\in L^2(\mathbb{R}^d)\)
Let \(f=u+iv\), where \(u,v\in L^2\) are real. Then for given \(\varepsilon>0\), there exist simple functions \(\varphi_u,\varphi_v\) s.t. \(\|u-\varphi_u\|_{L^2}<\varepsilon/2\) & \(\|v-\varphi_v\|_{L^2}<\varepsilon/2\). Note that \(\varphi_f=\varphi_u+i\varphi_v\) is a simple function with complex coefficient.

Then, \[ \begin{aligned} \|f-\varphi_f\|_{L^2} &=\|(u+iv)-(\varphi_u+i\varphi_v)\|_{L^2} \\ &=\|(u-\varphi_u)+i(v-\varphi_v)\|_{L^2} \\ &\le \|u-\varphi_u\|_{L^2}+\|v-\varphi_v\|_{L^2}<\varepsilon. \end{aligned} \]

This holds for all \(\varepsilon>0\) and all \(f\in L^2\), so simple functions are dense in \(L^2(\mathbb{R}^d)\).

(a) Show that \(H_\eta\) is a Hilbert space and \(U:f\mapsto \eta^{1/2}f\) gives a unitary correspondence between \(H_\eta\) and \(L^2([a,b])\).

proof
① \(H_\eta\) is a vector space over \(\mathbb{C}\) \((f,g\in H_\eta,\ c\in\mathbb{C})\)

\(|f+g|^2\le (|f|+|g|)^2\le 2(|f|^2+|g|^2)\ \ (\because\ |f|^2+|g|^2-2|f||g|=(|f|-|g|)^2\ge 0)\), so \[ \int_a^b |f(t)+g(t)|^2\eta(t)\,dt <2\int_a^b |f(t)|^2\eta(t)\,dt+2\int_a^b |g(t)|^2\eta(t)\,dt<\infty\ \ (\because\ f,g\in H_\eta). \] So, \(f+g\in H_\eta\).

\(\int_a^b |cf(t)|^2\eta(t)\,dt=|c|^2\int_a^b |f(t)|^2\eta(t)\,dt<\infty\), so \(cf\in H_\eta\).

② \((f,g)_\eta\) is linear on \(H_\eta\), \((f,g)=\overline{(g,f)}\), and \((f,f)_\eta\ge 0\) \((f,g\in H_\eta)\)

1. Linear \[ \begin{aligned} (f_1+f_2,g)_\eta &=\int_a^b (f_1(t)+f_2(t))\overline{g(t)}\,\eta(t)\,dt \\ &=\int_a^b f_1(t)\overline{g(t)}\,\eta(t)\,dt+\int_a^b f_2(t)\overline{g(t)}\,\eta(t)\,dt \\ &=(f_1,g)_\eta+(f_2,g)_\eta, \end{aligned} \] \[ \begin{aligned} (c f,g)_\eta &=\int_a^b c f(t)\overline{g(t)}\,\eta(t)\,dt \\ &=c\int_a^b f(t)\overline{g(t)}\,\eta(t)\,dt=c(f,g)_\eta, \end{aligned} \] \[ \begin{aligned} (f,g_1+g_2)_\eta &=\int_a^b f(t)\overline{(g_1(t)+g_2(t))}\,\eta(t)\,dt \\ &=\int_a^b f(t)\overline{g_1(t)}\,\eta(t)\,dt+\int_a^b f(t)\overline{g_2(t)}\,\eta(t)\,dt \\ &=(f,g_1)_\eta+(f,g_2)_\eta, \end{aligned} \] \[ \begin{aligned} (f,cg)_\eta &=\int_a^b f(t)\overline{cg(t)}\,\eta(t)\,dt \\ &=\overline{c}\int_a^b f(t)\overline{g(t)}\,\eta(t)\,dt=\overline{c}(f,g)_\eta. \end{aligned} \]

2. \((f,g)_\eta=\overline{(g,f)_\eta}\) Note that \(\ f=u+iv,\ u,v\in L^1(\mathbb{R}^d),\ \text{then }\overline{\int f}=\int\overline{f},\) \[ \text{and }\overline{\int_{\mathbb{R}^d}(u+iv)} =\overline{\int_{\mathbb{R}^d}u+i\int_{\mathbb{R}^d}v} =\int_{\mathbb{R}^d}u-i\int_{\mathbb{R}^d}v =\int_{\mathbb{R}^d}(u-iv) =\int_{\mathbb{R}^d}\overline{f}. \] \[ \begin{aligned} (g,f)_\eta &=\int_a^b g(t)\overline{f(t)}\,\eta(t)\,dt \\ &=\overline{\int_a^b \overline{g(t)}f(t)\,\eta(t)\,dt} =\overline{\int_a^b f(t)\overline{g(t)}\,\eta(t)\,dt}\quad(\because \eta\in\mathbb{R}) \\ &=\overline{(f,g)_\eta}. \end{aligned} \]

3. \((f,f)_\eta\ge 0\ \ :\ \ \int_a^b f(t)\overline{f(t)}\,\eta(t)\,dt=\int_a^b |f(t)|^2\eta(t)\,dt\ge 0\ \ (\because |f(t)|^2\ge 0,\ \eta>0)\)

③ \(\|f\|_\eta=0\ \ \text{if }f=0\)

Note that two functions in \(H_\eta\) are equivalent if they differ on a set of measure \(0\). Then \(\int_a^b |f(t)|^2\eta(t)\,dt=0\) implies \(|f(t)|^2=0\ \ (\because \eta>0)\), and thus \(f(t)=0\).

\(f(t)=0\) obviously leads to \(\int_a^b |f(t)|^2\eta(t)\,dt=0\).

④-1 \(|(f,g)_\eta|\le \|f\|_\eta\|g\|_\eta\ \ (f,g\in H_\eta)\) Cauchy ineq

If \(\|f\|_\eta\) (or \(\|g\|_\eta\)) \(=0\), \(f(t)=0\) and \(\int_a^b f(t)\overline{g(t)}\,\eta(t)\,dt=(f,g)_\eta=0\).

Consider \(f,g\) such that \(\|f\|_\eta\ne 0\) and \(\|g\|_\eta\ne 0\), and let \(f_0=\dfrac{f}{\|f\|_\eta}\), \(g_0=\dfrac{g}{\|g\|_\eta}\).

\(|f_0-\overline{g_0}|^2\ge 0\) leads to \(2|f_0\overline{g_0}|\le |f_0|^2+|\overline{g_0}|^2=|f_0|^2+|g_0|^2\ \ (|\overline{g_0}|=|g_0|)\), so \[ \begin{aligned} 2|(f_0,g_0)_\eta| &=2\left|\int_a^b f_0(t)\overline{g_0(t)}\,\eta(t)\,dt\right| \\ &\le 2\int_a^b |f_0(t)\overline{g_0(t)}|\,\eta(t)\,dt \\ &\le \int_a^b |f_0(t)|^2\eta(t)\,dt+\int_a^b |g_0(t)|^2\eta(t)\,dt \\ &=\frac{1}{\|f\|_\eta^2}\int_a^b |f(t)|^2\eta(t)\,dt+\frac{1}{\|g\|_\eta^2}\int_a^b |g(t)|^2\eta(t)\,dt =2. \end{aligned} \]

Thus, \(\left|\left(\dfrac{f}{\|f\|_\eta},\dfrac{g}{\|g\|_\eta}\right)_\eta\right|\le 1\) and \((f,g)_\eta\le \|f\|_\eta\|g\|_\eta\).

④-2 \(\|f+g\|_\eta\le \|f\|_\eta+\|g\|_\eta\ \ (f,g\in H_\eta)\) triangle ineq.

\[ \begin{aligned} \|f+g\|_\eta^2 &=(f+g,f+g)_\eta \\ &=\|f\|_\eta^2+(f,g)_\eta+(g,f)_\eta+\|g\|_\eta^2 \\ &=\|f\|_\eta^2+2\operatorname{Re}\{(f,g)_\eta\}+\|g\|_\eta^2 \\ &\le \|f\|_\eta^2+2|(f,g)_\eta|+\|g\|_\eta^2 \\ &\le \|f\|_\eta^2+2\|f\|_\eta\|g\|_\eta+\|g\|_\eta^2 \qquad(\because ④\text{-}1) \\ &=(\|f\|_\eta+\|g\|_\eta)^2. \end{aligned} \] Thus, \(\|f+g\|_\eta\le \|f\|_\eta+\|g\|_\eta\).

⑤ \(H_\eta\) is complete in the metric \(\|f-g\|_\eta\) use continuity of \(\eta\)

1. \(\exists\,b_1,b_2\in[a,b]\ \text{s.t. }\eta(b_1)=\sup\eta(t)\ \&\ \eta(b_2)=\inf\eta(t)\) \(\eta\) is continuous on a compact set \([a,b]\), so there exist \(b_1,b_2\in[a,b]\) such that \(M_1\overset{\mathrm{def}}{=}\eta(b_1)=\sup_{t\in[a,b]}\eta(t)\) and \(M_2\overset{\mathrm{def}}{=}\eta(b_2)=\inf_{t\in[a,b]}\eta(t)\). \((M_1,M_2>0)\)

2. \(g\in H_\eta\), then \(g\in L^2\) For \(g\in H_\eta\), \[ \|g\|_\eta^2=\int_a^b |g(t)|^2\eta(t)\,dt \ge M_2\int_a^b |g(t)|^2\,dt = M_2\|g\|_{L^2([a,b])}^2. \] Since \(\|g\|_\eta<\infty\), \(g\in L^2([a,b])\). (let \(\|g\|_{L^2([a,b])}=\|g\|_2\))

3. \(\{f_n\}\) Cauchy in \(H_\eta\), and Cauchy in \(L^2([a,b])\) Consider a Cauchy sequence \(\{f_n\}\subset H_\eta\); for given \(\varepsilon>0\) there is \(N_\varepsilon\in\mathbb{N}\) such that \(m,n\ge N_\varepsilon\) implies \(\|f_m-f_n\|_\eta<\varepsilon\). Then, \[ \|f_m-f_n\|_\eta^2=\int_a^b |f_m(t)-f_n(t)|^2\eta(t)\,dt \ge M_2\int_a^b |f_m(t)-f_n(t)|^2\,dt = M_2\|f_m-f_n\|_2^2. \] This leads to \(\|f_m-f_n\|_2<\varepsilon/\sqrt{M_2}\) for \(m,n\ge N_\varepsilon\), and this holds for all \(\varepsilon>0\). So, \(\{f_n\}\) is also a Cauchy sequence in \(L^2([a,b])\).

4. \(f\in L^2([a,b])\ \to\ f\in H_\eta\) \(\int_a^b |f(t)|^2\eta(t)\,dt \le M_1\int_a^b |f(t)|^2\,dt <\infty\ (f\in L^2([a,b]))\), so \(f\in H_\eta\) as well.

5. \(\|f_n-f\|_2\to 0\), then \(\|f_n-f\|_\eta\to 0\) \(L^2([a,b])\) is complete, so there is \(f\in L^2([a,b])\) such that \(\lim_{n\to\infty}\|f_n-f\|_2=0\). Then, \[ \|f_n-f\|_\eta^2=\int_a^b |f_n(t)-f(t)|^2\eta(t)\,dt \le M_1\int_a^b |f_n(t)-f(t)|^2\,dt = M_1\|f_n-f\|_2^2. \] \(\lim_{n\to\infty}\|f_n-f\|_\eta=0\) and \(f\in H_\eta\ (\because 4)\), thus \(H_\eta\) is complete.

⑥ \(H_\eta\) is separable : needs continuity of \(\eta\)

\(L^2([a,b])\) is separable, so there is \(\{f_k\}\subset L^2([a,b])\) such that their finite linear combinations are dense in \(L^2([a,b])\) : For \(f\in L^2([a,b])\) and given \(\varepsilon>0\), there is \(\psi_\varepsilon=\sum_{k=1}^n a_k f_k\) such that \(\|f-\psi_\varepsilon\|_2^2<\varepsilon^2\).

Note that \(f\in H_\eta\) and \(f_k\in H_\eta\ \forall k\in\mathbb{N}\) (\(\because ⑤\)), so \(\psi_\varepsilon\in H_\eta\). Then \[ \int_a^b |f(t)-\psi_\varepsilon(t)|^2\eta(t)\,dt \le M_1\int_a^b |f(t)-\psi_\varepsilon(t)|^2\,dt < M_1\varepsilon^2. \]

This holds for all \(\varepsilon>0\) and all \(f\in H_\eta\), so \(H_\eta\) is separable.

⑦ \(U\) is unitary

1. Linear \((f,g\in H_\eta,\ \alpha,\beta\in\mathbb{C})\) \(U(\alpha f+\beta g)=\sqrt{\eta}(\alpha f+\beta g)=\alpha\sqrt{\eta}f+\beta\sqrt{\eta}g=\alpha\cdot U(f)+\beta\cdot U(g)\)

2. bijection Consider the case \(U(f)=U(g)\); that is \(\sqrt{\eta(t)}f(t)=\sqrt{\eta(t)}g(t)\). Since \(\eta(t)>0\), \(f(t)=g(t)\). Thus, \(U\) is injective (1-1). Let \(h\in L^2([a,b])\); \(\int_a^b |h(t)|^2\,dt<\infty\). Let \(f=h/\sqrt{\eta}\). Since \(\int_a^b |f(t)|^2\eta(t)\,dt=\int_a^b |h(t)|^2\,dt<\infty\), \(f\in H_\eta\). Since \(U(f)=h\), \(U\) is surjective (onto)

3. \(\|U(f)\|_2=\|f\|_\eta\ \ \forall f\in H_\eta\) \(\|f\|_\eta^2=\int_a^b |f(t)|^2\eta(t)\,dt=\int_a^b |\sqrt{\eta(t)}f(t)|^2\,dt=\|\sqrt{\eta}f\|_2^2=\|U(f)\|_2^2\)

\(\therefore\ U\) is unitary.

(b) Generalize (a) to the case where \(\eta\) is not continuous.

proof
It suffices to prove \(H_\eta\) is complete & separable
(\(\because\) only ⑤ & ⑥ in (a) use the continuity)

⑤ \(H_\eta\) is complete

1. \(f\in H_\eta \Rightarrow \sqrt{\eta}\,f\in L^2([a,b])\) \[ \|\sqrt{\eta}f\|_2^2=\int_a^b \eta(t)|f(t)|^2\,dt=\int_a^b |f(t)|^2\eta(t)\,dt=\|f\|_\eta^2<\infty, \] so \(\sqrt{\eta}f\in L^2([a,b])\).

2. \(\{f_n\}\) Cauchy in \(H_\eta \to \{\sqrt{\eta}f_n\}\) Cauchy in \(L^2([a,b])\) Let \(\{f_n\}\) be a Cauchy sequence in \(H_\eta\); for given \(\varepsilon>0\) there is \(N_\varepsilon\in\mathbb{N}\) such that \(m,n\ge N_\varepsilon\) implies \(\|f_m-f_n\|_\eta^2<\varepsilon^2\). Again, \(m,n\ge N_\varepsilon\) implies \[ \|f_m-f_n\|_\eta^2=\int_a^b |f_m(t)-f_n(t)|^2\eta(t)\,dt =\int_a^b \left|\sqrt{\eta(t)}f_m(t)-\sqrt{\eta(t)}f_n(t)\right|^2\,dt<\varepsilon^2. \] This holds for all \(\varepsilon>0\) and \(N_\varepsilon\), so \(\{\sqrt{\eta}f_n\}\) is a Cauchy sequence in \(L^2([a,b])\).

3. \(g\in L^2([a,b])\Rightarrow g/\sqrt{\eta}\in H_\eta\) \[ \|g/\sqrt{\eta}\|_\eta^2=\int_a^b \frac{1}{\eta(t)}|g(t)|^2\eta(t)\,dt =\int_a^b |g(t)|^2\,dt<\infty\ \ (\because g\in L^2), \] so \(g/\sqrt{\eta}\in H_\eta\).

4. \(\|\sqrt{\eta}f_n-g\|_2\to 0\) then \(\|f_n-g/\sqrt{\eta}\|_\eta\to 0\) There is \(g\in L^2([a,b])\) such that \(\lim_{n\to\infty}\|\sqrt{\eta}f_n-g\|_2=0\) (\(L^2([a,b])\) is complete). Let \(f=g/\sqrt{\eta}\), then \(f\in H_\eta\) (\(\because 3\)). \[ \begin{aligned} \|f_n-f\|_\eta^2 &=\int_a^b |f_n(t)-f(t)|^2\eta(t)\,dt =\int_a^b |\sqrt{\eta(t)}f_n(t)-g(t)|^2\,dt \\ &=\|\sqrt{\eta}f_n-g\|_2^2, \end{aligned} \] so \(\lim_{n\to\infty}\|f_n-f\|_\eta=\lim_{n\to\infty}\|\sqrt{\eta}f_n-g\|_2=0\).

Thus, \(H_\eta\) is complete.

⑥ Separable

\(L^2([a,b])\) is separable, so there is \(\{f_k\}\subset L^2([a,b])\) such that their finite linear combinations are dense in \(L^2([a,b])\) : For \(f\in L^2([a,b])\) and given \(\varepsilon>0\), there is \(\psi_\varepsilon=\sum_{k=1}^n a_k f_k\) such that \(\int_a^b |f-\psi_\varepsilon|^2<\varepsilon^2\).

Note that \(f/\sqrt{\eta}\in H_\eta\), \(f_k/\sqrt{\eta}\in H_\eta\ \forall k\in\mathbb{N}\) and \(\psi_\varepsilon/\sqrt{\eta}\in H_\eta\ \ (\because ⑤)\). Then,

\[ \begin{aligned} \left\| \frac{f}{\sqrt{\eta}}-\frac{\psi_\varepsilon}{\sqrt{\eta}} \right\|_\eta^2 &=\int_a^b \frac{1}{\eta(t)}|f(t)-\psi_\varepsilon(t)|^2\,\eta(t)\,dt \\ &=\int_a^b |f(t)-\psi_\varepsilon(t)|^2\,dt =\|f-\psi_\varepsilon\|_2^2<\varepsilon^2. \end{aligned} \]

This means \(\left\{\frac{f_k}{\sqrt{\eta}}\right\}_k\) is dense in \(H_\eta\), so \(H_\eta\) is separable.

(a) the correspondence \(U:F\mapsto f\) with \(f(x)=\frac{1}{\sqrt{\pi}(i+x)}F\!\left(\frac{i-x}{i+x}\right)\) gives a unitary mapping of \(H_1\) to \(H_2\).

proof
① Linearity \((F_1,F_2\in H_1,\ \alpha,\beta\in\mathbb{C})\)

Let \(G(x)\overset{\mathrm{def}}{=}\alpha F_1(x)+\beta F_2(x)\), then \(U(G(x))=\frac{1}{\sqrt{\pi}(i+x)}G\!\left(\frac{i-x}{i+x}\right)\).

\(G\!\left(\frac{i-x}{i+x}\right)=\alpha F_1\!\left(\frac{i-x}{i+x}\right)+\beta F_2\!\left(\frac{i-x}{i+x}\right)\),

so \(U(G(x))=\frac{1}{\sqrt{\pi}(i+x)}\left\{\alpha F_1\!\left(\frac{i-x}{i+x}\right)+\beta F_2\!\left(\frac{i-x}{i+x}\right)\right\}=\alpha\cdot U(F_1)+\beta\cdot U(F_2)\).

② Bijection

1. 1-1 (injective)
   Consider the case \(U(F_1)=U(F_2):\ \frac{1}{\sqrt{\pi}(i+x)}F_1\!\left(\frac{i-x}{i+x}\right)=\frac{1}{\sqrt{\pi}(i+x)}F_2\!\left(\frac{i-x}{i+x}\right)\). Letting \(\frac{i-x}{i+x}=e^{i\theta}\), \(F_1(e^{i\theta})=F_2(e^{i\theta})\), so \(U\) is injective (1-1).

2. \(\frac{i-x}{i+x}=e^{i\theta}\ \Rightarrow\ x=\tan\frac{\theta}{2}\)
   \[ \begin{aligned} i-x=e^{i\theta}(i+x) &\iff (1+e^{i\theta})x=i(1-e^{i\theta}) \\ &\iff x=\frac{i(1-e^{i\theta})}{1+e^{i\theta}} =\frac{i(1-\cos\theta-i\sin\theta)}{1+\cos\theta+i\sin\theta} =\frac{\sin\theta+i(1-\cos\theta)}{1+\cos\theta+i\sin\theta} \\ &=\frac{\{\sin\theta+i(1-\cos\theta)\}\{1+\cos\theta-i\sin\theta\}}{(1+\cos\theta)^2+\sin^2\theta} \\ &=\frac{\sin\theta(1+\cos\theta)-i\sin^2\theta+i(1-\cos\theta)(1+\cos\theta)+(1-\cos\theta)\sin\theta}{1+2\cos\theta+\cos^2\theta+\sin^2\theta} \\ &=\frac{2\sin\theta+i(1-\cos^2\theta-\sin^2\theta)}{2(1+\cos\theta)} =\frac{2\cdot 2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2\cdot 2\cos^2\frac{\theta}{2}} =\frac{\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}} =\tan\frac{\theta}{2}. \end{aligned} \]

3. onto (surjective) Consider \(g\in H_2\) and let \(\sqrt{\pi}(i+x)g(x)=G\!\left(\frac{i-x}{i+x}\right)\). \(\frac{i-x}{i+x}=e^{i\theta}\ \Rightarrow\ G(e^{i\theta})=\sqrt{\pi}\left(i+\tan\frac{\theta}{2}\right)g\!\left(\tan\frac{\theta}{2}\right).\) \[ \begin{aligned} \|G\|_{H_1}^2 &=\frac{1}{2\pi}\int_{-\pi}^{\pi}\pi\left(i+\tan\frac{\theta}{2}\right)\left(-i+\tan\frac{\theta}{2}\right)\left|g\!\left(\tan\frac{\theta}{2}\right)\right|^2\,d\theta \\ &=\frac{1}{2}\int_{-\pi}^{\pi}\left(\tan^2\frac{\theta}{2}+1\right)\left|g\!\left(\tan\frac{\theta}{2}\right)\right|^2\,d\theta \quad\left(\text{let }\tan\frac{\theta}{2}=x,\ dx=\frac{1}{2}\sec^2\frac{\theta}{2}\,d\theta\right) \\ &=\int_{\mathbb{R}}|g(x)|^2\,dx \quad\left(\because\ \tan^2\frac{\theta}{2}+1=\sec^2\frac{\theta}{2}\right) \\ &<\infty\quad(\because\ g\in H_2). \end{aligned} \] This means \(G\in H_1\), so \(U\) is surjective (onto).

③ \(\|UF\|_{H_2}=\|F\|_{H_1}\)

Also, ② shows \(\|F\|_{H_1}=\|UF\|_{H_2}\).

(b) As a result, \(\left\{\frac{1}{\sqrt{\pi}(i+x)}\left(\frac{i-x}{i+x}\right)^n\right\}_n\) is an orthonormal basis for \(H_2\)

proof
① \(\{e^{in\theta}\}_n\) : orthonormal basis for \(H_1\)

It is clear that \(\{e^{in\theta}\}\) is an orthonormal set in \(H_1\).

Consider \(F\in H_1\) such that \(\frac{1}{2\pi}\int_{-\pi}^{\pi}F(e^{i\theta})\cdot e^{-in\theta}\,d\theta=0\) for all \(n\in\mathbb{N}\). Since the Fourier coefficients of \(F\) are all \(0\), \(F=0\) (Thm 4.3.1). Thus, \(\{e^{in\theta}\}\) is an orthonormal basis for \(H_1\) (Thm 4.2.3).

② \(\left\{\frac{1}{\sqrt{\pi}(i+x)}\left(\frac{i-x}{i+x}\right)^n\right\}_n\) let \(\{f_n\}_n\) : orthonormal basis for \(H_2\)

\(\frac{i-x}{i+x}=e^{i\theta}\), so \(e^{in\theta}=\left(\frac{i-x}{i+x}\right)^n\).

Then, \(U(e^{in\theta})=U\!\left(\left(\frac{i-x}{i+x}\right)^n\right)=\frac{1}{\sqrt{\pi}(i+x)}\left(\frac{i-x}{i+x}\right)^n\ \overset{\text{let}}{=}f_n(x)\).

1. \(\|f_n\|_{H_2}=\|e^{in\theta}\|_{H_1}=1\) (\(U\) is unitary), so \(\|f_n\|_{H_2}=1\ \ \forall n\in\mathbb{N}\).

2. \((f_n,f_m)=(e^{in\theta},e^{im\theta})=0\ (n\ne m)\), so \(\{f_n\}\) is orthogonal.

So, 1) & 2) implies \(\{f_n\}\) is an orthonormal set in \(H_2\).

3. \((f,f_n)=0\ \ \forall n,\ \text{then }f=0\in H_2\)

Consider \(f\in H_2\) such that \((f,f_n)=0\) for all \(n\). \(U\) is bijective, so there uniquely exists \(F\in H_1\) s.t. \[ f(x)=\frac{1}{\sqrt{\pi}(i+x)}F\!\left(\frac{i-x}{i+x}\right)=U(F(e^{i\theta}))\quad\left(\frac{i-x}{i+x}=e^{i\theta}\right). \] Then, \[ (f(x),f_n(x))_{H_2}=(F(e^{i\theta}),e^{in\theta})_{H_1} =\frac{1}{2\pi}\int_{-\pi}^{\pi}F(e^{i\theta})e^{-in\theta}\,d\theta=0\ \ \forall n. \] The Fourier coefficients of \(F\) are all \(0\), so \(F=0\) and \(f=0\).

\(\therefore\ \{f_n\}\) is an orthonormal basis for \(H_2\).