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\newcommand{\supp}{\operatorname{supp}} %%%Analysis%%% \newcommand{\Lip}{\operatorname{Lip}} %%%Probability%%% \newcommand{\lebg}{(\mb R,\mc R, \lambda)} \newcommand{\nlebg}{(\mb R^{n},\mc R^{n}, \lambda)} \newcommand{\pbs}{(\Omega,\mc F,\mb P)} \newcommand{\msp}{(\Omega,\mc F,\mu)} \newcommand{\msk}[1]{(\Omega_{#1},\mc F_{#1},\mu_{#1})} \newcommand{\var}{\operatorname{var}} \newcommand{\dev}{\operatorname{dev}} \renewcommand{\vec}[1]{\overrightarrow{#1}} \newcommand{\tr}{\operatorname{tr}} \newcommand{\Sp}{\operatorname{span}} \newcommand{\nullity}{\operatorname{nullity}} \newcommand{\rank}{\operatorname{rank}} \newcommand{\cof}{\operatorname{cof}} \newcommand{\adj}{\operatorname{adj}} \newcommand{\ipd}[2]{\langle #1|#2 \rangle} % inner product \newcommand{\ran}{\operatorname{ran}} \newcommand{\cod}{\operatorname{codim}} \newcommand{\Ber}{\operatorname{Ber}} \newcommand{\str}{\operatorname{str}} \newcommand{\im}{\operatorname{Im}} \newcommand{\ad}{\operatorname{ad}} \newcommand{\der}{\operatorname{der}} \newcommand{\mb}[1]{\mathbb{#1}} % Field \newcommand{\stab}{\operatorname{stab}} \newcommand{\Hom}{\operatorname{Hom}} \newcommand{\rel}{\operatorname{rel}} %%%Functional Analysis%%% \newcommand{\hs}{\mathcal{H}}% Hilbert space \newcommand{\bs}{\mathcal{X}}% Banach space \newcommand{\bdo}[1]{\mathcal{B}(#1)}% The set of bounded operators \newcommand{\fro}[1]{\mathcal{F}(#1)}% Finite rank operators \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mk}[1]{\mathfrak{#1}} \newcommand{\tsp}{(X,\tau)} \newcommand{\Div}{\operatorname{Div}} %Dimensions \textwidth = 6in \oddsidemargin = 0.25in \evensidemargin = 0.25in \textheight = 8.7in \topmargin = -0.2in \begin{document} \pagestyle{empty} \begin{center} \textbf{201B, Winter '11, Professor John Hunter} \textbf{Homework 6 Solutions} \end{center} \bigskip \bigskip\noindent \textbf{1.} Let $X$ be a (real or complex) linear space and $P, Q : X \to X$ projections. \smallskip\noindent (a) Show that $I-P$ is the projection onto $\ker P$ along $\ran P$. \begin{proof} To show that $I-P$ is a projection we need to show that $(I-P)^2=I-P$. It is easy to see that \begin{equation*} (I-P)(x)=x-Px, \end{equation*} and therefore \begin{equation*} \begin{aligned} (I-P)^2(x)=&(I-P)(x-Px)\\ =&I(x-Px)-P(x-Px) \\ =&x-Px-Px+P^2(x)\\ =&x-Px-Px+Px\\ =&x-Px\\ =&(I-P)(x). \end{aligned} \end{equation*} Note that we used that $P$ is a projection, which by definition means in particular that $P^2=P$. We still have to prove $I-P$ is the projection onto $\ker P$ along $\ran P$. $\bullet$ Let $x \in \ker P$. Then $Px = 0$. So $$(I - P)x = x - Px = x.$$ \\ $\bullet$ Let $y \in \mbox{ran}\,(I - P)$. Then $y = (I - P)v = v - Pv$ for some $v \in X$. So $$Py = P(v - Pv) = 0.$$ Therefore, ran$\,(I - P) = \ker P$. \\ $\bullet$ Similarly, we can show that ran $P = \ker (I - P)$. \end{proof} \smallskip\noindent (b) The projections $P$, $Q$ are orthogonal, written $P\perp Q$, if $PQ=QP=0$. Show that $P + Q$ is a projection if and only if $P\perp Q$. \begin{proof} First let's assume that $P\perp Q$. We want to prove that $P + Q$ is a projection and for this we will use that $P^2=P$ and that $Q^2=Q$: \begin{equation*} \begin{aligned} (P+Q)^2(x)=&(P+Q)(Px+Qx)\\ =&P(Px+Qx)+Q(Px+Qx)\\ =&P^2(x)+PQ(x)+QP(x)+Q^2(x)\\ =&P(x)+Q(x). \end{aligned} \end{equation*} Now, let's assume we know that $P+Q$ is a projection and we want to prove that this implies that $P\perp Q$. Hence, since we assumed that $P+Q$ is a projection , we know that this equality holds $(P+Q)^2=P+Q$, i.e., \begin{equation*} \begin{aligned} (P+Q)^2(x)=&(P+Q)(Px+Qx)\\ =&P(Px+Qx)+Q(Px+Qx)\\ =&P^2(x)+PQ(x)+QP(x)+Q^2(x)\\ =&P(x)+PQ(x)+QP(x)+Q(x)\\ =&(P+Q)(x)+PQ(x)+QP(x). \end{aligned} \end{equation*} This means that \begin{equation*} \begin{aligned} 0=PQ(x)+QP(x). \end{aligned} \end{equation*} Hence, $PQ = -QP$.\\ Let $x \in \mbox{ran}\, PQ$. We have that $$PQx = x .$$ Applying $P$ to both sides of this equation, we get $Px = x$. \\ Analougus, $$-QPx = x.$$ Applying $Q$ to both sides gives us $Qx = x$. \\ Therefore \[ x = -QPx = -Qx = -x.\]\\ Hence, $x = 0$, which implies $PQ = 0 = QP$. \\ \\ \ \end{proof} \smallskip\noindent (c) If the projections $P$, $Q$ commute, show that $PQ$ is the projection onto $\ran P \cap \ran Q$ along $\ker P + \ker Q$. \begin{proof} We know that $PQ=QP$. To shwo that $PQ$ is the projection onto $\ran P \cap \ran Q$ along $\ker P + \ker Q$, let's apply the definition: \begin{equation*} \begin{aligned} (PQ)^2(x)=&(PQ)(P(Qx))\\ =&(PQ)(Q(Px))\\ =&P(Q^2(Px))\\ =&P(Q(Px))\\ =&P(P(Qx))\\ =&P^2(Qx)\\ =&(PQ)(x). \end{aligned} \end{equation*} Hence, we proved that $PQ$ is the projection. Remains to show that is a projection onto $\ran P \cap \ran Q$ along $\ker P + \ker Q$.\\ $\bullet$ Let $x \in \mbox{ran}\,P \cap \mbox{ran}\,Q$. Then $Px = x = Qx$. So $PQx = Px = x$ i.e., $x \in \mbox{ran}\, PQ$.\\ $\bullet$ Let $y \in \mbox{ran}\,PQ$. Then $PQy = y = QPy$. So, applying the samse stratergy as in part (b) we see that $y \in \mbox{ran}\, P$ and $y \in \mbox{ran}\, Q$. \\ Therefore, $\mbox{ran}\, PQ = (\mbox{ran}\,P \cap \mbox{ran}\, Q)$.\\ $\bullet$ Let $x \in \ker P + \ker Q$ and $y \in \ker PQ$. Writing $x$ as $x = u+v$ where $u \in \ker P$ and $v \in \ker Q$ we have \[ PQx = PQ(u+v) = PQu+PQv = PQu = QPu = 0. \] So $\ker P + \ker Q \subset \ker PQ$. \\ $\bullet$ Let $x \in \ker PQ$. Then, either $x \in \ker Q$ or $x \in (\ker Q)^C$. If $x \in \ker Q$, then $x = 0 + x$. If $x \in (\ker Q)^C$, then $Qx \in \ker P$. \\ Hence $x$ can be expressed as $x = x +Qx - Qx$. Observe that $x - Qx \in \ker Q$. \\ Therefore, $\ker PQ = \ker P + \ker Q$. \\ \ \end{proof} \smallskip\noindent (d) Give an example (or examples) to show that $P+Q$ need not be a projection if $PQ=0$ but $QP\ne 0$, and $PQ$ need not be a projection if $P$,$Q$ do not commute. \begin{proof} Consider the prijection given by the matrices as follows: \[ A = \left( \begin{array}{ll} 0 & 0 \\ 0 & 1 \end{array} \right) \hspace{2pc} \mbox{and} \hspace{2pc} A = \left( \begin{array}{ll} 0 & 1 \\ 0 & 1 \end{array}\right). \] Please check that these matrices are projections, but their sums are not. \end{proof} \bigskip\noindent \textbf{2.} Let $\mathcal{H} = L^2(\mathbb{R})$. For any Lebesgue measurable set $A \subset \mathbb{R}$, define \[ P_A : \mathcal{H} \to \mathcal{H} \] by $P_A f = \chi_A f$ where $\chi_A$ is the characteristic function of $A$. (We define $P_\emptyset = 0$.) Show that $P_A$ is an orthogonal projection. What are its range and kernel? Show that $P_A$, $P_B$ commute. What is $P_A P_B$? When is $P_A \perp P_B$? What is $P_A + P_B$ in that case? \begin{proof} $\bullet$ $P_A f = \chi_A f$ where $\chi_A$ is the characteristic function of $A$. >From the following computations: \begin{equation*} \begin{aligned} P_A(P_A f) = P_{A}(\chi_A f)=\chi_A (P_A f)=\chi_A (\chi_A f)=\chi_A f=P_A f, \end{aligned} \end{equation*} we get that $P_A f$ is a projection on $\mathcal{H}$. To prove that indeed is a orthogonal projection we need to show that \begin{equation*} \begin{aligned} \left\langle P_A f,g\right\rangle =\left\langle P_A g,f\right\rangle \ \mbox{for any} \, f, \, g \in L^2(\mathbb{R}). \end{aligned} \end{equation*} But this translates to: \begin{equation*} \begin{aligned} \left\langle P_A f,g\right\rangle =&\int _{\mathbb{R}}\chi_{A}f\bar{g}\, dx\\ =&\int _{\mathbb{R}}\chi_{A}f\, g\, dx\\ =&\int _{\mathbb{R}}\chi_{A} g \bar{f}\, dx\\ =&\int _{\mathbb{R}}\chi_{A} g f\, dx\\ =&\left\langle P_A g,f\right\rangle. \end{aligned} \end{equation*} Hence we proved that $P_A f$ is a orthogonal projection on $\mathcal{H}$. \\ $\bullet$ $\mbox{Ker}(P_{A})=\left\lbrace f\in L^2(\mathbb{R}) \, \mid \, P_{A}f=0 \right\rbrace $. Hence \begin{equation*} \ker P_A = \left\{ f \in L^2(\mathbb{R}) : f = 0 \mbox{ a.e. on } A \right\} \end{equation*} $\bullet$ From the above rationament we conclude that the $ \mbox{ran}\, P_A = \left\{ f \in L^2(\mathbb{R}) : f = 0 \mbox{ a.e. on } A^C \right\}$ .\\ $\bullet$ Let $A$ and $B \subset \mathbb{R}$, then $P_A f = \chi_A f$ where $\chi_A$ is the characteristic function of $A$, and $P_B f = \chi_B f$ where $\chi_B$ is the characteristic function of $B$. For any $f \in \mathcal{H}$ \begin{equation*} \begin{aligned} (P_{A}P_{B})(f)=&P_{A}(P_{B}(f))\\ =&P_{A}(\chi_{B}f)\\ =&\chi_{A}\chi_{B}f\\ =&\chi_{B}\chi_{A}f\\ =&P_{B}(\chi_{A}f)\\ =&P_{B}(P_{A}(f))\\ =&(P_{B}P_{A})(f). \end{aligned} \end{equation*} Hence, $P_{A}$ and $P_{B}$ commute.\\ $\bullet$ Looking at the expression of $P_{A}P_{B}$ we've wrote above , we see that \begin{equation*} \begin{aligned} (P_{A}P_{B})(f)=\chi_{B}\chi_{A}f=\chi_{A\cap B}f \end{aligned} \end{equation*} $\bullet$ When $P_{A}\perp P_{B}$?\\ \quad \emph{Answer}: $P_A \perp P_B$ if $A \cap B = \emptyset$.\\ $\bullet$ In this case, i.e., when $P_{A}\perp P_{B}$ then $P_{A}+P_{B}$?\\ \quad \emph{Answer}:$P_A + P_B = P_{A \cup B}$. \\ \ \end{proof} \ \\ \ \bigskip\noindent \textbf{3.} Suppose that $\mathcal{H}$ is a separable Hilbert space with ON basis $\left\lbrace e_n : n\in \mathbb{N}\right\rbrace $. Let $M$ be the closed linear span of \[ e_1,\quad e_3,\quad e_5,\quad e_7, \quad\dots \] and $N$ the closed linear span of \[ e_1 + \frac{1}{2} e_2,\quad e_3 + \frac{1}{2^2} e_4,\quad e_5 + \frac{1}{2^3} e_6,\quad e_7 + \frac{1}{2^3} e_8\quad \dots. \] (a) Show that $M \cap N = \left\lbrace 0\right\rbrace $. If $X = M\oplus N$, show that \[ \overline{X} = \mathcal{H},\qquad X\ne \mathcal{H}. \] (Thus, $X$ is an inner-product space when equipped with the $\mathcal{H}$-inner-product.) \begin{proof} $\bullet$ Suppose $x \in M \cap N$ and let's take $x \neq 0$ because it is obviously that $0 \in M \cap N$. Then $x$ may be expressed in terms of the elements of the basis of $M$ as follows \begin{equation*} \begin{aligned} x=\sum _{i \in \mathbb{N}}x_{i}e_{i} \end{aligned} \end{equation*} and also in terms of the elements of the basis $N$ as follows \begin{equation*} \begin{aligned} x=\sum _{i \in \mathbb{N}}\tilde{x}_{i}\left( e_{i} +\frac{1}{2^{\frac{i+1}{2}}}e_{i+1}\right). \end{aligned} \end{equation*} There must be an $i$, which I will denote by $j \in \mathbb{N}$, such that $x_{j}\neq 0$. In this case, we must have $\tilde{x}_{j}\neq 0$, which in particular means that the coefficient of $e_{j+1}$ is nonzero. But $e_{j+1}$ is an element in $N \setminus M$. Therefore, $M\cap N =\left\lbrace 0 \right\rbrace $.\\ $\bullet$ Now, I want to prove that given $X=M\oplus N$, then $\bar{X}=\mathcal{H}$ and also $X \neq \mathcal{H}$. We know $\mathcal{H}$ is a Hilbert space, so in particular it is complete; this implies that $X \subset \mathcal{H}$. Hence, we need to prove that there exists an element $m$ that belongs to $\mathcal{H} \setminus X$. Since $\left\lbrace e_{n} \, \mid \, n\in \mathbb{N}\right\rbrace $ is an ON basis for $\mathcal{H}$, then we can write $m$ in terms of the elements of the basis: \begin{equation*} m=\sum_{i\in \mathbb{N}}x_{i}e_{i}. \end{equation*} But also we can express the element $m$ in terms of the basis of $N$ and also in terms of the basis of $M$. Matching $i$=even (which are the terms in $N$) and $j$=odd (which are the terms in $M$) we get: \begin{equation*} \begin{aligned} &x_{i}2^{\frac{i}{2}}\left( e_{i-1}+\frac{1}{2^{\frac{i}{2}}}e_{i}\right)=x_{i}2^{\frac{i}{2}}e_{i}+x_{i}e_{i} \ \mbox{in}\ N\\ &x_{i}e_{i}-x_{i+1}2^{\frac{i+1}{2}}e_{i}=\left( x_{i}-x_{i+1}2^{\frac{i+1}{2}} \right) e_{i}\ \mbox{in}\ M\\. \end{aligned} \end{equation*} Having this said we can ``approximate'' $m$ by the following sequence of elements: \begin{equation*} \begin{aligned} m_{n}=\sum _{i= even\, \in \mathbb{N}}^n x_{i}2^{\frac{i}{2}}\left( e_{i-1}+\frac{1}{2^{\frac{i}{2}}}e_{i}\right)+\sum _{i= odd\, \in \mathbb{N}}^n\left( x_{i}-x_{i+1}2^{\frac{i+1}{2}} \right) e_{i}. \end{aligned} \end{equation*} Note that the RHS has two terms, and it is trivial to see that the first term belongs to $N$ and the second term belongs to $M$. To be more clear the reason those partial sums are in $N$, respectively in $M$, is because partial sums with finite terms are always convergent. Take the limit as $n$ goes to $\infty$ from $m_{n}$ and set $m = \lim_{n\rightarrow \infty}$ and also set $x_{i}=\frac{1}{i}$. We can see that using Parceval's Theorem, we get that \begin{equation*} \sum_{n}^{\infty} \frac{e_{i}}{i}=m. \end{equation*} But, $m \in \mathcal{H}$ because the series $\sum_{n}^{\infty}\frac{1}{i^2}$ is convergent. Therefore, $m\not \in M\oplus N$. This is true because of the partial sum of the terms of $N$,which made us observe that \begin{equation*} \sum _{i= odd\, \in \mathbb{N}}\frac{2^{i}}{i} e_{i}+\frac{1}{i}e_{i} \end{equation*} doesn't converge since by the $n$-th therm test we can see that $\dfrac{2^{\frac{i}{2} } }{i}$ goes to $\infty$ as $i$ goes to $\infty$. \end{proof} \smallskip\noindent (b) Let $P : X\to X$ be the projection of $X$ onto $M$ along $N$. Show that $P$ is unbounded. \begin{proof} Let's compute the norm of $P$ where \begin{equation*} \Vert P\Vert =\sup_{\Vert x\Vert =1}\Vert Px\Vert . \end{equation*} Take the sequence $x_{n}:= e_{2n}$ and see how $P$ is acting on the elements of this sequence. \\ Note that the norm of $x_{n}$ is $1$, i.e., $\Vert x_{n}\Vert =1$. We have the following ways of writing the elements of $x_{n}$ sequence . The idea is to write them in terms of the basis of $M$ and $N$ respectively; \begin{equation*} e_{2} =2 (e_{1}+\frac{1}{2}e_{2})-2e_{1}. \end{equation*} Therefore \begin{equation*} Px_{1} =P e_{2}=\Vert 2e_{1}\Vert =2. \end{equation*} Continuing we get \begin{equation*} e_{2n} =2 ^n(e_{2n-1}+\frac{1}{2^n}e_{2n})-2^ne_{2n-1}. \end{equation*} Therefore \begin{equation*} Px_{n} =P e_{2n}=\Vert 2e_{2n-1}\Vert =2^n. \end{equation*} Hence, as $n \rightarrow \infty$, $\Vert Px_{n}\Vert \rightarrow \infty$. So $P$ is unbounded. \end{proof} \newpage \bigskip\noindent \textbf{4.} Let $\mathcal{H} = H^1(\mathbb{T})$ denote the Sobolev space of $2\pi$-periodic functions in $L^2(\mathbb{T})$ whose weak derivative belongs to $L^2(\mathbb{T})$ with inner product \[ \langle u, v \rangle_{\mathcal{H}} = \int_{\mathbb{T}} \left(\bar{u} v + \bar{u}^\prime v^\prime\right)\, dx. \] For $f\in L^2(\mathbb{T})$, define $F : \mathcal{H} \to \mathbb{C}$ by \[ F(v) = \int_{\mathbb{T}} \bar{f} v\, dx. \] Show that $F\in \mathcal{H}^\ast$ and find the element $u\in \mathcal{H}$ such that \[ F(v) = \langle u, v \rangle_{\mathcal{H}}. \] What is $\|F\|_{\mathcal{H}^\ast}$? \begin{proof} To show that $F\in \mathcal{H}^\ast$ we need to show that the functional is bounded in $H^1(\mathbb{T})$: \begin{equation*} \begin{aligned} \vert F(v)\vert=&\vert \int_{\mathbb{T}}\bar{f}v\vert \, dx\\ \leq &\int_{\mathbb{T}}\vert \bar{f}\vert \vert v\vert \, dx\\ \end{aligned} \end{equation*} Applying Cauchy-Schwartz inequality, we get \begin{equation*} \begin{aligned} \vert F(v)\vert \leq &\int_{\mathbb{T}}\vert \bar{f}\vert \vert v\vert \, dx\\ \leq &\left( \int_{\mathbb{T}}\vert \bar{f}\vert ^2 \, dx\right) ^{\frac{1}{2}} \left( \int_{\mathbb{T}}\vert v \vert ^2 \, dx\right) ^{\frac{1}{2}} \\ =&\Vert f\Vert_{L^2({\mathbb{T}})}\Vert v\Vert_{L^2({\mathbb{T}})}\\ \leq&\Vert f\Vert_{L^2({\mathbb{T}})}\Vert v\Vert_{H^1({\mathbb{T}})}.\\ \end{aligned} \end{equation*} By Riesz representation theorem, since $F$ is a bounded linear functional on the Hilbert space $\mathcal{H}$, then there is a unique vector $u \in \mathcal{H}$ such that \[ F(v) = \langle u, v \rangle_{\mathcal{H}}. \] Let's find $u$ such that \begin{equation*} \begin{aligned} \int_{\mathbb{T}}\bar{f} v \, dx=\int_{\mathbb{T}} \left(\bar{u} v + \bar{u}^\prime v^\prime\right)\, dx. \end{aligned} \end{equation*} We can rewrite the expression above as follows: \begin{equation*} \begin{aligned} \int_{\mathbb{T}}\left( \bar{f} v-\bar{u} v - \bar{u}^\prime v^\prime \right) \, dx=0 , \end{aligned} \end{equation*} which is equivalent to (integration by parts): \begin{equation*} \begin{aligned} \int_{\mathbb{T}}\left( \bar{f} v-\bar{u} v + \bar{u}^{\prime \prime }v \right) \, dx&=0 , \ \mbox{for} \ \forall v \in \mathcal{H}\\ \int_{\mathbb{T}}\left( \bar{f} -\bar{u} + \bar{u}^{\prime \prime} \right)v \, dx&=0 , \ \mbox{for} \ \forall v \in \mathcal{H}. \end{aligned} \end{equation*} But this means that $u$ is a weak solution of the ODE: $$\bar{f} -\bar{u} + \bar{u}^{\prime \prime} =0.$$ Rearranging the terms, and taking the complex conjugate , we get that the ODE can be written as: $$ u^{\prime \prime}-u =-f,$$ where $f \in L^2 (\mathbb{T})$. This is a second order inhomogeneous constant coefficient ODE, which can be solved as follows:\\ 1) You solve the homogeneous ODE: $ u^{\prime \prime}-u =0$ and find the homogeneous solution $u_{h}$\\ 2) You look for a particular solution $u_{p}$ which can be found using the \emph{variation of parameters},\\ 3) The solution is given by $u=u_{h}+u_{p}.$ This way is much harder, in the sense that the computations get messier than expected. I have managed to finish them, but doesn't deserve the time to type them up. So better let's try something easier. \ \\ \ \textbf{ Easier way to compute the norm!} \\ We will solve that ODE writing $u$ by its Fourier series. Since $u \in H^1(\mathbb{T})$, then $f$, $u^{'}$, and $u \in L^2(\mathbb{T})$. \\ Hence, \begin{equation*} \begin{aligned} &f= \sum_{n \in \mathbb{Z}} \hat{f}(n)e^{inx},\\ &u = \sum_{n \in \mathbb{Z}} \hat{u}(n)e^{inx}, \\ &u = -\sum_{n \in \mathbb{Z}} n^2\hat{u}(n)e^{inx}.\\ \end{aligned} \end{equation*} Plugging in in $ u^{\prime \prime}-u =-f$ we get: \begin{equation*} \begin{aligned} -\sum_{n \in \mathbb{Z}} \hat{f}(n)e^{inx}=-\sum_{n \in \mathbb{Z}} n^2\hat{u}(n)e^{inx}- \sum_{n \in \mathbb{Z}} \hat{u}(n)e^{inx}. \end{aligned} \end{equation*} This is equivalent to \begin{equation*} \hat{f}(n) = \hat{u}(n) +n^2\hat{u}(n) \end{equation*} or \begin{equation*} \hat{u}(n) = \frac{\hat{f}(n)}{1+n^2}. \end{equation*} Thus, \begin{equation*} u = \sum_{n \in \mathbb{Z}}\frac{\hat{f}(n)}{1+n^2}e^{inx}. \end{equation*} Note that we didn't really know much about $u^{''}$, meaning we didn't know its regularity, but we can prove that it belongs to $L^2(\mathbb{T})$, by showing that its Fourier coefficients are square summable. This is easy to see: \begin{equation*} \begin{aligned} \sum_{n \in \mathbb{Z}} \vert \hat{u}^{''}(n)\vert ^2=&\sum_{n \in \mathbb{Z}} \vert \frac{n^2\hat{f}(n)}{1+n^2}\vert ^2\\ \leq &\sum_{n \in \mathbb{Z}} \vert \hat{f}(n)\vert ^2\\ \leq & \infty, \end{aligned} \end{equation*} because $f \in L^2(\mathbb{T})$. Therefore $u^{''} \in L^2(\mathbb{T})$ Now, applying the Riesz representation theorem and Parseval's identity, we see that \begin{equation*} \begin{aligned} \|F\|_{\mathcal{H}^*} =& \|u\|_{\mathcal{H}} \\ =& \vert\langle u, u \rangle_{\mathcal{H}} \vert ^{\frac{1}{2}}\\ =&\left( \int_{\mathbb{T}} (\bar{u}u+\bar{u^{'}}u^{'})\, dx\right)^{\frac{1}{2}}\\ =& \left( 2\pi\sum_{n \in \mathbb{Z}} \vert \hat{u}(n)\vert ^2 +2\pi \sum_{n \in \mathbb{Z}} \vert \hat{u}^{'}(n)\vert ^2\right)^{\frac{1}{2}}\\ =& \left( 2\pi\sum_{n \in \mathbb{Z}} \vert \frac{\hat{f}(n)}{1+n^2}\vert^2 +2\pi \sum_{n \in \mathbb{Z}} \vert \frac{n\hat{f}(n)}{1+n^2}\vert ^2\right)^{\frac{1}{2}}\\ =& \left( 2\pi\sum_{n \in \mathbb{Z}} \frac{(1+n^2)\vert \hat{f}(n)\vert ^2}{(1+n^2)^2}\vert^2 \right)^{\frac{1}{2}}\\ =& \sqrt{2\pi }\left( \sum_{n \in \mathbb{Z}} \frac{\vert \hat{f}(n)\vert ^2}{(1+n^2)}\right)^{\frac{1}{2}}.\\ \end{aligned} \end{equation*} We are done! \end{proof} \end{document}