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\begin{document}
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\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}