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\fancyhead[L]{Hand-in 8 Functional Analysis}
\fancyhead[R]{Thomas van Maaren (9825827)}


\begin{document}


\textbf{Exercise 2.} For each $N \in \mathbb{N}$ we define the functional $M_{N}: \ell^{\infty}(\mathbb{Z}) \rightarrow \mathbb{C}$ by

$$
M_{N}(x)=\frac{1}{2 N+1} \sum_{|n| \leq N} x_{n}, \quad \forall x \in \ell^{\infty}(\mathbb{Z})
$$

For each $a \in \mathbb{Z}$ we define the operator $T_{a}: \ell^{\infty}(\mathbb{Z}) \rightarrow \ell^{\infty}(\mathbb{Z})$ by

$$
\left(T_{a} x\right)_{n}=x_{n+a}, \quad \forall x \in \ell^{\infty}(\mathbb{Z}) .
$$

Let also $e \in \ell^{\infty}(\mathbb{Z})$ be defined by $e_{n}=1$ for all $n \in \mathbb{Z}$, and let $V \subset \ell^{\infty}(\mathbb{Z})$ be the subspace of all $x$ that can be written as a finite sum:

$$
x=\lambda_{0} e+\sum_{j=1}^{p} \lambda_{j}\left(f_{j}-T_{a_{j}} f_{j}\right)
$$

for some ( $x$-dependent) $p \in \mathbb{N}, \lambda_{0}, \ldots, \lambda_{p} \in \mathbb{C}, a_{1}, \ldots, a_{p} \in \mathbb{Z}$, and $f_{1}, \ldots, f_{p} \in \ell^{\infty}(\mathbb{Z})$.

(i) [1 p.] Show that for all $x \in \ell^{\infty}(\mathbb{Z})$ and $a \in \mathbb{Z}$,

$$
\left|M_{N}\left(x-T_{a} x\right)\right| \leq \frac{2|a|}{2 N+1}\|x\|_{\infty} .
$$

(ii) [0,5 p.] For each $x \in V$, show that $\lim _{N \rightarrow \infty} M_{N}(x)$ equals the coefficient $\lambda_{0}$ in any decomposition (1). In consequence, we can define a map $L: V \rightarrow \mathbb{C}$ by

$$
L(x)=\lambda_{0} .
$$

Show that $L \in V^{\prime}$.
(iii) $[0,5$ p. $]$ Show that the functional $L$ has norm $\|L\|=L(e)=1$.

(iv) $[1,5$ p. $]$ Show that there exists a functional $M \in\left(\ell^{\infty}(\mathbb{Z})\right)^{\prime}$ such that:

(a) $M(x)=L(x)$ for all $x \in V$,

(b) $M(e)=\|M\|=1$,

(c) $M(x)=M\left(T_{a} x\right)$ for all $a \in \mathbb{Z}$ and $x \in \ell^{\infty}(\mathbb{Z})$.

(v) $[0,5$ p. $]$ Show that there exists no $y \in \ell^{1}(\mathbb{Z})$ such that

$$
M(x)=\sum_{n \in \mathbb{N}} x_{n} y_{n}, \quad \forall\left(x_{n}\right)_{n \in \mathbb{Z}} \in \ell^{\infty}(\mathbb{Z})
$$

(vi) $[0,5$ p. $]$ Using the result from the previous question show that $\ell^{1}(\mathbb{Z})$ is not reflexive.

In the next exercises, all vector spaces are over $\mathbb{C}$.


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\begin{enumerate}[(i)]
	\item Observe that if $a\geq 0$

		\begin{align}
			\left|M_N(x-T_ax)\right| &= 
			\frac{1}{2N+1}\left|\sum_{|n|\leq N} (x-T_ax)_n\right|=
			\frac{1}{2N+1}\left|\sum_{n=-N}^N x_n - \sum_{n=-N+a}^{N+a}x_n \right|\\
			&= \frac{1}{2N+1} \left| \sum_{n=-N}^{-N+a-1} x_n+ \sum_{n=-N+a}^N x_n
			-\sum_{n=-N+a}^{N} x_n - \sum_{n=N+1}^{N+a} x_n\right|\\
			&= \frac{1}{2N+1} \left| \sum_{n=-N}^{-N+a-1}x_n-\sum_{n=N+1}^{N+a} x_n\right|\leq \frac{1}{2N+1}\left( \sum_{n=-N}^{-N+a-1} |x_n| + \sum_{n=N+1}^{N+a} |x_n|\right)\\
			&\leq \frac{1}{2N+1}\left( \sum_{n=-N}^{-N+a-1} \|x\|_\infty + \sum_{n=N+1}^{N+a} \|x\|_\infty\right)=\frac{1}{2N+1}\left( a \|x\|_\infty + a \|x\|_\infty\right)\\
			&= \frac{2a\|x\|_\infty}{2N+1} = \frac{2|a|\|x\|_\infty}{2N+1}
		\end{align}

	Now we need to prove it is true if $a<0$. It is easy to prove that $T_a$ is an
		isometry. For any $a\in \Z$ and $x\in \ell^\infty$ we see that
		$\|T_a x\|_\infty = \sup_{n\in \Z} |x_{n+a}| = \sup_{n\in \Z} |x_n| =
		\|x\|_\infty$. If $a<0$ we know that $-a>0$, so

		\begin{align}\left|M_n(x-T_ax)\right|&=\left|-M_N(T_a x-x)\right|
			=|M_n(T_ax - T_{-a}T_ax)|\\
			&\leq \frac{2(-a) \|T_ax\|_\infty}{2N+1} =\frac{2|a| \|x\|_\infty}{2N+1} 
		\end{align}

		in the last equality we used that $T_a$ is an isometry.
	\item We will first determine $\lim_{N\to \infty} M_n(e)$
		\begin{align}
			\lim_{N\to \infty} M_n(e) = \lim_{N\to \infty} \frac{1}{2N+1} \sum_{|n|\leq N} 1 = \lim_{N\to \infty} \frac{2N+1}{2N+1} = 1
		\end{align}
		Because of (i) we know that for any $1\leq j\leq p$
		\begin{align}
			\lim_{N\to \infty} |M_N(f_j - T_{a_j}f_j| \leq \lim_{N\to \infty}\frac{2|a|}{2N+1}\|x\|_\infty = 0
		\end{align}
		hence
		\begin{align}
			\lambda_0 &= \lambda_0\cdot 1 + \sum_{j=1}^p \lambda_j \cdot 0=
			\lambda_0\cdot \lim_{N\to \infty} M_N(e) + \sum_{j=1}^p \lambda_j \cdot \lim_{N\to \infty} M_N(f_j-T_{a_j}f_j)\\
			&= \lim_{N\to \infty} M_N\left(\lambda_0 e+\sum_{j=1}^p\lambda_j(f_j-T_{a_j}f_j)\right) = \lim_{N\to \infty} M_N(x)
		\end{align}
		To show that $L\in V'$ we need to show that $L$ is linear and
		bounded. In (iii) we will prove that $\|L\|=1$, hence $L$ is
		bounded and therefore we only need to show that $L$ is linear.
		For this we need to prove that $L(x+\kappa x') = L(x) + \kappa
		L(x')$ for all $x,x'\in V$ and $\kappa\in \F$. We can write $x$
		and $x'$ as $\lambda_{0} e+\sum_{j=1}^{p}
		\lambda_{j}\left(f_{j}-T_{a_{j}} f_{j}\right)$ and
		$\lambda_{0}' e+\sum_{j=1}^{p'}
		\lambda_{j}'\left(f_{j}'-T_{a_{j}'} f_{j}'\right)$. Now define
		$\mu_0=\lambda_0 + \kappa \lambda_0'$ and $\mu_i = \lambda_i,b_i=a_i,g_i=f_i$
		for $1\leq i\leq p$ and $\mu_i = \kappa \lambda_{i-p}',b_i=a_{i-p}',g_i=f_{i-p}$ for
		$p<i\leq p+p'$. We see that $x+\kappa y= \mu_0 e + \sum_{i=1}^{p+p'} \mu_i (g_i - T_{b_i})$, so $L(x+\kappa y) = \mu_0 = \lambda_0 + \kappa \lambda_0' = L(x) + \kappa L(x')$.
	\item We saw in (ii) that $L(e) = 1$ and we know that $\|e\|_\infty = 1$, hence $\|L\|\geq 1$. For all $x\in V$ we see that
		\begin{align}
			|L(x)| &= |\lambda_0| = \lim_{N\to \infty} |M_N(x)| \leq 
			\lim_{N\to \infty} \frac{1}{2N+1} \sum_{|n|\leq N} |x_n|
			\leq \lim_{N\to \infty} \frac{1}{2N+1} \sum_{|n|\leq N} \|x\|_\infty\\
			&= \lim_{N\to \infty} \frac{2N+1}{2N+1} \|x\|_\infty = \|x\|_\infty
		\end{align}

		hence $\|L\|=1$.
	\item We know that $V$ is a linear subspace of $\ell^\infty(\Z)$ and that $L\in V'$. Because of theorem 5.19 we know there exists $M\in \left(\ell^\infty\right)'$ such that $\|M\| = \|L\|$. Because it is an extension we know that $M(x) = L(x)$ for all $x\in V$, Also we see that $M(e) = L(e) = 1 = \|L\| = \|M\|$. Finally we also see that $x-T_ax\in V$ and also $L(x-T_ax)=0$ for all $x\in \ell^\infty(\Z)$ and $a\in\Z$. Thus

		$$M(x) = M(x-T_ax)+M(T_ax) = L(x-T_ax)+M(T_ax)=M(T_ax)$$
	\item Assume that there exists an $y\in \ell^1(\Z)$ such that $$M(x)=\sum_{n\in \Z} x_ny_n$$ for all $(x_n)_{n\in \Z} \in \ell^\infty(\Z)$.
		We know $M(e) = 1$, therefore $\sum_{n\in \Z} y_n =1$, so there must be a $n\in \N$ such that $y_n\neq 0$. Now define $e^n\in \ell^{\infty}(\Z)$
		as $e^n_n =1$ and $e^n_i=0$ for all $i\in \Z\backslash \{n\}$. We see because of (iv)(c) that $y_{n'} = M(e^{n'}) = M(T_{n'-n}e^{n}) = M(e^{n}) = y_n$ for all $n'\in \Z$.
		Therefore $(y_n)_{n\in \Z}$ is a constant sequence unequal to 0. Because of this $y\notin \ell^1(\Z)$. This is a contradiction.
	\item Define $S : \ell^\infty(\Z) \to (\ell^1(\Z))'$ as $S(x)(y) = \sum_{n\in \Z}x_ny_n$. We will show that $S$ is well-defined and bijective.
		\textbf{well-defined:} Clearly $S(x)$ is linear for all $x\in
		\ell^\infty(\Z)$. We see that $$|S(x)(y)| = \left|\sum_{n\in
		\Z}x_ny_n \right|\leq \sum_{n\in \Z}|x_n| |y_n|\leq
		\|x\|_\infty\sum_{n\in \Z} |y_n|=\|x\|_\infty \|y\|_1$$, hence
		$\|S(x)\|\leq \|x\|_\infty$, also $$\|S(x)\| = \sup_{\|y\|_1=1}|S(x)(y)| \geq \sup_{n\in \Z}|S(x)(e^n)| = \sup_{n\in \Z}|x_n| = \|x\|_\infty$$ and therefore
		$\|S(x)\|= \|x\|_\infty$, so
		$S(x)\in(\ell^1(\Z))'$  for all $x\in \ell^\infty(\Z)$.
		\textbf{injective:}
		Let $x,x'\in \ell^\infty(\Z)$ and assume $S(x) = S(x')$. We see that $x_n = S(x)(e^n) = S(x')(e^n) = x_n'$ for all $n\in \Z$, hence $x=x'$.
		\textbf{surjective:}
		Let $f\in \ell^1(\Z)$. Choose $x_n:=f(e^n)$ for all $n\in \Z$. We see that $\|x\|_\infty = \sup_{n\in\Z} |f(e^n)| \leq \sup_{n\in\Z} \|f\|\|e^n\| =\|f\|$, therefore
		$x\in \ell^\infty(\Z)$. Also because $f$ is continuous $S(x)(y)
		= \sum_{n\in \Z} x_ny_n = \sum_{n\in \Z} f(e^n)y_n = \sum_{n\in
		\N} f(y_ne^n) = \lim_{N\to \infty} \sum_{|n|\leq N} f(y_ne^n) =
		f(\lim_{N\to \infty} \sum_{|n|\leq N} y_ne^n) = f(y)$ for all
		$y\in \ell^1(\Z)$.
		
		Because of this we can define $\Psi : \left(\ell^1(\Z)\right)' \to \F$ as $\Psi(S(x)) = M(x)$ for all $x\in \ell^\infty(\Z)$. We show that $\Psi$ is linear.
		Let $S(x),S(x')\in \left(\ell^1(\Z)\right)'$ and $\lambda \in \R$ then 
		$$\Psi(S(x)+\lambda S(x')) = \Psi(S(x+\lambda x')) = M(x+\lambda x') = M(x) + \lambda M(x') = \Psi(S(x)) + \lambda \Psi(S(x'))$$

		also it is bounded
		$$\left| \Psi(S(x))\right| = |M(x)| \leq \|M\|\|x\|_\infty = \|M\|\|S(x)\|$$

		hence $\|\Psi\|\leq \|M\|$ and therefore $\Psi \in
		\left(\ell^1(\Z)\right)''$. Because of (v) we know that there
		is no $y\in \ell^1(\Z)$ such that $\Psi(S(x)) = S(x)(y) =
		J_{\ell^1(\Z)}(y)(S(x))$ for all $x\in \ell^\infty(\Z)$, hence
		$\Psi \neq J_{\ell^1(\Z)}(y)$ for all $y\in \ell^1(\Z)$, hence
		$J_{\ell^1(\Z)}(\ell^1(\Z)) \neq \left(\ell^1(\Z)\right)''$ and
		therefore $\ell^1(\Z)$ is not reflexive.
\end{enumerate}

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