Please check my proof of a bounded linear operator and to show it a finite rank operator

Question

There is an exercise I am trying to solve, but I am not sure if my solution is 100% correct. Please help me to check it carefully.

Let $A = (A_{n,m})_{n,m \in \mathbb{N}}$ with $A_{n,m} \in \mathbb{C}$ for $n,m \in \mathbb{N}$ be a matrix with the norm \begin{align} \|A\| := \sup_{m \in \mathbb{N}} \sum_{n \in \mathbb{N}} |A_{n,m}| < \infty \end{align}

1. Show that there is a well defined bounded linear operator \begin{align} T: l^1(\mathbb{N}) \to l^1(\mathbb{N}), \qquad (Tf)(n) := \sum_{m \in \mathbb{N}} A_{n,m} f(m) \end{align} with $\|T\| = \|A\|$.
2. $T$ is a finite rank operator if $\lim_{n \to \infty} \sum_{m \in \mathbb{N}} |A_{n,m}| = 0$.

My solution:

We first show $T$ is linear. For any $f_1, f_2 \in l^1(\mathbb{N})$ and $\alpha, \beta \in \mathbb{C}$, we have \begin{align} [T(\alpha f_1 + \beta f_2)](n) &= \sum_{m \in \mathbb{N}} A_{n,m} (\alpha f_1 + \beta f_2)(m) \\ &= \sum_{m \in \mathbb{N}} A_{n,m} [\alpha f_1 (m) + \beta f_2(m)] \\ &= \sum_{m \in \mathbb{N}} A_{n,m} \alpha f_1 (m) + \sum_{m \in \mathbb{N}} A_{n,m} \beta f_2(m) \\ &= \alpha \sum_{m \in \mathbb{N}} A_{n,m} f_1 (m) + \beta \sum_{m \in \mathbb{N}} A_{n,m} f_2(m) \\ &= \alpha(T f_1)(n) + \beta (T f_2)(n) \end{align}

We then show $T$ is bounded. \begin{align} |(Tf)(n)| &= \left|\sum_{m \in \mathbb{N}} A_{n,m} f(m)\right| \\ &\le \sum_{m \in \mathbb{N}} \left|A_{n,m}\right| \left|f(m)\right| \\ &\le \sup_{m \in \mathbb{N}} \left|A_{n,m}\right| \sum_{m \in \mathbb{N}} \left|f(m)\right| \\ &\le \sup_{m \in \mathbb{N}} \sum_{n \in \mathbb{N}} |A_{n,m}| \sum_{m \in \mathbb{N}} \left|f(m)\right| \\ &= \|A\| \|f\|_{l^1} \end{align} Therefore, the norm of $T$ is bounded by $\|T\| := \sup_{f \in l^1(\mathbb{N})} \{\|Tf\|: \|f\|_{l^1} = 1\} = \sup_{f \in l^1(\mathbb{N})} \{|(Tf)(n)| : \|f\|_{l^1} = 1, n \in \mathbb{N}\} \le \|A\|$

Now, can we say that $T$ is well defined? Actually, I am not sure about the exact definition of "well-defined".

To show the equality $\|T\| = \|A\|$ holds, I need some idea to construct a specific $f$. I am always not able to find such a $f$ due to the lack of experience. I tried but I am not sure if the following two ideas work.

Let $f = [f(m)]_{m \in \mathbb{N}}$ where \begin{align} f(m) := \sup_{n \in \mathbb{N}}\frac{|A_{m,n}|}{A_{n,m}}, \qquad \|f\| = 1 \end{align} Because I hope to cancel out the $A_{n,m}$ in the summation by $f(m)$. Then \begin{align} \|T\| \|f\| \ge |(Tf)(n)| &\ge \sum_{m \in \mathbb{N}} A_{n,m} f(m) \\ &= \sum_{m \in \mathbb{N}} A_{n,m} \sup_{n \in \mathbb{N}}\frac{|A_{m,n}|}{A_{n,m}} \\ &= \sup_{n \in \mathbb{N}} \sum_{m \in \mathbb{N}} |A_{m,n}| \\ &= \sup_{m \in \mathbb{N}} \sum_{n \in \mathbb{N}} |A_{n,m}| \\ &=: \|A\| \end{align} Thus $\|T\| \ge \|A\|$ in addition to $\|T\| \le \|A\|$ implies the equality. Is this a good idea?

Alternatively, I am thinking about \begin{align} f(m) := \sum_{n \in \mathbb{N}}\frac{|A_{m,n}|}{A_{n,m}}, \qquad \|f\| = ? \end{align} which also makes $|(Tf)(n)| \ge \|A\|$ for sure, but I am not sure if $f \in l^1(\mathbb{N})$. Will this work?

Does the construction of such a specific $f$ usually require $\|f\|=1$?

As to the 2nd question, I have no idea to show $T$ a finite rank operator. The only thing I know is a finite rank operator requires the range $T(V)$ a finite dimensional subspace.

Answer 1

So far I only thought about 1. - your proof of linearity is correct. To show that $T$ is well-defined one has to show $T(\ell_1(\mathbb N))\subseteq \ell_1(\mathbb N)$, i.e., if $f\in\ell_1(\mathbb N)$, then $Tf\in\ell_1(\mathbb N)$. As you stated correctly, $$\|T\|=\sup_{\|f\|_{\ell_1}=1}\|Tf\|_{\ell_1}$$ so in this case, $T$ being well-defined is directly connected to $T$ being bounded. For this one checks $$ \|Tf\|_{\ell_1}=\sum_{n\in\mathbb N} |(Tf)(n)|=\sum_{n\in\mathbb N} \Big|\sum_{m\in\mathbb N}A_{n,m}f(m)\Big|\leq\sum_{n\in\mathbb N}\sum_{m\in\mathbb N}|A_{n,m}f(m)|=\sum_{m\in\mathbb N}\Big(\sum_{n\in\mathbb N}|A_{n,m}|\Big) |f(m)|\leq \Big(\sup_{m\in\mathbb N}\sum_{n\in\mathbb N}|A_{n,m}|\Big)\sum_{m\in\mathbb N} |f(m)|=\|A\|\|f\|_{\ell_1} $$ so $\|T\|\leq\|A\|$. Note that at one point we interchanged the summation over $m$ and $n$ which is allowed because all summands are non-negative.

To show that the norm inequality actually is an equiality (your ideas don't work for various reasons), we have to distinguish two cases. First assume, there exists $m_0\in\mathbb N$ such that $\|A\|=\sum_{n\in\mathbb N}|A_{n,m_0}|$ (so the supremum taken for the norm of $A$ is actually a maximum). Then choose $$ f_{m_0}(m):=\begin{cases} 1&\text{ if }m=m_0\\0&\text{ else}\end{cases} $$ as $\|f_{m_0}\|_{\ell_1}=\sum_{m\in\mathbb N}|f_{m_0}(m)|=1$ and $$ \|Tf_{m_0}\|_{\ell_1}=\sum_{n\in\mathbb N} |(Tf_{m_0})(n)|=\sum_{n\in\mathbb N} |A_{n,m_0}|=\|A\| $$ so $\|T\|\geq\|A\|$ (and in total $\|T\|=\|A\|$ as you pointed out). But the supremum taken in $\|A\|$ might be an actual supremum (''which happens infinitely far back'') so there exists a sequence $(n_k)_{k\in\mathbb N}$ in $\mathbb N$ such that $\|A\|=\lim_{k\to\infty}\sum_{n\in\mathbb N}|A_{n,k}|$. Then the same calculation as above shows $$ \|Tf_{n_k}\|_{\ell_1}=\sum_{n\in\mathbb N}|A_{n,k}|\overset{k\to\infty}\longrightarrow\|A\| $$ using $\|f_{n_k}\|_{\ell_1}=1$ so the supremum (now actual supremum and not maximum) of $\|T\|$ is lower bounded by $\|A\|$ which as above implies $\|T\|=\|A\|$.