Diagonal sequence trick: Uniform bound necessary?

There is a treatment of the "diagonal sequence trick" in Reed and Simon (Functional Analysis Vol.1) stated there as follows:

Let $f_n(m)$ be a sequence of functions on the positive integers which is uniformly bounded, i.e. $|f_n(m)| \le C$ for all $n,m$. Then there is a subsequence $\{f_{\hat n(i)}(m)\}^\infty_{i = 1}$ so that for each fixed $m$, the sequence $f_{\hat n(i)}(m)$ converges as $i \to \infty$.

I was wondering whether the uniform bound is actually necessary. I am not sure where it is essential in the proof, outlined below:

Consider the sequence $f_n(1)$. It is a bounded set of numbers, so we can find a subsequence $f_{n(i)}$ such that $f_{n_1(i)}(1) \to f_\infty(1)$, for some number $f_\infty(1)$. Now consider the sequence $f_{n_1(i)}(2)$. We can find a subsequence $f_{n_2(i)}(2) \to f_\infty(2)$ as $i \to \infty$. Proceeding inductively, we find successive subsequences $f_{n_k(i)}$ so that $f_{n_{k+1}(i)}$ is a subsequence of $f_{n_k(i)}$, and $f_{n_k(i)} \to f_\infty(k)$ as $i \to \infty$. Thus, in particular, $f_{n_k(i)}(j) \to f_\infty(j)$ as $i \to \infty$ for $j = 1,2,\dots,k$. To get a subsequence $f_{\hat n(i)}$ converging for each $j$, take the diagonal sequence $\hat n(k) = n_k(k)$. Then $f_{\hat n(k)}, f_{\hat n(k+1)}, \dots$ is a subsequence of $f_{n_k(i)}$ so $f_{\hat n(i)}(k) \to f_\infty(k)$ as $i \to \infty$ for each $k$.

Solution 1:

The proof needs that $n\mapsto f_n(m)$ is bounded for each $m$ in order to find a convergent subsequence. But it is indeed not necessary that the bound is uniform in $m$ as well. For example, you might have something like $f_n(m) = \sin(nm) e^m$ and the argument still works. Intuitively, we don't need to obtain convergence of the diagonal sequence afresh but obtain it already from it being essentially a subsequence of all subsequences introduced.