Nested uniform convergence

Question

Suppose $f^{m,n}$ converges to $f^{n}$ uniformly as $m\to\infty$ and $f^{n}$ converges to $f$ uniformly as $n\to\infty$. Is it true that $f^{m_n,n}$ converges to $f$ uniformly as $n\to\infty$ where $m_n$ is strictly increasing in $n$?

Answer 1

Yes, for some sequence $\{m_n\}%$ strictly increasing to $\infty$; false if $\{m_n\}$ is given in advance. In either case the arguments boil down to the numbers $||f^{m_{n},n}-f||$ etc so the question does not have much to do with functions and uniform limits. Some details: for each $n$ choose $m_n$ such that $||f^{m,n}-f^{n}||<\frac 1 n$ for $m \geq m_n$. Then $||f^{m_{n},n}-f||\to 0$. [ Norm is the sup norm]. For a counterexampe to show that this does not hold for any $m_n$ take $f^{m,n}\equiv \frac m {m+n}$,$m_n=n$.

Answer 2

No. Take$$\begin{array}{rccc}f^{m,n}\colon&[0,1]&\longrightarrow&\mathbb R\\&t&\mapsto&\begin{cases}t^n&\text{ if }t\leqslant 1-\frac1m\\\left(1-\frac1m\right)^n&\text{ otherwise.}\end{cases}\end{array}$$Then $(f^{m,n})_{m\in\mathbb N}$ converges uniformly $f_n(t)=t^n$. However. $(f^{n,n})_{n\in\mathbb N}$ doesn't converge uniformly.