Do we have the following uniform convergence?

Question

We are given a sequence $f_N$ such that $f_N(x) \xrightarrow{N\to\infty} f(x)$ uniformly in $x$. $f$ is continuous, but the $f_N$'s are not necessarily continuous. Moreover we know that $y_N = \frac{\lfloor Ny \rfloor}{N} \xrightarrow{N\to\infty} y$, thus also $f_N(y_N) \xrightarrow{N\to\infty} f(y)$ uniformly. Also given are bounded and smooth functions $\alpha_N$ and $\alpha$ such that $\alpha_N(x) \xrightarrow{N\to\infty} \alpha(x)$ uniformly in $x$. Question: What can we say about the convergence of $f_N(y_N)\alpha(y_N)$? I would think it converges uniformly to $f(y)\alpha(y)$.

Answer 1

Three points :

• You never mentioned that the convergence of $\alpha_N \to \alpha$ is uniform. I hope so, otherwise you definitely can't get your result.
• Assuming that, you essentially have a correct proof in your comment, except that you don't know whether $f_N(y_N)$ is bounded. But of course it is : $y_N$ is a convergent sequence, so you can find a closed interval $[a,b]$ that contains it. On this interval, $y$ is bounded in absolute value (by say $K$) because it is continuous. Moreover, for $N$ large enough, $|f_n - f| \leq 1 $ because of uniform convergence. Hence on $[a,b]$, $|f_N| \leq K+ 1$ for $N$ large enough. Hence $|f_N(y_N)| \leq 1$ for $N$ large enough.
• You should not say $f_N(x) \xrightarrow{N\to\infty} f(x)$ uniformly. It is meaningless for a convergence of numbers to be uniform. You should say $f_N \xrightarrow{N\to\infty} f$ uniformly. Don't mix up the function $f$ with the number $f(x)$.