A question on point-wise/uniform convergence of a sequence of functions

Question

Let $\{f_n(\cdot)\}_{n \in \mathbb{N}}$ be a sequence of real-valued functions defined on $[0, k]$ for some $0 < k < 1$. Suppose that for every fixed $x \in [0,k]$, we have the point-wise convergence \begin{equation} \lim_{n \to \infty} f_n(x) = 0. \end{equation} The question is: if I let $\{x_n\}_{n \in \mathbb{N}}$ be a sequence of arbitrary points in $[0,k]$. Then does it hold that \begin{equation} \sup_{n \in \mathbb{N}} | f_n(x_n) | < \infty? \end{equation}

Answer 1

The answer is no.

Here is a counter example: $f_n(x):[0,\frac{2}{3}] \to \mathbb R$ with $f_1(x)=0$ and for $n \geq 2$ $$f_n(x)=\left\{ \begin{array}{lc} n^2x &\mbox{ if } 0 \leq x \leq \frac{1}{n} \\ 2n-n^2x &\mbox{ if } \frac{1}{n} < x < \frac{2}{n} \\ 0 &\mbox{ if } x >\frac{2}{n}\\ \end{array} \right.$$

Note that $x_n=\frac{1}{n}$ gives you what you want.

Answer 2

No, let $f_{n}:[0,k]\rightarrow\mathbb{R}$ be given by $$f_{n}(x)=\begin{cases}n&\text{ if }x=\frac{1}{n}\\0&\text{ else}\end{cases}.$$ Clearl pointwise $f_{n}$ converges to $0$, but $$\lim_{n\rightarrow\infty}f_{n}(\frac{1}{n})=\lim_{n\rightarrow\infty}n=\infty.$$

Answer 3

No. Suppose that:

• $f_n$ is an affine function on $\left[0,\frac k{4n}\right]$ with $f_n(0)=0$ and $f_n\left(\frac k{4n}\right)=n$;
• $f_n$ is an affine function on $\left[\frac k{4n},\frac k{2n}\right]$ with $f_n\left(\frac k{4n}\right)=n$ and $f_n\left(\frac k{2n}\right)=0$;
• $f_n$ is the null function on $\left[\frac k{2n},k\right]$.

Then $f_n\left(\frac k{4n}\right)=n$ for each $n$.

Answer 4

The question has been answered. To answer a question the OP asks in comments on the answers: It's easy to see that $\sup_nf_n(x_n)<\infty$ for every sequence $x_n$ if and only if the $f_n$ are uniformly bounded ($f_n(x)\le C$ for every $n$ and every $x$.)

And, since it seems possible that this is what you really want to know: It's also easy to see that $f_n(x_n)\to0$ for every sequence $(x_n)$ if and only if $f_n\to0$ uniformly.