Pointwise convergence does not imply $f_n(x_n)$ converges to $f(x)$

Question

I have been given a sequence of real valued continuous functions $(f_1,f_2,...,f_n)$, and a real valued sequences $(x_1,x_2,...,x_n)$, where $x_n$ converges to $x$. Also $f$ is a continuous real valued function.

Then if $f_n$ converges to $f$ uniformly on $\mathbb R$, I think I have shown that this implies that $f_n(x_n)$ converges to f(x).

But does this hold when $f_n$ converges to $f$ pointwise on $\mathbb R$? I am trying to come up with a counterexample, but I haven't had any luck so far.. Does anyone know of such an example, or does pointwise convergence imply the same as uniform convergence?

Thank you!

Answer 1

Even on a compact interval instead of $\mathbb R$, this cannot be true...

Try $f_n$ defined on $[0,1]$ by $$f_n(x)=nx^n(1-x), $$ and $$x_n=1-1/n. $$ Then $x_n\to x$ with $x=$ $___$, $f_n\to f$ pointwise with $f=$ $___$, but $f_n(x_n)\to$ $___$ while $f(x)=$ $___$ hence the convergence cannot be uniform.

Note that the regularity of each $f_n$ is not an issue.

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Another example: pick some nonzero continuous function $g$ such that $g(x)\to0$ when $|x|\to\infty$, say $g:x\mapsto\mathrm e^{-x^2}$, and define $f_n$ by $f_n(x)=g(n^2x-n)$ for every $x$.

Then $f_n\to f$ pointwise, where $f=$ $___$, but $f_n((n+x_0)/n^2)$ does not converge to $___$, for some well chosen $x_0$.

Answer 2

If you have any function $g\colon [0,1]\to\mathbb R$, a sequence $(a_n)$ and you define functions $f_n\colon [0,1]\to\mathbb R$ with $f_n(x)=g(x)$ for $x\ge\frac2n$, $f_n(0)=g(0)$ and $f_n(\frac1n)=a_n$ then $f_n\to g$ pointwise, but $f_n(\frac1n)=a_n$.

If $g$ is continuous, you can make the $f_n$ continuous. If $g\in C^\infty$ you can arrange that $f_n\in C^\infty$.