0
$\begingroup$

Consider the sequence of functions $f_n(x)=\frac{x}{nx+1}, x\in(0,1).$ Does it converge uniformly?

If $f_n$ are differentiable $\forall n$ and $f_n(x)$ converges pointwise to $f(x)$, this technique is often used. I' ve tried to apply it:

Pointwise convergence: if $x\in(0,1), \lim_{x \rightarrow \infty} \frac{x}{nx+1} = 0.$

Uniform convergence: $f_n$ are differentiable and $f_n$ converges pointwise to $0$, so we now proceed to find $\sup_{0\leq x\leq 1} |f_n(x)-f(x)|$. In order to find it, we maximize $|f_n(x)-f(x)|=|\frac{x}{nx+1}|$;

$(\frac{x}{nx+1})' =\frac{nx+1-nx}{(nx+1)^2}=\frac{1}{(nx+1)^2}=0$. There is no $x\in(0,1)$ such that $\frac{1}{(nx+1)^2}=0$ so $|f_n(x)-0|$ doesn' t reach it's maximum for $x \in (0,1)$. My question is, is this enough to conclude that $f_n$ does not converge to $f$ uniformly? And if so, why?

$\endgroup$
0
$\begingroup$

Your sequence converges uniformly to the null function. Note that each $f_n$ is increasing and that therefore$$(\forall n\in\mathbb{N})(\forall x\in(0,1)):f_n(x)\leqslant\lim_{x\to1}f_n(x)=\frac1{n+1}$$and that $\lim_{n\to\infty}\frac1{n+1}=0$.

$\endgroup$
  • $\begingroup$ Thanks José Carlos. Do you know why the technique I used failed? I read that it's equivalent to the Cauchy condition, which is necessary and sufficient for the uniform convergence of the sequence... $\endgroup$ – Yagger Apr 8 '18 at 12:19
  • $\begingroup$ @Yagger You proved correctly that $f_n'$ is never $0$. And indeed $f_n$ has no maximum. But how do expect to conclude from that that the sequence doesn't converge uniformly? $\endgroup$ – José Carlos Santos Apr 8 '18 at 12:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.