0
$\begingroup$

Suppose you want to show a series does not converge uniformly on some interval. If you know the point wise limit is $f$, and you can show the $\sup |f_{n} - f|$ does not go to zero on your interval, then that does it, the convergence can't be uniform.

Is this correct?

$\endgroup$
2
  • $\begingroup$ That's correct. $\endgroup$ Aug 10 '13 at 13:25
  • $\begingroup$ So if the series converges uniformly, it has to converge uniformly to the pointwise limit? $\endgroup$
    – Joe B.
    Aug 10 '13 at 13:33
1
$\begingroup$

That is correct.

Uniform convergence implies pointwise convergence. So, given the fact that $f_n\rightarrow f$ pointwise as $n\rightarrow\infty$, the only possible uniform limit of $(f_n)$ is $f$.

But, as you say: uniform convergence of $f_n$ to $f$ on the set $X$ is equivalent to saying that $\sup_{x\in X}\lvert f_n(x)-f(x)\rvert\rightarrow0$ as $n\rightarrow\infty$. So, if this is not the case, then your convergence cannot be uniform.

In some cases, there are also other properties that you can look for; for instance, the uniform limit of continuous functions is continuous, and so if your $f_n$ are all continuous and your $f$ is not, then convergence cannot be uniform.

$\endgroup$
0
$\begingroup$

Claim: Suppose $f_n(x) \to f(x)$ as $n \to \infty$ for all $x \in X$ and set $M_n = \sup_{x \in X}|f_n(x) - f(x)|$, then $f_n \to f$ uniformly on $X$ if and only if $\lim_{n \to \infty} M_n = 0$.

Proof: The proof pretty much just uses the definition of uniform convergence of a sequence of functions.

($\implies$) Suppose $f_n \to f$ uniformly on $X$. Then given $\varepsilon > 0$ there exists an $N \in \mathbb{N}$ such that for all $n \geq N$ and for all $x \in X$ we have that $|f_n(x) - f(x)| < \varepsilon$. Hence, $M_n \leq \varepsilon$. As $\varepsilon$ is arbitrary, we are done.

The other direction is basically the same but it would probably be instructional for you to write out any details you don't see.

$\endgroup$

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.