Showing a series is not uniformly convergence

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?

• That's correct. Aug 10 '13 at 13:25
• So if the series converges uniformly, it has to converge uniformly to the pointwise limit? Aug 10 '13 at 13:33

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.

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.