# Pondering examples and counterexamples for uniform convergence

As practice for our analysis final, my prof suggested we come up with examples and counterexamples (where one of the conditions isn't satisfied so the result is contradicted) for different theorems on uniform convergence, namely those concerning continuity, differentiability and integrability (outlined here: http://en.wikipedia.org/wiki/Uniform_convergence#Applications).

I've come up with examples and counterexamples and I was hoping someone could confirm they're correct since I might need them on the final.

CONTINUITY:

Example:

$f_n(x) = xe^{-nx}$ is a sequence of continuous functions that converges uniformly on $[0,1]$ to a function $f = 0$, hence f is continuous on $[0,1]$

Counterexample:

$f_n(x) = e^{-nx}$ is a sequence of continuous functions that converges on $[0,1]$ to a function $f$

$$f(x) := \begin{cases} 1 &\text{if x=0}\\ 0 &\text{x\in} (0,1] \end{cases}$$ So $f_n(x)$ converges to a discontinuous function f, which can only happen if convergence is not uniform.

DIFFERENTIABILITY:

Example:

$f_n(x) = \frac{x}{x+n}$ converges uniformly on $[0,1]$ to a function $f = 0$

$f'_n(x) = \frac{n}{(x+n)^2}$ converges uniformly on $[0,1]$ to a function $g = 0$

Hence f is differentiable and f' = g.

Counterexample:

$f_n(x) = \frac{1}{n}\sin{nx}$ converges uniformly on $[0,1]$ to a function $f = 0$

$f'_n(x) = \cos{nx}$ converges non-uniformly on $[0,1]$ to a function $g = 1$

Hence f is differentiable but f' ≠ g.

INTEGRABILITY: Only Riemann integrals.

Example:

Example 4.2 on page 8 of this PDF: http://www.mai.liu.se/~pehor/kurser/TATA57/Uniform%20Convergence.pdf

Counterexample:

I'm not sure here either. I struggle with integration. I found this example in one of the practice problems from my book:

$f_n(x) = \frac{nx}{1+nx}$ because it converges non-uniformly on $[0,1]$.

$$f(x) := \begin{cases} 0 &\text{if x=0}\\ 1 &\text{x\in} (0,1] \end{cases}$$

$$\lim\limits_{n\to\infty} \int_{0}^{1} f_n(x)dx = \int_{0}^{1} f(x)dx = 1$$

Does this work as a counterexample? Why or why not? I don't think there are any examples where $\lim\limits_{n\to\infty} \int_{0}^{1} f_n(x)dx ≠ \int_{0}^{1} f(x)dx$, right?

For you counterexample to differentiability, note that $f_n'(x)=\cos(nx)$ does not converge to any limit on $[0,1]$. (Draw the graph of $\cos(10x)$!) You may still use it as a counterexample, though, since $f_n(x)$ converges uniformly to $f(x)$, which is differentiable, but $f_n'(x)$ does not converge to $f'(x)$.
Your counterexample on integration does not really work, since it would make you believe you did not need the uniform convergence to get the desired limit result. Indeed, you would really like to have an example where $f_n(x)$ converges to $f(x)$ non-uniformly on $[0,1]$, but where $\lim_{n\to\infty}\int_0^1 f_n(x)\,{\rm d}x\neq \int_0^1f(x)\,{\rm d}x$. For a simple example, let $f_n(x)$ be a function with $$f_n(0)=0,\qquad f_n\Bigl({1\over n}\Bigr)=n,\qquad f_n\Bigl({2\over n}\Bigr)=0\qquad f_n(1)=0$$ and which is linear on each of the three intervals $[0,{1\over n}]$, $[{1\over n},{2\over n}]$, and $[{2\over n},1]$. Draw the graphs of $f_n(x)$ for $n=3,4,5$, and determine $f(x)$ and $\int_0^1 f_n(x)\,{\rm d}x$.