## May 2000#5

Problem Statement: For every $n\in\mathbb{N}$ and $x\in(0,1)$ define $f_n(x)=\dfrac{1}{nx+1}$. Find the function $f(x), f:(0,1)\rightarrow\mathbb{R}$ that $\{f_n(x)\}$ converges to pointwise. Is the converges uniform?

Proof: Fix $x\in(0,1)$. Then $\lim_{n\rightarrow\infty}\dfrac{1}{nx+1}=0$ and so $\{f_n(x)\}$ converges to $0$ pointwise.

Claim: This convergence is not uniform.

If the convergence is uniform then it follows that for every $\varepsilon>0$ we may find an $N\in\mathbb{N}$ such that $|f_n(x)|<\varepsilon$ for every $x\in (0,1)$ and for every $n>N$. Let $\varepsilon=\dfrac{1}{3}$ and $n>N$. Since $x=\dfrac{1}{n}\in (0,1)$ this implies that $|f(n)(\dfrac{1}{n})|=|\dfrac{1}{n\dfrac{1}{n}+1}|=|\dfrac{1}{1+1}|=\dfrac{1}{2}<\dfrac{1}{3}=\varepsilon$. This is clearly a contradiction and so we have found a counterexample such that the sequence breaks uniform continuity. Thus, $\{f_n(x)\}$ converges to $0$ pointwise but not uniformly.

$\Box$

Reflection: This one tripped me up a bit. I was able to get the piecewise convergence but I could not decide whether it converged uniformly or not. My intuition told me that it didn’t, but I couldn’t think of a counter example like above where it broke uniform convergence. While trying to find a counter example I saw that as $x$ got closer to $0$ the $N\in\mathbb{N}$ needed to make $f_n(x)$ converge got larger and larger. This made me think that we couldn’t find a universal $N$ that would work for all $x\in(0,1)$. After seeing the above counter example it makes a lot more sense. I think what I need to keep in mind when trying to find a counter example for these kind of things is that I don’t necessarily want to find a specific $x$ value where it breaks, but that I should tailor my $s$ value to what I know, i.e. the $N$.