## May 1999 #5

Problem Statement: Let $f$ be a real valued differentiable function on $(0,2)$ whose derivative $f'(x)$ is bounded on $(0,2)$.  Show that $\{f(\dfrac{1}{n})\}$ converges.

Proof: First note that $\dfrac{1}{n}\in (0,2)$ for every $n\in\mathbb{N}$. Furthermore we know that $f'(x)$ is bounded so there exists a real number $M$ such that $|f'(x)|\leq M$ for every $x\in (0,2)$. Since $f$ is differentiable on $(0,2)$ we may apply the Mean Value Theorem to any closed interval contained in $(0,2)$.

Let $n\in\mathbb{N}$. Since $[\dfrac{1}{n+1},\dfrac{1}{n}]$ is contained in $(0,2)$ we know $f$ is differentiable and continuous on the interval. Then by the Mean Value Theorem there exists a $c\in (\dfrac{1}{n+1},\dfrac{1}{n})$ such that $|f'(c)|=|\dfrac{f(\dfrac{1}{n+1})-f(\dfrac{1}{n})}{\dfrac{1}{n+1}-\dfrac{1}{n}}|\leq M$. This implies that $|f(\dfrac{1}{n})-f(\dfrac{1}{n+1})|\leq M |\dfrac{1}{n}-\dfrac{1}{n+1}|$.

Let $\varepsilon >0$. Then since $\{\dfrac{1}{n}\}$ converges it is also Cauchy. Thus,  we know there exists an $N\in\mathbb{N}$ such that $|\dfrac{1}{n}-\dfrac{1}{n+1}|<\dfrac{\varepsilon}{M}$ for every $n\geq N$.

This implies that for $n\geq N$ it holds that $|f(\dfrac{1}{n})-f(\dfrac{1}{n+1})|\leq M|\dfrac{1}{n}-\dfrac{1}{n+1}|. Thus, $\{f(\dfrac{1}{n})\}$ is Cauchy and so it converges.

$\Box$

Reflection:  When I first attacked this problem I tried to use that $f$ is continuous when $x=0$, but we are not given that information. After further thought I realized that I was being asking information about the function after having been given information about the derivative, which should be a huge red flag to use the MVT. I then tried to use the interval $(0,\dfrac{1}{n})$ in my MVT inequality but Jeremy pointed out that we can’t do that since we must have continuity on the closed interval in order to use the MVT. Then I realized I could show convergence by showing Cauchy.

At first glance I didn’t realize all of the little nuances of this problem, but it’s those little nuances that can make or break a qual. I need to keep in mind the hypotheses when I’m taking the qual (in 12 days!!!).