## Fall 1991 #5

This begins my study of sequences and series of functions. These bad boys have a tendency of tripping me up so I’ve decided to spend some quality time learning their ways.

Problem Statement: Let $A\subseteq \mathbb{R}$ and let $\{f_n\}$ be a sequence of real-valued functions on $A$ converging uniformly to a bounded real-valued function $f$ on $A$. Show there exist constants $K, N$ such that $|f_n(x)|\leq K$ for every $x\in A$ when $n\geq N$.

Proof: Let $\{f_n\}$ be such a sequence converging uniformly to $f$ on $A$ where $f$ is bounded. Since $f$ is bounded there exists a $K>0$ such that $|f(x)|\leq K$ for every $x\in A$.

Let $\varepsilon >0$, then there exists an $N\in \mathbb{N}$ such that $|f_n(x)-f(x)|<\varepsilon$ for every $n\geq N, x\in A$ since $\{f_n\}$ converges to $f$ uniformly. Consider $|f_n(x)|$ for $n\geq N$, then it follows that $|f_n(x)|=|f_n(x)-f(x)+f(x)|\leq |f_n(x)-f(x)|+|f(x)|<\varepsilon+K$ for every $x\in A$. Since $\varepsilon$ can be made arbitrarily small it follows that $|f_n(x)|\leq K$ for every $x\in A$ and $n\geq N$.

$\Box$

Reflection: This one wasn’t too bad. The key rests in the fact that $\{f_n\}\rightarrow f$ uniformly, so we may say that for every $\varepsilon>0$ there exists an $N\in \mathbb{N}$ such that $|f_n(x)-f(x)|<\varepsilon$ for $n\geq N$ and for every, $x\in A$. Since we may say this about every x in $A$ we are able to make the bounding argument. If we did not have uniform convergence then we would not be able to make that argument since we would need a different $N$ for each $x\in A$.

You get to pick $K$, so it doesn’t need to be the bound on $f$. It could be larger, like $M+11$ where $M$ is the bound on $f$. Then you can say that for $\varepsilon = 1$, there is an $N$ with $|f_n(x)| \le M + 1 < M+11 = K$ when $n \ge N$.