Suppose $f_n$ are continuous functions converging uniformly to $f$ on the compact interval $[a,b]$. Show that for any natural power $k$, the sequence $f_n^k$ converges uniformly to $f^k$ on $[a,b]$.
I tried to understand the problem by starting with k=2 but i got stuck. Below is my work:
$|f_{n}^2 – f^2| = |f_n – f|\cdot|f_n + f|$. Then, $||f_{n}^2-f^2||_\infty \leq ||f_{n}-f||_\infty\cdot(||f_n||_\infty + ||f||_\infty)$. Since $f_n$ converges uniformly to $f$ then $||f_n-f||_\infty \leq \epsilon$. Now, $||f_{n}^2-f^2||_\infty \leq \epsilon\cdot(||f_n||_\infty + ||f||_\infty)$
I don't know what to do next from here. Also I don't know how to show it for any natural power k. Any help would be greatly appreciated.
Best Answer
Hint: First show that the $f_n$ are uniformly bounded, meaning there exists an $M$ so that $|f_n|_\infty<M$ for all $n\ge 1$. Use the fact that for large $n$, $f_n$ is at most $f+\epsilon$, and there are only finitely many small $n$.
Assuming you can do this, WLOG you also have $|f|_\infty<M$ (possibly by choosing a larger $M$). Then, $$ |f_n^2-f^2|_\infty\le |f_n-f|_\infty\cdot (|f_n|_\infty+|f|_\infty)\le |f_n-f|_\infty\cdot 2M $$ Now, you can choose $n$ large enough so $|f_n-f|_\infty$ is less than any $\delta$. Let $\delta=\epsilon/2M$.
For $k\ge 3$, use $$ |f_n^k-f^k|=|f_n-f|\cdot |f_n^{k-1}+f_n^{k-2}f^1+f_n^{k-3}f^2+\dots+f_nf^{k-2}+f^{k-1}|\le |f_n-f|\cdot kM^{k-1} $$