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Consider the normed space $(C([0,1]),||\cdot||_{\infty})$. Define $F:C([0,1])\to C([0,1])$ by


Show that $F$ is continuous with respect to $||\cdot||_{\infty}$.

I've attempted to show this like so:

Let $f_{0}\in C([0,1])$. We want to show that $F$ is continuous at $F_{0}$.
Let $\epsilon>0$. Choose $\delta= ?$

Then for $f\in C$ such that $||f-f_{0}||_{\infty}<\delta$, we have: $||F(f)-F(f_{0})||_{\infty}\le ||f^2-f_{0}^2||_{\infty}\le ||(f-f_{0})(f+f_{0})||_{\infty}$

I seem to have hit a dead end, and don't even know if the last part is useful. Any tips?

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What is $g$? It appears in the question but not in your solution... – user12477 Mar 7 '12 at 21:57
In your computation if seems that $F(f)=f^2$ but at the beginning it's written $F(f)=f+g$. – Davide Giraudo Mar 7 '12 at 21:57
My apologies, I've fixed that (the $f+g$ was from the previous question in the book). – dplanet Mar 7 '12 at 22:00
Thanks for the fix. Your attempt is moving in the right direction: the key now is to make sure that $||f+f_0||_\infty$ doesn't get too big. Try comparing with the proof that the function $s:\mathbf{R}\to\mathbf{R}$ with $s(x)=x^2$ is continuous. – user12477 Mar 7 '12 at 22:04
up vote 1 down vote accepted

We have if $\lVert f-f_0\rVert\leq \delta$ that $$\lVert f+f_0\rVert\leq \lVert f-f_0\rVert +2\lVert f_0\rVert\leq \delta+2\lVert f_0\rVert$$ and we can assume for example that $\delta\leq 1$, so if $\lVert f-f_0\rVert\leq \delta$ we have $$\lVert F(f_0)-F(f)\rVert\leq (1+2\lVert f_0\rVert)\lVert f-f_0\rVert\leq (1+2\lVert f_0\rVert)\delta,$$ so we can choose $\delta:=\frac 1{1+2\lVert f_0\rVert}\varepsilon$.

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Why can we assume that $\delta \le 1$? – dplanet Mar 7 '12 at 22:12
If $\delta$ work$, $\min(\delta,1)$ will work too. – Davide Giraudo Mar 7 '12 at 22:13

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