# Show that a function $\phi:(V, \|\cdot\|_{C^1}) \rightarrow (W, \|\cdot\|_{\infty})$, $f \rightarrow f'$ is continuous

I have the following:

Show that the function $$\phi:(V, \|\cdot\|_{C^1}) \rightarrow (W, \|\cdot\|_{\infty}),~~f \rightarrow f'$$ is continuous. With

$$\|\cdot\|_\infty = \sup\{|f(x)| ~ \big| ~x\in [0,1]\}$$

and

$$\|\cdot\|_{C^1}= \sup\{|f(x)| + |f'(x)| \}$$

Together with $a,b \in \mathbb{R},~ a < b,~V = C^1[a,b];$ the vector space of all once-continouously differentiable functions and $W = C^0[a,b]$ the vector soace of continuous functions from $[a,b] \rightarrow \mathbb{R}$.

I want to prove that saying the following:

A map $\phi:(V, \|\cdot\|_{C^1}) \rightarrow (W, \|\cdot\|_{\infty})$ between two metric spaces is continuous in $x \in (V, \|\cdot\|_{C^1})$, if

$$\lim_{n \rightarrow \infty} \phi(x_n) = \phi(x) \in (W, \|\cdot\|_{\infty})$$

is true for a sequence $(x_n)_{n\in\mathbb{N}}$ that approaches $x$.
$\phi$ is continuous, if it is continouous in all points $x$.

Now let $f_n \in (V, \|\cdot\|_{C^1})$ be a sequence that approaches $f$. Thus we have

$$\|f_n - f\| = \sup\{|f_n - f| + |f'_n - f'|\} < \epsilon ~~~~\text{for large enough n}$$

$$\Rightarrow |f_n - f| + |f'_n - f'| < \epsilon$$ $$\Rightarrow |f_n - f| < \frac{\epsilon}{2} ~~~~\text{and} ~~~~|f'_n - f'| < \frac{\epsilon}{2}$$

So that means that for the function itself one gets:

$$\|\phi(f_n) - \phi(f)\|_\infty = \sup\{|\phi(f_n) - \phi(f)|\} = \sup\{|f'_n - f'|\} < \frac{\epsilon}{2}$$

So I have just hopefully proven that

$$\text{if }~~ \lim_{n \rightarrow \infty} f_n = f ~~\text{ then }~~ \lim_{n \rightarrow \infty} \phi(f_n) = \phi(f)$$

Is that correct? Especially because $\|\cdot\|_\infty$ is only defined for $[0,1]$ this makes me doubtful. Do I have to make any restrictions for $f_n$?

Thank you very much for the help.

FunkyPeanut

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I changed several instances of \text{sup} to \sup. Among the effects are that something like a\sup b looks like this $a\sup b$, i.e. proper spacing, which won't be there if you write a\text{sup}b. Another is that in a "displayed" setting you can write $\displaystyle a\sup_{b\in\mathcal{C}}f(b)$ and the subscript is directly under $\sup$. MathJax is based on the math notation system used in $\TeX$. If I keep seeing things like this, I might even start to suspect that some people don't know that $\TeX$ was created by Donald Knuth, not by idiots. – Michael Hardy May 6 '14 at 16:08
What are $V$ and $W$? Also, you cannot simply conclude that $\|f_n-f\|_{C^1}<\epsilon\implies \|f'_n-f'\|_{\infty}<\frac{\epsilon}{2}$. It does follow that $\|f'_n-f'\|_{\infty}<\epsilon$, since $\|f'\|_{\infty}\le\|f\|_{C^1}$. (If $V$ and $W$ are what I think they are.) – Vincent Boelens May 6 '14 at 17:22
Oh sorry I forgot that. I will edit it in. So is it really enough to just say that $(||f_n - f||_{C^1} < \epsilon \land ||f'||_\infty < ||f||_{C^1}) \Rightarrow ||f'_n - f||_\infty < \epsilon$? Well it makes sense.... – FunkyPeanut May 6 '14 at 17:29
Well, you'd have to prove that $\|f'\|_{\infty}\le\|f\|_{C^1}$ for all $f\in V$. But after that it follows immediately. – Vincent Boelens May 6 '14 at 20:01