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the problem

I have to show that a function $D: C^1([a, b]) \mapsto C^0([a, b]): f \mapsto f'$ is continuous given a metric $\| \cdot \|_{C^1([a, b])}$.

The metric $\| \cdot \|_{C^0([a, b])}$ is defined as:

$$ \| f \|_{C^0([a, b])} = \max_{x \in [a, b]} |f(x)| $$

And the metric $\| \cdot \|_{C^1([a, b])}$ is defined as:

$$ \| f \|_{C^1([a, b])} = \| f \|_{C^0([a, b])} + \| f' \|_{C^0([a, b])} $$

my attempts

This is what I have myself so far:

$\epsilon$-$\delta$

That criterion basically says:

$$ \forall p \in C^1([a, b]): \forall \epsilon > 0: \exists \delta > 0: \forall x: \| x-p \|_{C^1([a, b])} < \delta \Rightarrow \| f(x) - f(p) \|_{C^1([a, b])} < \epsilon $$

After a little simplification I ended up with:

$$ \| x-p \|_0 < \delta \Rightarrow \| x'' - p'' \|_0 < \epsilon $$

I do not see how I could choose $\delta$ to make that work.

sequences

The other approach is via a limit of a sequence:

$$ \forall \epsilon > 0: \exists n_0 \in \mathbb N: \forall n > n_0: \| D(x_n) - D(x) \|_{C^1([a, b])} < \epsilon $$

I used the definition of $D$ and the metric and ended up with:

$$ \| x_n' - x' \|_0 + \| x_n'' - x'' \|_0 < \epsilon $$

Given that $x_n$ converges to $x$, “it just has to be that way”, but I am not sure how to show that.

The problem also states that with the plain $\| \cdot \|_0$ metric, the function $D$ is not continuous. But that would mean that $\| x_n' - x' \|_0 < \epsilon$ does not hold. With the $\| \cdot \|_1$ metric, the left side of the inequality just gets larger, so I think it would be even stricter and not work if the first one did not even work.


How can I show that this $D$ is continuous?


So with the correct metric, I see the answer:

$$ \forall p \in C^1([a, b]): \forall \epsilon > 0: \exists \delta > 0: \forall x: \| x-p \|_{C^1([a, b])} < \delta \Rightarrow \| D(x) - D(p) \|_{C^0([a, b])} < \epsilon $$

\begin{align*} \| x-p \|_{C^1([a, b])} &< \delta &\Rightarrow& &\| D(x) - D(p) \|_{C^0([a, b])} &< \epsilon \\ \| x-p \|_{C^0([a, b])} + \| x'-p' \|_{C^0([a, b])} &< \delta &\Rightarrow& &\| x' - p' \|_{C^0([a, b])} &< \epsilon \end{align*}

Since $\| x-p \|_{C^0([a, b])} \geq 0$ can I choose $\delta := \epsilon$, to fulfill the $\Rightarrow$ statement.

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3  
$Df$ is not in $C^1$, it is in $C^0$, so you are using the wrong metric in the range of $D$. You need to show that $\| D(x_n) - D(x) \|_{C^0([a, b])} < \epsilon$ –  Thomas Andrews Apr 24 '12 at 16:30
    
The caveat about $\|\cdot\|_0$ is in terms of the domain. That is, if you use $\|\cdot\|_0$ as your metric on the domain of $D$, then $D$ is not continuous. –  Thomas Andrews Apr 24 '12 at 16:33
    
Well, if I use epsilon-delta and use the 1-metric for delta and the 0-metric for epsilon, the answer screams right at me: $\delta := \epsilon$ :-) –  queueoverflow Apr 24 '12 at 16:43
    
I updated my question with that. –  queueoverflow Apr 24 '12 at 16:46
    
Yes, it is essentially "obvious" if you get the right metrics here. Basically, the metric in $C^1$ is defined precisely to make sure that $Df$ is close to $Dg$ if $f$ is close to $g$ in $C^1$. –  Thomas Andrews Apr 24 '12 at 17:21

1 Answer 1

up vote 3 down vote accepted

$D$ is linear so all you need to show is that there exists a constant $K>0$ such that $\|D(f)\|_{C^0} \leq K\|f\|_{C^1}$ for all $f$. By considering the definitions of the norms, you'll see that $K=1$ works.

You can then ask yourself more, is $K=1$ the smallest constant that works? and for which $f$ do we have $\|D(f)\|_{C^0} = K\|f\|_{C^1}$?

share|improve this answer
    
I know that it is linear since all the “rules of differentiation” are linear? –  queueoverflow Apr 24 '12 at 16:44
    
I don't know how differentiation was taught to you as it varies from country to country and teacher to teacher. But anyways, you must have encountered the properties: $(f + g)' = f' + g'$ and $(\lambda f)' = \lambda f'$ for any $f,g \in C^1$, $\lambda \in \mathbb{R}$. This is precisely the statement that $D$ is a linear mapping. –  YBL Apr 24 '12 at 19:50
    
If you're not familiar with the linear maps and normed spaces yet, just replace $f$ by $f-f_n$ and $D(f)$ by $D(f) - D(f_n)$ in my answer. You'll get the same result in a language that may seem more familiar. –  YBL Apr 24 '12 at 20:02
    
We had a theorem in the lecture that states that every linear function is smooth if it has this $K$ property. I now have it in two ways, one with your solution and the other with epsilon-delta. –  queueoverflow Apr 25 '12 at 7:22

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