# Differential operator is not continuous between this metric spaces

Let $\mathbb{D}$ be the set of functions $f:[0,1]\to \mathbb{R}$ of class $C^1$ (differentiable with continuous derivative). Let $\mathcal{C}[0,1]$ be the set of continuous functions in $[0,1]$ ($\to \mathbb{R}$). Let $d_1,d_2$ be the metrics given by:

$d_1(f,g)=\displaystyle\int_0^1 |f(x)-g(x)| dx$

and

$d_2(f,g)=\sup_{x\in[0,1]} |f(x)-g(x)|$

Let $D:\mathbb{D}\to \mathcal{C}[0,1]$ be such that $D(f)=f'$.

a) Is $D:(\mathbb{D},d_1) \to (\mathcal{C}[0,1],d_1)$ a continuous function?

b) Is $D:(\mathbb{D},d_2) \to (\mathcal{C}[0,1],d_2)$ a continuous function?

DISCLAIMER: $D:(\mathbb{D},d_1)$ represents the topologic space given by the metric topology induced by $d_1$ (and so for $d_2$).

Well, this is a bit spoiler, but I'm quite sure that in both cases $D$ is not continuous. The idea would be to think of continuity in the "$\varepsilon-\delta$" sense. Some functions $F,G$ could be very "close", and however $F'$ and $G'$ be very far away. It's not hard to imagine examples of what i'm saying. However, it's a bit difficult to describe the problem precisely.

• Welcome to math stackexchange. I'm glad to see your question is nicely typeset (well done) this adds a lot to the quality of the question. Since this is a learning community, we do expect you to include your own thoughts on the problem (and definitions that may be less familiar or defined differently in various textbooks). Please consider an edit to include what you've thought about and/or tried and where you are stuck. – TravisJ Apr 29 '15 at 1:09
• Looks great. I'll see if I can come up with any hints/suggestions for you. – TravisJ Apr 29 '15 at 2:16

Hint: in both cases, you can find a sequence of functions which stay close to $0$, but whose derivatives go further and further away from $0$.
Look at the functions of the form $x^{-p}$ for various $p>0$.
(In fact, you may see that both of these metrics are induced by norms and $D$ is linear, so $D$ is continuous iff it is bounded.)
• I was thinking about these, but are they differentiable on $[0,1]$ since the derivative is undefined at $x=0$? – TravisJ Apr 29 '15 at 2:29
• @TravisJ: I have not said that they are already good enough. I'm not solving your exercise, just hinting. You need to fix them (both to be defined and differentiable in $[0,1]$ and to stay close to $0$). – tomasz Apr 29 '15 at 2:34
• @TravisJ: yes, that works. You can also just cut off at some point before $0$ and then use some kind of mollifier or another "smoothing" method. That would give you a global, smooth function on ${\bf R}$, but it takes some more work to do that formally. Shifting is easier if you just want a function on $[0,1]$. – tomasz Apr 29 '15 at 17:33