# Bounded operator

1. Aug 24, 2009

### foxjwill

1. The problem statement, all variables and given/known data
Is the derivative operator $$D:L^2(0,1)\to L^2(0,1)$$ bounded? In other words, is there a c>0 such that for all $$f\in L^2(0,1)$$,
$$\|Df\|\leq c\|f\|?$$​

2. Relevant equations
For all $$f\in L^2(0,1)$$,
$$\|f\| = \int_0^1 |f|^2\,dx.$$​

3. The attempt at a solution
I'm pretty sure the answer is no. Here's my work:

Suppose $$c^2>0$$ satisfies the above requirements. Define $$f(x)=e^{(c+1)x}$$. Then
$$\|Df\| = \int_0^1 (c+1)^2e^{2(c+1)x}\,dx = (c+1)^2\|f\| > c^2\|f\|.$$​
But this contradicts the fact that $$\|Df\|\leq c^2\|f\|.$$ Thus, D is unbounded. Q.E.D.

Is this correct?

2. Aug 24, 2009

### Dick

You have the right kind of example but the logic is a little overcomplicated. If you define f(x)=e^(dx) then ||Df||=d^2||f||. At this point you are done since d^2 can be chosen as large as you want. So there can't be a c such that ||Df||<c||f|| for any f.

3. Aug 24, 2009

### g_edgar

No, just d||f|| ... provided d>0 ... but that is just as good.

4. Aug 24, 2009

### Dick

Right. There should have been a square root on the definition of ||f|| in the original post. Thanks for catching that.