Bounded Operator: Is D:L^2(0,1) Bounded?

[foxjwill]

Homework Statement

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 $$f\in L^2(0,1)$$,

$$\|Df\|\leq c\|f\|?$$​

Homework Equations

For all $$f\in L^2(0,1)$$,

$$\|f\| = \int_0^1 |f|^2\,dx.$$​

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?

 

[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.

 

[g_edgar]

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||.

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

 

[Dick]

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

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