Functionnal analysis (norm in Sobolev space)

[quasar987]

[SOLVED] Functionnal analysis (norm in Sobolev space)

Homework Statement

In relations to the problem of finding the eigenvalues of the operator -d²/dt², it can be shown (Lax-Milgram) that given f in L²[0,1], there exists a unique u in $$H^1_0[0,1]$$ such that

$$\int_0^1u'v' = \int_0^1fv \ \ \forall v \in H^1_0[0,1]$$

Consider the map S:L²[0,1]-->$$H^1_0[0,1]$$ defined by S(f)=u.

In arguing that S in continuous, my book says to take v=u. The equation characterizing u is then

$$\int_0^1(u')^2 = \int_0^1fu$$

We deduce from this that $$||u'||^2_{L^2}\leq ||f||_{L^2}||u||_{L^2}$$ (Hölder). And here my book simply says "It results that $$||u||_{H^1}\leq C||f||_{L^2}$$, for a constant C."

How does that follow?

Homework Equations

$$||u||_{H^1}=\sqrt{||u'||_{L^2}^2+||u||_{L^2}^2}$$

The Attempt at a Solution

Well, we can use the inequality to write

$$||u||_{H^1}^2=||u'||_{L^2}^2+||u||_{L^2}^2\leq |f||_{L^2}||u||_{L^2}+||u||_{L^2}$$, but then what?

 

[quasar987]

That last equation should read

$$||u||_{H^1}^2=||u'||_{L^2}^2+||u||_{L^2}^2\leq ||f||_{L^2}||u||_{L^2}+||u||_{L^2}^2$$

 

[quasar987]

Perhaps slightly closer would be to use

$$ab\leq \frac{a^2+b^2}{2}$$

on Hölder's inequality to obtain instead

$$||u'||^2_{L^2}\leq \frac{||f||_{L^2}^2+||u||_{L^2}^2}{2}$$

and thus

$$||u||_{H^1}\leq \frac{3}{2}||u||_{L^2}^2+\frac{1}{2}||f||_{L^2}^2$$

But can we relate ||u||_2 to ||f||_2 ? :grumpy: