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Functionnal analysis (norm in Sobolev space)

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quasar987
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[SOLVED] Functionnal analysis (norm in Sobolev space)

1. 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 [tex]H^1_0[0,1][/tex] such that

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

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

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

[tex]\int_0^1(u')^2 = \int_0^1fu [/tex]

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

How does that follow? :confused:


2. Homework Equations
[tex]||u||_{H^1}=\sqrt{||u'||_{L^2}^2+||u||_{L^2}^2}[/tex]


3. The Attempt at a Solution

Well, we can use the inequality to write

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

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quasar987
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That last equation should read

[tex]||u||_{H^1}^2=||u'||_{L^2}^2+||u||_{L^2}^2\leq ||f||_{L^2}||u||_{L^2}+||u||_{L^2}^2[/tex]
 
quasar987
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Perhaps slightly closer would be to use

[tex]ab\leq \frac{a^2+b^2}{2}[/tex]

on Hölder's inequality to obtain instead

[tex]||u'||^2_{L^2}\leq \frac{||f||_{L^2}^2+||u||_{L^2}^2}{2}[/tex]

and thus

[tex]||u||_{H^1}\leq \frac{3}{2}||u||_{L^2}^2+\frac{1}{2}||f||_{L^2}^2[/tex]

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

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