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[SOLVED] Functionnal analysis (norm in Sobolev space)
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?
[tex]||u||_{H^1}=\sqrt{||u'||_{L^2}^2+||u||_{L^2}^2}[/tex]
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?
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?
Homework Equations
[tex]||u||_{H^1}=\sqrt{||u'||_{L^2}^2+||u||_{L^2}^2}[/tex]
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?