Let H be a real Hilbert space, $l\in H'$, which is the dual space of H, and $B:H\times H\rightarrow \mathbb{R}$ a symmetric, bounded and coercive Bilinear form. We define $J:H\rightarrow \mathbb{R}$ as $J(x):=B(x,x)-2l(x)$ Show:

i) J is continuous and bounded below.

ii) J possesses a minimum in exactly one point.

I already showed the continuity. For the bounded part, I used that $B$ is coercive. Re(B(x,x))=B(x,x) is true, because H is a real Hilbert space, so $B(x,x)\ge a||x||^2$ for an $a>0$. Now B(x,x) is bounded below. I just have to show that $2l(x)$ is also bounded. But I only know that $l$ is linear and continuous. Can someone help me?

For ii) I got: Since $B$ is coercive and bounded below and $l$ linear:$B(x,x)\ge 0$ . I don't know how to continue here. Can someone give me a hint?

  • $\begingroup$ Continuous = Bounded for linear operators on normed vector spaces. This resolves your first problem. For the second part, I believe the proof usually goes like this: take a minimizing sequence, prove the sequence converges in some sense (possibly weakly, possibly along a subsequence), prove that the limit is the unique minimizer. $\endgroup$ – User8128 Nov 30 '17 at 21:17

$B(x,x)-2l(x)\geq a\|x\|^2-2\|l\|\|x\|$, the function $f(x)=ax^2-2\|l\|x$ define on the real numbers is bounded below, this implies that $J$ is bounded below.

For ii) the derivative of $J$ at $x$ evaluated at $u$ is $J'_x(u)=2B(x,u)-2l(u)$, $J$ is an extrema implies that $J'_x=0$ i.e $B(x,.)=l$, there is just one element which satisfies that equation since $B$ is coercive.

  • $\begingroup$ How do you get from $2l(x)$ to $2||x||$? Then why is $f(x)=ax^2-x$ did you forget the 2? $\endgroup$ – Tobi92sr Nov 30 '17 at 21:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.