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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?

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  • $\begingroup$ 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
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$B(x,x)-2l(x)\geq a\|x\|^2-2\|l\|\|x\|$, 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.

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  • $\begingroup$ 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

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