In a paper on solving PDE's I recently encountered following setting. Let $X$ be Hilbert space and $a:X\times X \to \mathbb{R}$ be a bounded symmetric positive definite bilinearform and $L:X \to \mathbb{R}$ a bounded linearform. Now the claim is there exists a unique $u \in X$ such that $$a(u,v)=L(v)\quad \text{for all }v \in X $$ and $||u||_a=||L||_{a*}$. Where $||v||_a:=\sqrt{a(v,v)}$ and $||\cdot||_{a*}$ is the dual norm induced by $a$.
I have been trying to prove this without success and I'm starting to doubt the claim. The Riesz rep. thm. requires completeness of the space, but boundedness of $a$ only implies that $||\cdot||_a$ is coarser than $||\cdot||_X$. For norm equivalence, which would give use completeness of $X$ with respect to $||\cdot||_a$, we would require the additional property of coercivity for $a$.
Also continuity of $L$ with respect to $||\cdot||_X$ does not imply continuity w.r.t. $||\cdot||_a$, since that is only the case if $||\cdot||_a$ is finer than $||\cdot||_X$. Thus it is also not clear if $L$ is in the dual of $X$ w.r.t. $||\cdot||_a$, i.e. if $||L||_{a*}$ is well-defined.
Any hints on the proof or rebuttle to my argument would be appreciated.
The claim is false as stated. Let $X = l^2(\Bbb{N}^+, \Bbb{R})$, and $a, L$ defined by $$a(u, v) := \sum_{n=1}^\infty \frac{u_n v_n}{n}$$ $$L(v) := \sum_{n=1}^\infty \frac{v_n}{n}$$ If you had a $u^0$ such that $\forall v\; a(u^0, v) = L(v)$, by putting in basis vectors, you'd have to have $\forall n\;u^0_n = 1$, which is not in $X$.