Let $a(u,v)$ be a bilinearform on a hilbert space $\mathcal{H}$ which satisfies all conditions for the Lax-Milgram Lemma.

Furthermore, $$a(u,v)=F(v),\ \forall v\in\mathcal{H}$$ for a bounded functional $F$ on $\mathcal{H}$.

I don't understand why a solution $u$ has to be unique.

Using Lax-Milgram we obtain a unique linear operator $T$ such that $$\langle Tu,v\rangle = F(v)$$ Then using the Riesz Theorem I can obtain another unique $w\in\mathcal{H}$ with $$\langle Tu,v\rangle = F(v)=\langle v,w\rangle$$

So we have that, since $w$ is unique that $Tu=w$. Why is $u$ then unique?

  • $\begingroup$ Is the statement $a(u,v) = F(v)$ for all $v \in H$? $\endgroup$ – Umberto P. Jan 19 at 12:55
  • $\begingroup$ Yes. I will make an edit. $\endgroup$ – EpsilonDelta Jan 19 at 13:37

Coercivity of $a$ implies that if $a(u,u) = 0$ then $u = 0$.

Suppose that $a(u,v) = F(v)$ and $a(w,v) = F(v)$ both hold for all $v \in H$. Then $$a(u-w,v) = 0$$ for all $v \in H$ and in particular $$a(u-w,u-w) = 0.$$ Thus $u-w = 0$.

  • $\begingroup$ I also know the LM-Lemma not involving coercivity, and if I am not mistaken, coercivity also implies invertibility of $T$ and then we have a unique solution. $\endgroup$ – EpsilonDelta Jan 19 at 14:53

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