Uniqueness of solution for $a(u,v)=F(v)$

Question

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?

Answer 1

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 $u-w = 0$.