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

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?

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

## 1 Answer

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 $$u-w = 0$$.

• 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. – EpsilonDelta Jan 19 at 14:53