Question about bilinear form on Hilbert space

Solution 1:

Here's one direction:

Assume $A$ is surjective. That is, for every $f \in H'$, there exists a $u \in H$ such that $$ a(u, \cdot) = f(\cdot) $$ Now, suppose that $v,w \in H$ are such that $B(v) = B(w)$, that is, $a(\cdot,v) = a(\cdot,w)$.

That is, for every $u$, we have $a(u,v) = a(u,w)$. By surjectivity, this implies that for every $f \in H'$, we have $f(v) = f(w)$. This in turn implies that $v = w$ (why? we could use this, for example, but that's overkill. Riesz representation could work here). Thus, $B$ is injective.

Riesz representation approach: For every $f \in H'$, we have $f(v) = f(w)$. Thus, for every $x \in H$, we have $\langle v,x \rangle = \langle w,x \rangle$. So, we have $$ \langle v,x \rangle = \langle w,x \rangle \qquad \forall x \in H \implies\\ \langle v,x \rangle - \langle w,x \rangle = 0 \qquad \forall x \in H \implies\\ \langle v-w,x \rangle = 0 \qquad \forall x \in H \implies\\ \langle v-w,v-w \rangle = 0 \implies\\ \|v -w\|^2 = 0 $$