Prove or disprove that $φ_v:u\mapsto \langle\mathcal A u,v\rangle$ is in $V^*$

Question

Let us consider a linear and continuous operator on a Hilbert space $V$, $\mathcal A:V\rightarrow V$, such that: $$\|\mathcal A u\|\leq M \|u\|, \ \ \forall u\in V, M>0$$ and now consider $\langle\mathcal A u,v\rangle, \ \ u,v\in V.$ We have: $$\langle\mathcal A u,v\rangle\leq \|\mathcal A u\| \|v\|\leq M \|u\|\|v\|$$ Let $V$ be a Hilbert space, and let $V^*$ denote its dual space, consisting of all continuous linear functionals from $V$ into the field $\mathbb R$ or $\mathbb C$. Prove or disprove that there exists a function $φ_v(u)$, for all $u$ in $V$ defined by $$\varphi_v(u) = \langle\mathcal A u,v\rangle$$ and it is an element of $V^*$.

I think this is true, since linearity of the inner product and boundedness which is equivalent to continuity here, but I'm not very sure and I'm not able to provide a precise proof.

Hints are welcome.

Answer 1

Let $v\in V $ be fixed we defined $$\varphi_v(u) = \langle\mathcal A u,v\rangle$$

Claim $u\mapsto \varphi_v(u): V\to \Bbb C$ is linear and continuous that is $\varphi_v\in V^*$

Proof:

Since $\mathcal {A}$ linear we have it follows that

$$\varphi_v(u_1+\alpha u_2) = \langle\mathcal A(u_1+\alpha u_2) , v\rangle = \langle\mathcal A(u_1)+\alpha\mathcal A( u_2) ,v\rangle\\=\langle\mathcal Au_1,v\rangle+\alpha\langle\mathcal A u_2,v\rangle = \varphi_v(u_1)+\alpha \varphi_v(u_2)$$ this prove the linearity.

On other hand since, $\mathcal A:V\to V$ is linear and continuous we have,

$$|\varphi_v(u)|=|\langle\mathcal A u,v\rangle|\leq \|\mathcal A u\| \|v\|\leq \|\mathcal A \|\|v\|\|u\| = K\|u\|$$ with $K=\|\mathcal A \|\|v\|$. that for all $u\in V$ we have $$|\varphi_v(u)|\le K\|u\|$$

This prove the continuity of $\varphi_v$ and hence $\varphi_v\in V^*.$