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 $\mathbb C$. If $x$ is an element of $V$, then the function $φ_x$, for all $y$ in $V$ defined by $$\varphi_x(y) = \left\langle y , x \right\rangle, $$ where $\langle\cdot,\cdot\rangle$ denotes the inner product of the Hilbert space, is an element of $V^*$. The Riesz representation theorem states that every element of $V^*$ can be written uniquely in this form.

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 $\langle\mathcal A u,v\rangle, \ \ u,v\in V.$ Since: $$\langle\mathcal A u,v\rangle\leq \|\mathcal A u\| \|v\|\leq M \|u\|\|v\|$$ Can we infer $\varphi_x(u)=\langle\mathcal A u,v\rangle$ continuous and linear?

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    $\begingroup$ Linearity follows from the properties of the inner product (or conjugate linearity, depending on your flavour), and you have shown boundedness which is equivalent to continuity here. The corresponding $x$ is, of course, $A^* v$. $\endgroup$ – copper.hat Jun 15 '15 at 19:17
  • $\begingroup$ @copper.hat: thank you very much! $\endgroup$ – Mark Jun 15 '15 at 19:58
  • $\begingroup$ You did the work :-). $\endgroup$ – copper.hat Jun 15 '15 at 20:01

The function $u\mapsto \varphi_x(u)=\langle\mathcal A u,v\rangle$ is obviously a linear form on V. Given that $\mathcal{A} $ is linear and $\langle\cdot,v\rangle$ is linear too. Moreover, applying the Cauchy-Schwartz inequality and the continuity of $\mathcal{A} $ respectively, we obtain

$$|\varphi_x(u)|=|\langle\mathcal A u,v\rangle|\le \|\mathcal A u\| \|v\|\le\|\mathcal A \| \|u\|\|v\| = M \| u\| $$ with $M=\|\mathcal A \|\|v\| .$ This show the continuity of $\varphi_x.$


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