Prove linear operator $T: \mathcal{H} \rightarrow \mathcal{H}$ and prove $M=||T||$.

Question

Let $\mathcal{H}$ be a Hilbert space and $B: \mathcal{H} \times \mathcal{H} \rightarrow \mathbb{C}$ be a functional which is linear with respect to the first argument and conjugate linear with respect to the second argument. Assume, furthermore that there exists an $M > 0$ such that $|B(f,g)| \leq M||f||||g||$ for all $f,g \in \mathcal{H}$. Prove that there exists a unique linear operator $T: \mathcal{H} \rightarrow \mathcal{H}$ such that $B(f,g)=(Tf,g)$ for all $f,g \in \mathcal{H}$. Show also that one can take $M=||T||$.

...I firstly need to understand the context of this question better...I am supposing "linear with respect to the first argument and conjugate linear with respect to the second argument" means that the function is linear in $H \times H$, and it is conjugate linear in $\mathbb{C}$. (Is this what that means?) Then, since $M$ exists there is a maximum value...but how can we use this to find a unique linear operator?

Answer 1

In order to avoild confusion, I use $(\cdot|\cdot)$ to denote the inner product on $\mathcal{H}$.

For each $f\in \mathcal{H}$, define $$\phi_f:\mathcal{H}\to\mathbb{C},\quad \phi_f(g)=\overline{B(f,g)}.$$ Since $B$ is conjugate linear in the second argument, it follows immediately that $\phi_f$ is a linear operator. Furthermore $\phi_f\in \mathcal{H}^*$ where $\mathcal{H}^*$ is the continuous dual of $\mathcal{H}$ (i.e. the space of all continuous linear functionals on $\mathcal{H}$), this is because $$|\phi_f(g)|\leq M\|f\|\|g\|,\quad\forall g\in \mathcal{H}.$$ Thus by Riesz representation theorem there exists a unique $\eta_f$ such that $$\phi_f(g)=(g|\eta_f),$$ this shows that $$B(f,g)=\overline{\phi_f(g)}=(\eta_f|g).$$ Define the linear operator $T:\mathcal{H}\to\mathcal{H}$ via $$T(f)=\eta_f,$$ we see that $T$ is uniquele determined, and is linear as $B$ is linear in the first argument. Moreover Consider the family $\{\psi_f:f\in\mathcal{H},\|f\|\leq 1\}\subseteq\mathcal{H}^*$ where $$\psi_f:\mathcal{H}\to\mathbb{C},\quad \psi_f=B(f,g).$$ We see that $$|\psi_f(g)|\leq M\|g\|,\quad\forall g\in\mathcal{H}.$$ Thus by the uniformly boundedness principle: $$\sup_{\|f\|\leq 1,\|g\|\leq 1}|B(f,g)|=\sup_{\|f\|\leq 1,\|g\|\leq 1}|\psi_f(g)|<\infty.$$ Noting that $$\|T\|=\sup_{\|f\|\leq 1}\|T(f)\|=\sup_{\|f\|\leq 1,\|g\|\leq 1}|(T(f)|g)|=\sup_{\|f\|\leq 1,\|g\|\leq 1}|B(f,g)|,$$ We see that $T$ is bounded, and one may choose $M=\|T\|$.

Answer 2

First of all, $$ M=\sup_{\|\,f\|=\|g\|=1} |B(\,f,g)|. $$ Thus, for every $\varepsilon>0$, there exist $\|\,f_\varepsilon\|=\|g_\varepsilon\|=1$, such that $$ \|Tf_\varepsilon\|=\|Tf_\varepsilon\|\|g_\varepsilon\|\ge (\,Tf_\varepsilon,g_\varepsilon)|=|B(\,f_\varepsilon,g_\varepsilon)|> M-\varepsilon. $$ Therefore, $$ \|T\|=\sup_{\|\,f\|=1}\|Tf\|>M-\varepsilon, $$ for every $\varepsilon>0$, and thus $\|T\|\ge M$.

We have that $$ M\|\,f\|\|g\|\ge|B(\,f,g)|=|(\,Tf,g)| $$ and hence $$ M\|\,f\|\|Tf\|\ge|B(\,f,Tf)|=|(\,Tf,Tf)|=\|Tf\|^2 $$ and thus $$ \|Tf\|\le M\|\,f\|, $$ which implies that $\|T\|\le M$.