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Let H be a Hilbert space. Assuming $\{Tn\}_n$ is a sequence of bounded linear operators from H to H such that for all x, y $\in$ H we have $\langle T_n x, y\rangle$ converges as $n \to \infty$.Show that there exists a unique bounded linear operator T such that $\langle T_n x,y\rangle \to \langle Tx,y\rangle$. I tried to use a result of extension which states that we can extend in a unique way a bounded linear operator over a dense subset to an operator over the whole space. But I don't see how to use it and if it's a good idea.

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Fix some $x\in H$ and define the linear functional $f_x$ given by the convergence $$f_x(y)=\lim_{n\to \infty}\langle T_n x,\, y\rangle.$$ It's easy to check that $f_x$ is indeed linear (with respect to $y$ variable). Now we claim that $f_x$ is also continuous. To prove this claim we first prove that the sequence $(T_n)$ is uniformly bounded. For this, fix $x\in H$ and define $S_n:H\to \mathbb{R}$ the linear operators given by $$\tag{*}S_n(y)=\langle T_n x,\,y\rangle.$$ Obviously, $S_n$ are linear and bounded since $$|S_n(y)|=|\langle T_n x,\,y\rangle|\leq ||T_n||\cdot ||x||\cdot ||y||.$$ Now, since the limit $\lim_{n}S_n(y)$ exists it follows by the uniform boundedness principle that the operators $(S_n)$ are uniformly bounded. In other words, $\sup_{n}||S_n||<\infty$. If we plug $y=T_nx$ in $(*)$ and use the uniform bound we get $$||T_n x||^2\leq ||T_n x||\cdot \sup_{n}||S_n||<\infty.$$ In other words, $||T_n x||\leq \sup_{n}||S_n||<\infty$. As the latter inequality holds for every $n$ we obtain that for every $x$ the sequence $(T_n x)$ is bounded. Using the uniform boundedness principle once more we obtain that the sequence of operators (T_n) is indeed uniformly bounded (with respect to operator norm). Now, since $\sup_{n}||T_n||<\infty$ we have that $$\tag{**}\begin{align} |f_x(y)|&=|\lim_{n\to\infty}|\langle T_n x,\,y\rangle|\\ &\leq \sup_{n}||T_n||\cdot ||x||\cdot ||y||<\infty \end{align} $$ Therefore, $f_x$ is bounded and thereby it is also continuous. In other words, $f_x\in H^*$ for every $x$. By the Riesz - Representation theorem it follows that there exists a unique $y_x\in H$ such that $||y_x||=||f_x||$ and $$f_x(y)=\langle y,\,y_x\rangle,$$ for every $y\in H$. By $(**)$ and since $||y_x||=||f_x||$ we obtain $$\tag{***}||y_x||\leq \sup_{n}||T_n||\cdot ||x||<\infty. $$ Define the map $T:H\to H$ by $Tx= y_x$. $T$ is well defined by the uniqueness of $y_x$. By observing that $f_{x+z}=f_x+f_z$ and $f_{\lambda x}=\lambda f_x$ it follows that $T$ is linear. Now, for every $||x||=1$ by $(***)$ we have that $$ \begin{align} ||T(x)||^2&=|\langle y_x,\,y_x\rangle|=|f_x(y_x)|\\ &=\sup_{n}||T_n||\cdot ||x||\dot ||y_x||\\ &\leq \bigl(\sup_{n}||T_n||\cdot ||x||\bigr)^2. \end{align} $$ Therefore, $||T(x)||\leq \sup_{n}||T_n||\cdot ||x||$ and this proves that $T$ is bounded. Lastly, since $f_x(y)=\langle y,\,y_x\rangle = \langle y,\, T_ x\rangle$ it is easily seen that $$\lim_{n\to\infty}\langle T_n x,\, y\rangle=\langle Tx,\,y\rangle$$ as desired.

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  • $\begingroup$ May I ask in line 5 to 6, why could we define $S_n$ from $H$ to $\mathbb{R}$? By $S_n(y)=⟨T_nx,y⟩$, isn't there possibility of $⟨T_nx,y⟩$ being complex? $\endgroup$
    – math noob
    Jul 3, 2022 at 12:02

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