# Show that there exists a unique bounded linear operator T such that $\langle T_n x,y\rangle \to \langle Tx,y\rangle$

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.

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.
• 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? Jul 3, 2022 at 12:02