# Continuity of Linear Operator Between Hilbert Spaces

Note: Please do not give a solution; I am curious to understand why my solution is incorrect, and would prefer guidance to help me complete the question myself. Thank you.

Let $$\mathcal{H}$$ be a Hilbert space, and suppose that $$T\in\text{Hom}(\mathcal{H},\mathcal{H})$$. Suppose that there exists an operator $$\tilde{T}:\mathcal{H}\rightarrow\mathcal{H}$$ such that, \begin{align} \langle Tx,y\rangle =\langle x,\tilde{T}y\rangle, \end{align} $$\forall x,y\in\mathcal{H}$$. Show that $$T$$ is continuous.

My current solution is as follows:

Assume for all $$\delta>0$$ there exists $$n>N\in\mathbb{N}$$ such that, \begin{align} \|x_{n}-x\|<\delta. \end{align} Then, \begin{align} \langle Tx_{n}-Tx,Tx_{n}-Tx\rangle &= \|Tx_{n}-Tx\|^{2}\\ &\leq\|Tx_{n}-Tx\|=\|T(x_{n}-x)\|\\ &\leq\|T\|\|x_{n}-x\|\rightarrow 0\text{ as }n\rightarrow\infty. \end{align}

What am I doing wrong? I notice I do not use the existence of $$\tilde{T}$$.

Second Attempt:

Assume $$\langle x_{n},x\rangle \rightarrow \langle x,x\rangle$$ as $$n\rightarrow\infty$$. Then, given $$\langle Tx,y\rangle = \langle x,\tilde{T}y\rangle$$, \begin{align} \langle Tx_{n},y\rangle &= \langle x_{n},\tilde{T}y\rangle\rightarrow_{n\rightarrow\infty}\langle x,\tilde{T}y\rangle=\langle Tx,y\rangle. \end{align} Therefore, $$Tx_{n}\rightarrow Tx$$ as $$n\rightarrow\infty$$.

Third Attempt:

Assume $$\|x_{n}-x\|\rightarrow 0$$ as $$n\rightarrow\infty$$. Then, \begin{align} \langle Tx_{n}-Tx,Tx_{n}-Tx\rangle=\langle x_{n}-x,x_{n}-x\rangle=\|x_{n}-x\|^{2}. \end{align}

By assumption $$\|x_{n}-x\|^{2}\rightarrow 0$$ as $$n\rightarrow\infty$$. Hence, \begin{align} \langle Tx_{n}-Tx,Tx_{n}-Tx\rangle = \|Tx_{n}-Tx\|^{2}\rightarrow 0\text{ as }n\rightarrow\infty. \end{align} Therefore, $$T$$ is continuous.

Fourth Attempt

Assume $$x_{n}\rightarrow x$$ as $$n\rightarrow\infty$$. Then, given $$\langle Tx,y\rangle = \langle x,\tilde{T}y\rangle$$, \begin{align} \langle Tx_{n},y\rangle =\langle x_{n},\tilde{T}y\rangle\rightarrow_{n\rightarrow\infty}\langle x,\tilde{T}y\rangle =\langle Tx,y\rangle. \end{align} Since $$Tx_{n}\rightarrow Tx$$ weakly, we know $$Tx:=\lim_{n\rightarrow\infty}Tx_{n}$$ exists which implies that $$T\in\mathcal{L}(\mathcal{H},\mathcal{H})$$. Hence $$\|T\|<\infty$$. So, \begin{align} \langle Tx_{n}-Tx,Tx_{n}-Tx\rangle &= \|Tx_{n}-Tx\|^{2}\\ &\leq\|Tx_{n}-Tx\|=\|T(x_{n}-x)\|\\ &\leq\|T\|\|x_{n}-x\|\rightarrow 0\text{ as }n\rightarrow\infty. \end{align} Therefore $$T$$ is continuous.

Fifth Attempt

Assume $$x_{n}\rightarrow x$$ as $$n\rightarrow\infty$$. Then, given $$\langle Tx,y\rangle = \langle x,\tilde{T}y\rangle$$, \begin{align} \langle Tx_{n},y\rangle =\langle x_{n},\tilde{T}y\rangle\rightarrow_{n\rightarrow\infty}\langle x,\tilde{T}y\rangle =\langle Tx,y\rangle. \end{align} Hence $$Tx_{n}\rightarrow Tx$$ weakly. Define $$f_{x}(y):=\langle Tx,y\rangle$$, then $$f_{x}:\mathcal{H}\rightarrow\mathbb{R}$$ is a linear operator.

Now, \begin{align} f_{x_{n}}(y)=\langle Tx_{n},y\rangle\rightarrow_{n\rightarrow\infty}\langle Tx,y\rangle=f_{x}(y). \end{align} Further, \begin{align} |f_{x_{n}}(y)|=|\langle Tx_{n},y\rangle|\leq\|Tx_{n}\|\|y\|\implies\|f_{x_{n}}\|\leq\|Tx_{n}\|. \end{align} Also, \begin{align} \|f_{x_{n}}\|\geq\frac{|f_{x_{n}}(Tx_{n})|}{\|Tx_{n}\|}=\|Tx_{n}\|. \end{align} Hence, $$\|f_{x_{n}}\|=\|Tx_{n}\|$$. We now show $$\|f_{x_{n}}\|_{\infty}<\infty$$, \begin{align} |f_{x_{n}}(Tx)|=|\langle Tx_{n},Tx\rangle|\rightarrow_{n\rightarrow\infty}|\langle Tx,Tx\rangle|=\|Tx\|^{2}, \end{align} hence, $$\sup_{n\in\mathbb{N}}|f_{x_{n}}(Tx_{n})|<\infty$$ and so by the UBP $$\sup_{n\in\mathbb{N}}\|f_{x_{n}}\|<\infty$$. This implies $$\|Tx_{n}\|<\infty$$ for all $$x_{n}$$ and so $$\|T\|<\infty$$.

Hence, \begin{align} \langle Tx_{n}-Tx,Tx_{n}-Tx\rangle &= \|Tx_{n}-Tx\|^{2}\\ &\leq\|Tx_{n}-Tx\|=\|T(x_{n}-x)\|\\ &\leq\|T\|\|x_{n}-x\|\rightarrow 0\text{ as }n\rightarrow\infty. \end{align} Therefore $$T$$ is continuous.

• The last inequality basically implies that the norm of T is bounded or that it is continuous Mar 16, 2019 at 1:26
• Comment on the second attempt: you showed that $Tx_n \to Tx$ weakly, not in norm. Off-topic comment: I admire your tenacity. Keep trying! Mar 16, 2019 at 2:15
• Thank you. Do you have a hint? Mar 16, 2019 at 2:22
• Third attempt made. Although not sure if this holds either. Mar 16, 2019 at 2:35

The problem is that we can't assume that $$T$$ has a finite norm. Before we add that condition about having an adjoint map $$\tilde{T}$$, we're simply assuming that $$T$$ is a linear map.

In fact, a linear map between normed vector spaces is continuous if and only if it has a finite operator norm. You assumed the statement we were trying to prove.

Second attempt: The assumption here should have been that $$x_n\to x$$, as in the others. Then, yes, $$\langle Tx_n,y\rangle \to \langle Tx,y\rangle$$ for each $$y$$. This is real progress. But, as stated in the comments, it's weak convergence rather than convergence in norm. Not quite there.

Third attempt: No, $$\langle Tu,Tu\rangle$$ is not equal to $$\langle u,u\rangle$$ - it's equal to $$\langle u,\tilde{T}Tu\rangle$$, and you don't know what $$\tilde{T}T$$ does. This is not helpful.

All, right, lets go back to the attempt that made some progress. Are you familiar with the uniform boundedness principle? One consequence of that theorem is that any sequence of points in a Hilbert space that converges weakly is bounded. Can we use this to ensure that $$T$$ is a bounded operator?

Fourth attempt: No, saying that $$Tx_n\to Tx$$ weakly does not imply that $$Tx_n\to Tx$$ strongly (in norm). For example, in $$\ell^2$$, the standard basis vectors $$e_n=(0,0,\dots,0,1,0,\dots)$$ converge to zero weakly but not strongly; $$\|e_n\|=1$$.

Fifth attempt: This looks like it basically works. There's a key step near the end:

...and so once more by the UBP $$\sup_{n\in N}\|T\|<\infty$$

You need more than just the one sequence for that, and calling it a supremum over $$n$$ doesn't really make sense anymore. The argument I have is a proof by contradiction; if $$T$$ is an unbounded operator, we can find $$y_n$$ such that $$\|y_n\|\to 0$$ and $$\|Ty_n\|\to\infty$$, so then $$x+y_n\to x$$ and $$\|T(x+y_n)\|\to\infty$$.

• So is the idea for me to use $\tilde{T}$ to cancel out the operator norm in my final inequality? Mar 16, 2019 at 2:06
• @Jack no, that won't rescue the proof. Mar 16, 2019 at 2:12
• New proof attempt. Please check if you can. Mar 16, 2019 at 2:13
• I am not familiar with that. I am only aware that given a family of linear operators $(T_{\alpha})$ are such that $T_{\alpha}:E\rightarrow F$, where $E,F$ are Banach spaces. If $\sup_{\alpha\in A}\|T_{\alpha}x\|<\infty$ then $\sup_{\alpha\in A}\|T_{\alpha}\|<\infty$. Mar 16, 2019 at 6:44
• Do you have a link to the corollary of Uniform Boundedness which you refer to? Mar 16, 2019 at 6:44

Are you familiar with the Closed Graph Theorem?

Assume that $$x_n \to x$$ and $$Tx_n \to y$$. We claim that $$y = Tx$$.

For any $$z \in H$$ we have $$\langle y,z\rangle \xleftarrow{n\to\infty}\langle Tx_n, z\rangle = \langle x_n,\tilde{T}y\rangle \xrightarrow{n\to\infty} \langle x,\tilde{T}y\rangle = \langle Tx,z\rangle$$

Hence $$y - Tx \perp H$$ and so on.