Xrhrft

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 $Tintext{Hom}(mathcal{H},mathcal{H})$. Suppose that there exists an operator $tilde{T}:mathcal{H}rightarrowmathcal{H}$ such that,
begin{align}
langle Tx,yrangle =langle x,tilde{T}yrangle,
end{align}
$forall x,yinmathcal{H}$. Show that $T$ is continuous.

My current solution is as follows:

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

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

Second Attempt:

Assume $langle x_{n},xrangle rightarrow langle x,xrangle$ as $nrightarrowinfty$. Then, given $langle Tx,yrangle = langle x,tilde{T}yrangle$,
begin{align}
langle Tx_{n},yrangle &= langle x_{n},tilde{T}yranglerightarrow_{nrightarrowinfty}langle x,tilde{T}yrangle=langle Tx,yrangle.
end{align}
Therefore, $Tx_{n}rightarrow Tx$ as $nrightarrowinfty$.

Third Attempt:

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

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

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 $x_nto x$, as in the others. Then, yes, $langle Tx_n,yrangle to langle Tx,yrangle$ 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, $langle Tu,Turangle$ is not equal to $langle u,urangle$ - it's equal to $langle u,tilde{T}Turangle$, 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?