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This might be rather elementary but I can't seem to find what's wrong with the following argument. I suppose it's something rather trivial. For operators $T\in B(\mathcal{H})$

We say that $T_i\to T$ weakly if $\forall x,y\in \mathcal{H}$ we have $$\langle T_i x, y\rangle \to \langle T x, y\rangle$$

We say that $T_i\to T$ strongly if $\forall x\in \mathcal{H}$ we have $$\| T_ix-Tx\|\to 0$$

What is wrong with the following argument?

since $\langle T_i x, y\rangle \to \langle T_i x, y\rangle$ by linearity this is the same as saying $\langle (T_i-T) x, y\rangle \to 0$. Now since this holds for any $x$ and $y$ then it holds in particular in the special case when $x=y$. But this would mean that as a special case $$\langle (T_i-T) x, x\rangle=\|T_ix-Tx\|^2 \to 0$$. This would mean that weak convergence implies strong convergence which is obviously wrong (I know that in fact, it is the other way around).

I seem to be confusing something elementary. Where does my confusion stem from?

El Ruño
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2 Answers2

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Your confusion is the (false) equality $$ \langle Sx,x\rangle=\|Sx\|^2. $$ The actual equality is $$ \langle S^*Sx,x\rangle=\|Sx\|^2. $$ Both equalities agree with $S$ is a projection, i.e. $S=S^*S$, but not in general.

So you have things like the following: if $\{P_j\}$ is a net of projections and it converges weakly to a projection P, then $P_j\to P$ strongly. Compare this with the fact that for any positive contraction $T$ there exists a net $\{P_j\}$ of projections such that $P_j\to T$ weakly.

Martin Argerami
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The last equality you wrote is false. Take for example $X=\ell^{2}$ the square summable sequences. Then take $T_{n}$ with $n \in \mathbb{N}$ defined as $$ T_{n}(x)=(\frac{x_{1}}{n},\,x_{2},\dots,\,x_{n},\,\dots). $$ This sequence converges weakly to $T(x)=(0,x_{2},\dots,x_{n},\dots):$ $$ \langle \bigl( T-T_{n}\bigr)x,y \rangle = \dfrac{y_{1}x_{1}}{n} \to 0 \quad n \to +\infty. $$ Despite this fact, if you write your equality you get $$ \frac{x_{1}^{2}}{n^{2}}=\frac{x_{1}}{n}. $$

Matteo Aldovardi
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