The limit of a strongly convergent sequence of linear bounded operators from a Banach space to a normed space

Question

Let $X$ be a Banach space and $Y$ be a normed space. If the sequence $\{T_n\}$ of bounded linear operators from $X$ into $Y$ is strongly convergent. Then there exists a bounded linear bounded operator $T:x \rightarrow Y$ st

$lim_{n\rightarrow \infty} T_n(x)=T(x)$ $\forall x \in X$

The Proof.

I don't understand how the author deduced that $T$ is bounded. why did he write $\|Tx-T_nx\| \leq \|T-T_n\| \|x\| <\epsilon$ all that we so far know about the operator $(T-T_n)$ is that it is a linear operator, it is not bounded so we can write this inequality $\|Tx-T_nx\| \leq \|T-T_n\| \|x\| <\epsilon$ furthermore he writes $\|T-T_n\| \|x\| <\epsilon$ but the assumption said $T_n \rightarrow $T$ strongly not uniformly. I'm confused about this part

Can anyone help?

Answer 1

Me too I don't understand, you cannot want to show that an operator is bounded and in the argument, use the fact that it is bounded. Another argument: $T(x)=lim_nT_n(x)$ implies that $\|T(x)\|=lim_n\|T_n(x)\|\leq \|T_n\|\|x\|\leq k\|x\|$. The $k$ is the $k$ you have defined by using the uniform boundedness principle.

Answer 2

You are correct; this is a small (but fixable error).

At that point in the proof, you cannot conclude that $$\|T-T_n\|<\epsilon$$ for $n$ large enough. But you can hold $x$ fixed, and then conclude that $$\|Tx-T_nx\|<\epsilon$$ because $T_nx\to Tx$. The following sequence of inequalities then shows that $$\|Tx\|\leq\|T_nx\|+\epsilon\|x\|$$ Thus $$\frac{\|Tx\|}{\|x\|}\leq\sup_n{\|T_n\|}+\epsilon$$ which is independent of $x$. The boundedness of $T$ then follows.