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I have some questions about proof of the theorem below is written in my notes. If someone help me about it I will be appreciated. Thanks

Theorem : Let $(X,\|.\|_X)$ is a normed space and $(Y,\|.\|_Y)$ is a Banach space then $(B(X,Y),\|.\|_{op})$ is a Banach space

I have problem about some writtens :

1) $\|T_n(x)-T_m(x)\|_Y \leq \|T_n-T_m\|_{op}$

$\|T_n(x)-T_m(x)\|_Y \leq sup_{\|x\|_X \leq 1}\|T_n(x)-T_m(x)\|_Y$ it is true only for some $x \in X$ not for all. How can we use it?

2) $T_n(x) \to y , \exists y \in Y$ since $Y$ is a Banach space.It's OK for me but after this statement "Since $\forall x \in X$ $\exists y \in Y$ we can write $lim_{n\to \infty}T_n(x)=T(x)$" is written.

How can we write it?

Thanks in advance :)

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For the first point, in general you don't have $\|T_n(x)-T_m(x)\|_Y \leq \|T_n-T_m\|$. Rather we have $\|T_nx - T_mx\| \leq \|T_n - T_m\| \cdot \|x\|$.

This follows immediately from noting that for $\|x\| \leq 1$, $$\|T_n(x)-T_m(x)\|_Y \leq sup_{\|x\|_X \leq 1}\|T_n(x)-T_m(x)\|_Y$$ (consider $\frac{x}{\|x\|}$). This is enough to show that if $(T_n)$ is Cauchy in $B(X,Y)$ then for every fixed $x$, $(T_nx)$ is Cauchy in $Y$ which is what you want.

For the second point, all that they mean is that one can define a map $T:X \to Y$ by setting $Tx = \lim_\limits{n\to \infty} T_nx$. They aren't claiming that this $T$ is then surjective - notice that the statement is instead that for every $x$ we can define $Tx$. In general, $T$ need not be surjective. It is however always a bounded linear map which is what you want.

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  • $\begingroup$ First, thanks a lot for answer. I have confused for an instant in second question. There is nothing about surjectivity. But I am not clear about first part. Being Cauchy Sequence condition must be satisfied for every $x$, mustn’t it? If $\|x\|_X \gt 1$ it won’t be a Cauchy sequence $\endgroup$
    – usereb
    May 19, 2018 at 22:42
  • $\begingroup$ You do want $(T_nx)$ to be Cauchy for every $x$. That is for each $x$ and $\varepsilon > 0$ you want to find an $N$ (which can depend on $\varepsilon$ and on $x$) such that $n,m \geq N$ implies $\|T_nx - T_mx\| \leq \varepsilon$. Since $(T_n)$ is Cauchy in $B(X,Y)$ there exists $N$ such that for $n,m \geq N$, $\|T_n - T_m\| \leq \frac{\varepsilon}{\|x\|}$ (wlog $x \neq 0$ since $(T_n0)$ is clearly Cauchy). Then for $n,m \geq N$, $\|T_nx - T_mx\| \leq \|T_n - T_m\| \cdot \|x\| \leq \frac{\varepsilon}{\|x\|} \|x\| = \varepsilon$ as desired. $\endgroup$ May 20, 2018 at 7:26
  • $\begingroup$ The critical point is that since you fix $x$ and $\varepsilon$ first and then an $N$ it doesn't matter if the right hand side of $\|T_n x - T_m x\| \leq \|T_n - T_m \| \cdot \|x\|$ has a $\|x\|$ since once $x$ is fixed that might as well just be considered as some multiplicative constant and those are easy to deal with. $\endgroup$ May 20, 2018 at 7:28

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