Is it a mistake if I delete the condition in Banach - Steinhaus Theorem?

Question

I'm studying the Banach-Steinhaus theorem recently. There is a question to me, I'm sure I have made a mistake, but I can't find out where the mistake is.

The Banach - Steinhaus theorem says: Let X be a Banach space and Y a normed vector space. Suppose that F is a set of bounded linear operators from X to Y. If for all x in X one has

$$\sup_{T\in F}\|T(x)\|_Y<\infty,$$

then there is

$$\sup_{T\in F}\|T\|_{B(X,Y)}<\infty.$$

My problem is that if I delete the condition "$\sup_{T\in F}\|T(x)\|_Y<\infty$", can I attain the above conclusion? My reason is:

Since $F$ is a set of bounded linear operators, we have

there exists an $M_i\ge0$, such that $\|T_i(x)\|\lt M_i\|x\|$, here I just list the elements of $F$ as $T_i, i=1,2,3...$. This is from the definition of bounded operator.

Then let $M = max_i\{M_i\}$, I can have $$\|T_i(x)\|\lt M\|x\|, \forall i=1,2,3... $$

so there is $$\frac{\|T_i(x)\|}{\|x\|}\lt M, \forall i=1,2,3... $$ then have $$\sup_i{\|T_i\|}\lt M\lt \infty.$$

So where is my mistake? It really bothers me.

I can also change the statement of my thought to this:

Since $F$ is a set of bounded linear operators from $X$ to $Y$, any element $T_i$ of it should satisfy: $$\|T_i(x)\|\lt M_i\|x\|$$ where $0\le M_i \lt\infty$. Now consider each $x\in X$, when we select a $x$, we have $\|x\|\lt\infty$, so we get $$\|T_i(x)\|\lt M_i\|x\|\lt\infty$$ since each $x$ has above property, we can apply supremum to them, it won't change the true-false fact, that is $$\sup_{i}\|T_i(x)\|\lt M_i\|x\|\lt\infty$$ we finally derive the condition out!! what a mistake! so where is the mistake? I don't know. Could someone tell?

Answer 1

Two mistakes:

1. in your considerations $F$ is countable. Banach- Steinhaus works also for an uncountable family of operators.

2. the set $\{M_i: i \in \mathbb N\}$ can be unbounded.

Answer 2

The error is when you write $\max_iM_i$, it should be $\sup_iM_i$ and nothing guarantees that it's not infinite.