Let $T_{jk}$ be bounded operators from $X$ to $Y$, where $X$ and $Y$ are Banach spaces. Prove that if $\forall j$ $$\sup_{k} ||T_{jk}||=\infty$$
then there exists a vector $x \in X$ s.t. $$\forall j, \sup_{k} ||T_{jk}x||=+\infty.$$
I know by Banach-Steinhaus theorem that if I have the mentioned hypothesis then $\forall j$ there exists $x \in X$ that verifies the thesis, but I don't know how to show that there exists some $x$ that verifies the thesis $\forall j$. I have tried
if $$\forall j, \quad \sup_{k} ||T_{jk}||=+\infty,$$ then $$\sup_{j} \sup_{k} ||T_{jk}||=+\infty$$ and hence there exists $x$ s.t. $\sup_{j} \sup_{k} ||T_{jk}x||=+\infty$.
Firstly, i don't know if this is correct, then I don't know if this is useful and finally if so is, I don't know how to conclude.
If anyone could help me I will be very thankful.
Best Answer
Here is an idea: the Banach Steinhauss theorem actually gives you something stronger: If for fixed $j$ we have $\underset{k}{sup} \Vert T_{j,k} \Vert = \infty$, then there exists a $G_\delta$ dense set $S_j$ such that $$\forall x \in S_j, \underset{k}{sup} \Vert T_{j,k}(x) \Vert = \infty$$ You can deduce it by looking at the proof of the theorem itself(for example in Rudin's Real & Complex analysis).
Now By Baire's theorem, $\bigcap_{j=1}^\infty S_j$ is also dense, thus not empty, and gives you a vector you desired.