# Meagre set and sequence of operators in Banach Space [closed]

Let $X$ a Banach Space , $Y$ a Normed Space and $(T_n)_{n=0}^\infty$ a sequence in $L(X,Y)$ such that $\sup_n {\| T_n (x) \|} = \infty$

Show that:

• $Z=\{ x \in X |{ \sup_n\|T_n (x) \|} \leq \infty\}$ is meagre
• $Z^c$ is dense in X and is a intersection of dense open sets

Edit: With a hint I changed the notation for some more clear and asked more clearly that is a question. This exercise a professor of Functional Analysis created for us students, but I have some difficult in Analysis and need a little help to start in this beginning of the course.

## closed as off-topic by Nosrati, user91500, Lord_Farin, Cesareo, Riccardo.AlestraJan 17 at 10:20

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• Do you mean $Z = \{ x\in X\ | \ \sup_n ||T_n(x)|| < \infty \}$? – Vincent Aug 5 '16 at 10:19
• Yes, but I'm new here and don't know how put index in ' sup '. 'sup_n' it's a good idea, thx for your comment – 521124 Aug 5 '16 at 11:58
• You have a dangling free variable in the first sentence. What does it mean? – DanielWainfleet Aug 5 '16 at 17:45

What you are actually trying to show is the Banach-Steinhaus theorem. The first step consists in writing $Z = \bigcup_{M\in \mathbb{N}} Z_M$, with $Z_M = \{x\in X| \sup_n ||T_n(x)|| < M\}$. By contradiction, if $Z$ is not meagre, there is a small ball in some $Z_M$. Thanks to this ball, you can try to show that actually $||T_n(x)||$ is bounded by some constant independant of $n$ on the unit ball and thus conclude.