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I believe that what I am about to ask has a negative answer, but I can't seem to find a quick counterexample.

Consider a family of bounded operators $\{T_\alpha\}_{\alpha \in \mathcal{A}} \subset \mathcal{B}(X)$ where $X$ is a Banach space. The family has the property that for every $x \in X\setminus \{0\}$ there exists $M_x>0$ such that $\|T_\alpha x\| \geq M_x ,\ \forall \alpha \in \mathcal{A}$. Is it true then that there exists an $M>0$ such that $\|T_\alpha\| \geq M$?

I tried the approach used for proving the original Uniform Boundedness Principle, but it doesn't work.

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up vote 3 down vote accepted

I assume you know that for an operator $T$ the norm is defined to be $$ \| T \| := \sup_{\|x\| = 1} |T \; x| $$ Now pick an $x \neq 0$. We have $$ \| T_{\alpha} x \| \ge M_x $$ and therefore $$ \| T_{\alpha} \| \ge \left| T\Big(\frac{x}{\| x \|} \Big) \right| \ge \frac{M_x}{\| x \|} $$

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Looks like I was looking for a complicated proof of a simple fact. Thank you. – Beni Bogosel Nov 22 '11 at 13:05

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