# Reverse Uniform Boundedness Principle

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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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 $$\| T_{\alpha} \| \ge \left| T\Big(\frac{x}{\| x \|} \Big) \right| \ge \frac{M_x}{\| x \|}$$