Uniform lower bound of sequence of linear maps on a Banach space

Question

Suppose a sequence of $T_j:X\to\mathbb{R}$, with $X$ Banach, has the following property: $$\forall j:\|T_j\|\geq c>0$$ Then we have for all $n$, $$\exists x\in X\setminus \{0\}:\forall j\leq n:|T_jx|\geq\frac12c\|x\|$$ Does anybody know how to prove or disprove this? It seems like a reverse of the uniform boundedness principle somehow, but perhaps this analogy is incorrect.

Answer 1

This is not true even when $X = \mathbb R^n$. Let $T_i (x_1, \cdots ,x_n) = x_i$ for $i=1, \cdots, n$. Then $\|T_i\| =1$. On the other hand, for all $y\in \mathbb R^n\setminus \{0\}$,

$$\sum_{i=1}^n |T_j(y)|= \sum_{i=1}^n |y_i| \le \sqrt {n} \|y\| .$$

So when $n$ is large so that $\sqrt{n}<\frac{1}{2} n$, there is at least one $j$ so that

$$|T_j(y)| <\frac 12 \|y\|.$$

Thus such an $y$ cannot be found.