Norm operators bounded below implies almost uniform lower bound I have a hard time proving (or disproving) the following statement about continuous linear operators: 
$$(\exists c>0:\forall j:\|T_j\|\geq c)\Rightarrow(\exists\delta>0:\forall n:\exists x\in X:\forall 1\leq j\leq n:|T_jx|\geq\delta)$$
where $T_j$ is a sequence of continuous linear functionals on a Banach space $X$, so $T_j:X\to\mathbb{R}$. Does anybody have any thoughts on this? Thank you in advance!
 A: For an arbitrary $n$, if we have an $x_n\in X$ such that $T_j x_n \neq 0$ for $1 \leqslant j\leqslant n$, then we can scale by a large enough real number so that $\lvert T_j (tx_n)\rvert \geqslant \delta$ for $1\leqslant j \leqslant n$ whatever $\delta$ we prescribed.
So the problem reduces to the question whether we can always find such an $x_n$, or, in an equivalent formulation, whether
$$X \neq \bigcup_{j = 1}^n \ker T_j$$
for all $n$. The assertion $\lVert T_j\rVert \geqslant c$ for some $c > 0$ ensures that none of the $T_j$ is the zero functional, so $\ker T_j$ is a proper closed subspace of $X$ for all $j$. As a proper subspace, $\ker T_j$ has empty interior, so $\ker T_j$ is nowhere dense. Hence
$$N = \bigcup_{j = 1}^{\infty} \ker T_j$$
is a meagre subset of $X$, and by Baire's theorem $X \setminus N$ is dense, in particular $X\setminus N$ is nonempty.
Now take an $x_0 \in X\setminus N$, and for each $n$ choose $t_n \in \mathbb{R}$ so large that
$$\lvert T_j(t_n\cdot x_0)\rvert \geqslant \delta$$
for $1 \leqslant j \leqslant n$.
