Principle of Condensation of Singularities

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Theorem

Let $\struct {X, \norm {\,\cdot\,}_X}$ be a Banach space.

Let $\struct {Y, \norm {\,\cdot\,}_Y}$ be a normed vector space.

Let $\family {T_\alpha: X \to Y}_{\alpha \mathop \in A}$ be an $A$-indexed family of bounded linear transformations from $X$ to $Y$ such that:

$\ds \sup_{\alpha \in A} \norm {T_\alpha} = \infty$


Then:

$\ds \sup_{\alpha \in A} \norm {T_\alpha x} = \infty$ for some $x \in X$.


Proof

Aiming for a contradiction, suppose that:

$\ds \sup_{\alpha \in A} \norm {T_\alpha x} < \infty$ for each $x \in X$.

From the Banach-Steinhaus Theorem, we have:

$\ds \sup_{\alpha \in A} \norm {T_\alpha} < \infty$.

contradicting that:

$\ds \sup_{\alpha \in A} \norm {T_\alpha} = \infty$

So, we have:

$\ds \sup_{\alpha \in A} \norm {T_\alpha x} = \infty$ for some $x \in X$.

$\blacksquare$


Sources