Pointwise Cauchyness of Sequence of Continuous Linear Transformations on Non-Meager Set implies Everywhere Pointwise Cauchyness

Theorem
Let $\GF \in \set {\R, \C}$.

Let $X$ and $Y$ be topological vector spaces.

Let $\sequence {T_n}_{n \in \N}$ be a sequence of continuous linear transformations $T_n : X \to Y$ such that:
 * the set $C$ of $x \in X$ such that $\sequence {T_n x}_{n \in \N}$ is a Cauchy sequence is non-meager.

Then $C = X$, so that:
 * $\sequence {T_n x}_{n \in \N}$ is Cauchy for each $x \in X$.

Proof
Let $B$ be the set of all $x \in X$ such that:
 * $\map \Gamma x = \set {T_n x : n \in \N}$

is von Neumann-bounded in $Y$.

For each $x \in C$, we have that $\sequence {T_n x}_{n \in \N}$ is Cauchy.

From Image of Cauchy Sequence in Topological Vector Space is von Neumann-Bounded, it follows that:
 * $\set {T_n x : n \in \N} = \map \Gamma x$ is von Neumann-bounded in $Y$

for each $x \in C$.

So we have $C \subseteq B$.

Since $C$ is non-meager, we have that $B$ is non-meager by Subset of Meager Set is Meager Set.

From Banach-Steinhaus Theorem: Topological Vector Space, $\sequence {T_n}_{n \in \N}$ is equicontinuous and $B = X$.

We show that $C$ is a linear subspace of $X$.

Since $C$ is non-meager, $C \ne \O$.

From One-Step Vector Subspace Test, it is enough to show that $\alpha x + \beta y \in C$ for each $x, y \in C$ and $\alpha, \beta \in \GF$.

Let $x, y \in C$ and $\alpha, \beta \in \GF$.

Since each $T_n$ is linear:
 * $\map {T_n} {\alpha x + \beta y} = \alpha T_n x + \beta T_n y$

From Linear Combination of Cauchy Sequences in Topological Vector Space is Cauchy Sequence, we obtain that:
 * $\sequence {\map {T_n} {\alpha x + \beta y} }_{n \in \N}$ is Cauchy.

So $\alpha x + \beta y \in C$.

From One-Step Vector Subspace Test, we have that $C$ is a linear subspace of $X$.

By Non-Meager Linear Subspace of Topological Vector Space is Everywhere Dense, we have that:
 * $C$ is everywhere dense

since $C$ is non-meager.

We aim to deduce that $C = X$.

Let $x \in X$.

Let $W$ be an open neighborhood of ${\mathbf 0}_Y$.

From, there exists an open neighborhood $W_1$ of ${\mathbf 0}_Y$ such that:
 * $W_1 + W_1 \subseteq W$

Applying again there exists a open neighborhood $W_2$ of ${\mathbf 0}_Y$ such that:
 * $W_2 + W_2 \subseteq W_1$

Then, we have:
 * $W_2 + W_2 + W_2 + W_2 \subseteq W_1 + W_1 \subseteq W$

Since ${\mathbf 0}_Y \in W_2$, we have:
 * $W_2 + W_2 + W_2 \subseteq W$

Since $\sequence {T_n}_{n \in \N}$ is equicontinuous, there exists an open neighborhood $V$ of ${\mathbf 0}_X$ such that:
 * $T_n \sqbrk V \subseteq W_2$ for each $n \in \N$.

Replacing $V$ with $V \cap \paren {-V}$, we can take $V$ to be symmetric.

Since $C^- = X$, we have $x \in C^-$, and we can apply Condition for Point being in Closure: Topological Vector Space to obtain that:
 * there exists $x' \in C \cap \paren {x + V}$.

Since $\sequence {T_n x'}_{n \in \N}$ is a Cauchy sequence, there exists $N \in \N$ such that:
 * $\paren {T_n - T_m} x' \in W_2$ for $n, m \ge N$.

Now write:
 * $\paren {T_n - T_m} x = \map {T_n} {x - x'} + \paren {T_n - T_m} x' + \map {T_m} {x' - x}$

Since $x' - x \in V$, we have:
 * $\map {T_m} {x' - x} \in W_2$

Since $V$ is symmetric, we also have $x - x' \in V$, so that:
 * $\map {T_n} {x - x'} \in W_2$

So, for $n, m \ge N$ we have:
 * $\paren {T_n - T_m} x \in W_2 + W_2 + W_2 \subseteq W$

Since $W$ was an arbitrary open neighborhood of ${\mathbf 0}_X$, we have that $\sequence {T_n x}_{n \in \N}$ is Cauchy.

So $x \in C$, and we obtain $C = X$.