Pointwise Limit of Sequence of Bounded Linear Transformations from Banach Space to Normed Vector Space is Bounded Linear Transformation

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Let $\Bbb F \in \set {\R, \C}$.

Let $\struct {X, \norm \cdot_X}$ be a Banach space over $\Bbb F$.

Let $\struct {Y, \norm \cdot_Y}$ be a normed vector space over $\Bbb F$.

Let $T : X \to Y$ be a function.

Let $\sequence {T_n}_{n \mathop \in \N}$ be a sequence of bounded linear transformations such that:

$\ds T x = \lim_{n \mathop \to \infty} T_n x$

for each $x \in X$.

Then $T$ is a bounded linear transformation.


We first show that $T$ is a linear transformation.

Let $\alpha, \beta \in \Bbb F$ and $x, y \in X$.

Then, we have:

\(\ds \map T {\alpha x + \beta y}\) \(=\) \(\ds \lim_{n \mathop \to \infty} \map {T_n} {\alpha x + \beta y}\)
\(\ds \) \(=\) \(\ds \lim_{n \mathop \to \infty} \paren {\alpha T_n x + \beta T_n y}\) Definition of Linear Transformation
\(\ds \) \(=\) \(\ds \alpha \lim_{n \mathop \to \infty} T_n x + \beta \lim_{n \mathop \to \infty} T_n y\)
\(\ds \) \(=\) \(\ds \alpha T x + \beta T y\)

So $T$ is linear

Now we show that $T$ is bounded.

We use the Banach-Steinhaus Theorem.

From Modulus of Limit: Normed Vector Space, we have:

$\ds \norm {T x}_Y = \lim_{n \mathop \to \infty} \norm {T_n x}_Y$

From Convergent Real Sequence is Bounded, we have:

$\norm {T_n x}_Y$ is bounded in $n$ for each $x \in X$.

That is:

$\ds \sup_{n \in \N} \norm {T_n x}_Y$ is finite for each $x \in X$.

So, from the Banach-Steinhaus Theorem, we have:

$\ds \sup_{n \in \N} \norm {T_n}$ is finite, where $\norm \cdot$ is the norm on bounded linear transformation.

Pick $M > 0$ such that:

$\norm {T_n} \le M$

for each $n \in \N$.


$\norm {T_n x}_Y \le M \norm x_X$

for each $x \in X$ and $n \in \N$ from the definition of the norm on bounded linear transformation.

Then for each $x \in X$ we have:

\(\ds \norm {T x}_Y\) \(=\) \(\ds \lim_{n \mathop \to \infty} \norm {T_n x}_Y\)
\(\ds \) \(\le\) \(\ds M \norm x_X\) Limits Preserve Inequalities

So $T$ is bounded.