Bounded Linear Transformation preserves Cauchy Sequences

Theorem

Let $\struct {X, \norm {\, \cdot \,}_X}$ and $\struct {Y, \norm {\, \cdot \,}_Y}$ be normed vector spaces.

Let $T : X \to Y$ be a bounded linear transformation.

Let $\sequence {x_n}_{n \mathop \in \N}$ be a Cauchy sequence in $X$.

Then $\sequence {T x_n}_{n \mathop \in \N}$ is a Cauchy sequence in $Y$.

Proof

Since $T$ is a bounded linear transformation, there exists $M > 0$ such that:

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

for all $x \in X$.

So, since $T$ is linear, we have:

$\norm {T x_n - T x_m}_Y \le M \norm {x_n - x_m}_X$

Let $\epsilon > 0$.

Since $\sequence {x_n}_{n \mathop \in \N}$ is Cauchy, there exists $N \in \N$ such that:

$\ds \norm {x_n - x_m}_X < \frac \epsilon M$

for $n, m \ge N$.

Then:

$\ds \norm {T x_n - T x_m} < M \paren {\frac \epsilon M} = \epsilon$

for $n, m \ge N$.

So $\sequence {T x_n}_{n \mathop \in \N}$ is a Cauchy sequence in $Y$.

$\blacksquare$