# Continuity of Linear Transformation/Normed Vector Space

## Theorem

Let $X, Y$ be a normed vector spaces over $\R$.

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

The following statements are equivalent::

$(1): \quad T$ is continuous over $X$
$(2): \quad T$ is continuous at $\mathbf 0$
$(3): \quad \exists M > 0 : \forall x \in X : \norm {\map T x}_Y \le M \norm x_X$

## Proof

### $\paren 1 \implies \paren 2$

Let $T$ be continuous on $X$.

$X$ is a vector space.

By definition, $\exists \mathbf 0 \in X$.

Hence, $T$ is continuous at $\mathbf 0$.

$\Box$

### $\paren 2 \implies \paren 3$

Let $T$ be continuous at $\mathbf 0$.

Let $\epsilon := 1 > 0$.

By definition of continuity:

$\exists \delta > 0: \forall x \in X : \norm {x - \mathbf 0} = \norm x < \delta \implies \norm {\map T x - \map T {\mathbf 0} } = \norm {\map T x} < 1$

Let $y := \dfrac \delta {2 \norm x} x$.

By definition of norm:

$\norm x \in \R$

Hence:

$\dfrac \delta {2 \norm x} \in \R$

Because $X$ is a vector space:

$y \in X$

Furthermore:

 $\ds \norm y$ $=$ $\ds \norm {\frac \delta {2 \norm x} x}$ $\ds$ $=$ $\ds \frac \delta {2 \norm x} \norm x$ Norm Axiom $\text N 2$: Positive Homogeneity $\ds$ $=$ $\ds \frac \delta 2$ $\ds$ $<$ $\ds \delta$ $\ds \leadsto \ \$ $\ds \norm {\map T y}$ $<$ $\ds 1$

Then:

 $\ds \norm {\map T y}$ $=$ $\ds \norm {\map T {\frac \delta {2 \norm x} x} }$ $\ds$ $=$ $\ds \norm {\frac \delta {2 \norm x} \map T x}$ Definition of Linear Mapping $\ds$ $=$ $\ds \frac \delta {2 \norm x} \norm {\map T x}$ Norm Axiom $\text N 2$: Positive Homogeneity $\ds$ $<$ $\ds 1$ $\ds \leadsto \ \$ $\ds \norm {\map T x}$ $<$ $\ds \frac 2 \delta \norm x$

Suppose $x = \mathbf 0$.

Then:

 $\ds \norm {\map T {\mathbf 0} }$ $=$ $\ds \norm {\mathbf 0}$ Linear Transformation Maps Zero Vector to Zero Vector $\ds$ $=$ $\ds 0$ Norm Axiom $\text N 1$: Positive Definiteness $\ds$ $=$ $\ds \frac 2 \delta \norm {\mathbf 0}$

All together:

$\norm {\map T x} \le M \norm x$

where $M = \dfrac 2 \delta$.

$\Box$

### $\paren 3 \implies \paren 1$

Let $M > 0$ be such that:

$\forall x \in X: \norm {\map T x} \le M \norm x$

Let $x_0 \in X$.

Let $\epsilon > 0$.

Let $\delta := \dfrac \epsilon M > 0$.

Because $X$ is a vector space:

$x - x_0 \in X$

Suppose:

$\forall x \in X : \norm {x - x_0} < \delta$

Then:

 $\ds \norm {\map T x - \map T {x_0} }$ $=$ $\ds \norm {\map T {x - x_0} }$ Definition of Linear Mapping $\ds$ $\le$ $\ds M \norm {x - x_0}$ $\ds$ $<$ $\ds M \delta$ $\ds$ $=$ $\ds M \frac \epsilon M$ $\ds$ $=$ $\ds \epsilon$

By definition, $T$ is continuous at $x_0$.

But $x_0$ was arbitrary.

Hence, $T$ is continuous on $X$.

$\blacksquare$