Continuity of Linear Transformation/Normed Vector Space

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

Suppose $T : X \to Y$ is a linear mapping.

Then the following statments are equivalent:


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

$\paren 1 \implies \paren 2$
Let $T$ be continuous on $X$.

$X$ is a vector space.

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

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

$\paren 2 \implies \paren 3$
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:

Then:

Suppose $x = \mathbf 0$.

Then:

Altogether:


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

where $M = \dfrac 2 \delta$.

$\paren 3 \implies \paren 1$
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:

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

But $x_0$ was arbitrary.

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