Continuity of Linear Transformation/Normed Vector Space

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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 $X$.

$\blacksquare$


Sources

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