# Continuity of Linear Transformations

## Theorem

Let $H, K$ be Hilbert spaces, and let $A: H \to K$ be a linear transformation.

Then the following four statements are equivalent:

$(1): \quad A$ is continuous
$(2): \quad A$ is continuous at $\mathbf 0_H$
$(3): \quad A$ is continuous at some point
$(4): \quad \exists c > 0: \forall h \in H: \norm {\map A h}_K \le c \norm h_H$

## Proof

It is clear that $(1) \implies (2) \implies (3)$ and for $(4) \implies (2)$:

For any $\epsilon > 0$, there exists $\delta = \dfrac \epsilon c$, such that when $\norm {\mathbf 0_H - h}_H < \delta$:

$\norm {\map A h - \map A {\mathbf 0_H} }_K \le c \norm h_H < c\delta = \epsilon$

Now we prove $(3) \implies (1)$:

Let $A$ be continuous at some point $h_0$.

For any sequence $h_n \to h$ in $H$:

$h_n - h + h_0 \to h_0$

Hence:

$\displaystyle \lim_{n \mathop \to \infty} \map A {h_n - h + h_0} = \lim_{n \mathop \to \infty} \map A {h_n} - \map A h + \map A {h_0} = \map A {h_0}$

We see that:

$\displaystyle \lim_{n \mathop \to \infty} \map A {h_n} = \map A h$

Thus $A$ is continuous.

Now for the proof $(2) \implies (4)$.

We have that $A$ is continuous at $\mathbf 0_H$.

Hence there exists an open ball of positive real radius $a$, centred at $\mathbf 0_H$, such that its image under $A$ is included in the open ball of radius $1$, centred at $\mathbf 0_K$.

This is $\norm h_H < a$.

Then:

$\norm {\map A h}_K < 1$

Let $h$ be an arbitrary element in $H$.

Let $\epsilon > 0$.

We have:

$\norm a \dfrac h {\norm h_H + \epsilon}_H < a$

Hence:

$\norm {\map A {a \dfrac h {\norm h_H + \epsilon} }_K} = a \dfrac {\norm {\map A h}_K} {\norm h_H + \epsilon} < 1$

Therefore:

$\norm {\map A h}_K < \dfrac 1 a \norm h_H + \dfrac \epsilon a$

Let $\epsilon \to 0$.

Then:

$c = \dfrac 1 a$

$\blacksquare$