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:


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

We see that:
 * $\ds \lim_{n \mathop \to \infty} \map A {h_n} = \map A h$

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 $a$, centered at $\mathbf 0_H$, such that its image under $A$ is included in the open ball of radius $1$, centered 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:


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

Hence:


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

Therefore:


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

Let $\epsilon \to 0$.

Then:
 * $c = \dfrac 1 a$