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
 * $(4): \quad \exists c > 0: \forall h \in H: \left\Vert{Ah}\right\Vert_K \le c \left\Vert{h}\right\Vert_H$

Proof
It is clear that (1)=>(2)=>(3) and for (4)=>(2):

For any $\epsilon >0$, there exists $\delta=\frac{\epsilon}{c}$, such that when $||\mathbf 0_{H} -h||_H< \delta$


 * $||Ah-A\mathbf 0_{H}||_K\leq c||h||_H(1):

Assume $A$ is continuous at some point $h_0$. For any sequence $h_n\rightarrow h$ in $H$, then $h_n-h+h_0\rightarrow h_0$, hence

\[\lim_{n\to \infty}A(h_n-h+h_0)=\lim_{n\to \infty}Ah_n-Ah+Ah_0=Ah_0\]

We see $\lim_{n\to\infty}Ah_n=Ah$. Thus $A$ is continuous.

Now for the proof (2)=>(4)

Since $A$ is continuous at $\mathbf 0 _H$, hence there exists an open ball of radius of positive real $a$, centred at $\mathbf 0 _H$ that its image under $A$ is included in the open ball of radius $1$, centred at $\mathbf 0 _K$. Which is $||h||_H0$, we have

\[\left\|a\frac{h}{||h||_H+\epsilon}\right\|_H<a\]

Hence

\[\left\|A\left(  a\frac{h}{||h||_H+\epsilon}  \right) \right\|_K=  a\frac{\|Ah\|_K}{||h||_H+\epsilon} <1 \]

Therefore

\[ \|Ah\|_K<\frac{1}{a}\|h\|_H+\frac{\epsilon}{a} \]

Let $\epsilon\rightarrow 0$, we have $c=\frac{1}{a}$.