Continuity of Linear Transformations

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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 $||\mathbf 0_{H} -h||_H< \delta$

$||Ah-A\mathbf 0_{H}||_K\leq c||h||_H<c\delta=\epsilon$

Now we prove (3)=>(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 $||h||_H<a$, then $||Ah||_K<1$.

Let $h$ be an arbitrary element in $H$; let $\epsilon>0$, 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 $c=\frac{1}{a}$.

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