Linear Transformations

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

Properties The following are equivalent:

1. $T$ is continuous.
2. $T$ is continuous at zero.
3. $T$ is continuous at a point
4. $T$ is bounded; i.e., $\exists c > 0$ such that $\Vert T(h)\Vert \leq c \Vert h\Vert$ for all $h \in H$.

We define

$$\Vert T\Vert = \sum \{ \Vert T(h)\Vert : \Vert h\Vert \leq 1 \}$$

Monday, 8-29-05:

Definition 1.3.1   A linear functional on a vector space $V$ over $\mathbb{F}$ is a linear transformation

$$T : V \rightarrow \mathbb{F}$$

(where $\mathbb{F}$ is regarded as a vector space over itself).

Example: If $\mathcal{H}$ is a Hilbert space and $f \in \mathcal{H}$, then the map $T: \mathcal{H}\rightarrow \mathbb{F}$ given by $T(h) = \left\langle{ h, f }\right\rangle$ is a bounded linear functional. By the CBS inequality, $\Vert T \Vert \leq \Vert f \Vert$. In fact, $\vert T(f/\Vert f\Vert)\vert = \vert\left\langle{ f,f }\right\rangle \vert/\Vert f\Vert = \Vert f\Vert$. Thus, $\Vert T\Vert = \Vert f\Vert$.

Theorem 1.3.2   Riesz-Representation Theorem (for Hilbert spaces)
If $T: \mathcal{H}\rightarrow \mathbb{F}$ is a bounded linear functional, then there is a unique $g \in \mathcal{H}$ such that

$$T(h) = \left\langle{ h,g }\right\rangle$$

for all $h \in \mathcal{H}$.

Proof. If $T = 0$, clearly $f = 0$ (uses the fact we have an inner product, and not a semi-inner product).
Assume $T \neq 0$. Then $\ker T \neq H$. Pick $f \in (\ker T)^\perp$ such that $T(f) = 1$. For $h \in \mathcal{H}$, consider $h - T(h) f$. Then,

$$T(h - T(h) f) = T(h) - T(h) T(f) = 0$$

Thus,

$$\left\langle{ h - T(h) f, f }\right\rangle = 0$$

Distributing this out,

$$\left\langle{ h,f }\right\rangle = T(h) \Vert f\Vert^2$$

Thus,

$$T(h) = \left\langle{ h, \frac{f}{\Vert f\Vert^2} }\right\rangle$$

Let $$g = \frac{f}{\Vert f\Vert^2}$$.

If $k$ is another vector in $H$ with $T(h) = \left\langle{ h,k }\right\rangle$ for all $h \in H$, then

$$\left\langle{ h, g - k }\right\rangle = T(h) - T(h) = 0$$

for all $h$, thus $g - k \in H^\perp = \{\vec 0 \}$. $\qedsymbol$