Consider a continuous linear function $\lambda: H \to \mathbb{C}$, where $H$ is a Hilbert space. I want to show that there is $v \in H$ such that $$\lambda(h) = \langle h, v \rangle$$ for all $h \in H$, where $\langle, \rangle$ is the inner product on $H$. I tried choosing $v \in (\ker \lambda)^{\perp}$, but only got as far as showing $\lambda$ agrees with $\langle \cdot, v \rangle$ on $\ker \lambda$. Thanks for any advice.
Asked
Active
Viewed 80 times
0
-
3This is called Riesz representation theorem. – Julien Jun 23 '13 at 04:58
1 Answers
1
If $\ker \lambda = \mathbb{H}$, then choose $v = 0$ and you are finished.
So, suppose $\ker \lambda \neq \mathbb{H}$. Choose $v \in (\ker \lambda)^\bot \setminus \{0\}$. Note that $\operatorname{sp} \{ v \} = (\ker \lambda)^\bot$.
Let $\phi(x) = \frac{\lambda(v)}{\|v\|^2} \langle v, x \rangle$. By construction, we have $\phi(v) = \lambda(v)$, hence $\phi,\lambda$ agree on $\operatorname{sp} \{ v \}= (\ker \lambda)^\bot$. Now suppose $n \in \ker \lambda$. Then $\lambda(n) = 0$, and $\phi(n) = 0$, and hence $\lambda, \phi$ agree on $\ker \lambda$.
Since every $x \in \mathbb{H}$ can be written as $x=n+\alpha v$, with $n \in \ker \lambda$, we see that $\lambda(x) = \lambda(n+\alpha v)= \alpha \lambda (v) = \alpha \phi(v) = \phi(n+\alpha v) = \phi(x)$.
Note: Continuity of $\lambda$ appears implicitly in that it implies $\ker \lambda$ is closed, which allows the decomposition $\mathbb{H} = \ker \lambda \oplus (\ker \lambda)^\bot$.
copper.hat
- 172,524
-
You seem to use that $\dim (\ker \lambda)^{\perp} = 1$ in this solution. Can you explain why this should be the case? – Mr. Chip Jun 23 '13 at 08:11
-
Suppose $x,y \in (\ker \lambda)^\bot$. Then for some $\alpha$, you have $\phi(x+\alpha y) = 0$. Hence $x+\alpha \in \ker \lambda \cap (\ker \lambda)^\bot$, and so $x+\alpha y = 0$. It follows that $\dim (\ker \lambda)^\bot = 1$. – copper.hat Jun 23 '13 at 08:39