Riesz theorem without orthogonal projection theorem

Asked Jun 16 '17 at 09:23

Active Jun 16 '17 at 09:23

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Let $H$ be a Hilbert space. I am trying to prove the Riesz theorem, but without using the orthogonal projection theorem.

Let $f: H \longrightarrow K$ be a functional. The key is that since $f$ is continuous, $\ker f$ is closed and we can use the orthogonal projection theorem to find a (nonzero) element in $(\ker f)^\perp $.

But, could I find a (nonzero) element in $(\ker f)^\perp $ without that theorem?

My attempt: since in $(\ker f)$ is closed and has codim 1, it turns out that $H= \ker f \oplus K $. I suspect that I could find that $(\ker f)^\perp \neq 0$ from that, but I don't know how to do it.

asked Jun 16 '17 at 09:23

Minkowski

1,518

How do you know that the kernel has co-dimension $1$? – uniquesolution Jun 16 '17 at 09:28
@uniquesolution : $f(x_0)=1$ gives $x= (x-f(x)x_0)+f(x)x_0 \in \mbox{ker}(f)\oplus [{x_0}]$ – Disintegrating By Parts Jun 16 '17 at 11:33
@uniquesolution Because since $f: H \rightarrow K$ is a (nonzero) functional, it is onto, then $H/\ker f = K$. – Minkowski Jun 16 '17 at 12:25

Riesz theorem without orthogonal projection theorem

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