# John Conway's proof of Riesz representation theorem

I'm studying Functional Analysis from John Conway's "A Course in Functional Analysis" and I needed to go over some things from the first chapter and decided to reread the proof of Riesz representation theorem, it goes as follows:

Theorem: Let $$L: \mathcal{H} \rightarrow \mathbb{F}$$ be a bounded linear functional, then there is a unique vector $$h_0$$ such that $$L(x) = \langle x, h_0 \rangle \forall x \in \mathcal{H}$$.

Proof: Let $$M = Ker(L)$$. Because L is continuous M is a closed subspace of $$\mathcal{H}$$. We may assume that $$M \neq \mathcal{H}$$ and hence $$\mathcal{H}= M \oplus M^\perp$$. So we may choose $$f_0 \in M^\perp$$ such that $$L(f_0) = 1$$

From here and until here I understand the proof but I don't get what guarantees the existence of such a $$f_0$$, does every linear functional map something to 1? If so why is that? I am sorry if I'm missing something obvious.

We know that $$M \ne \mathcal{H}$$, so in particular $$M^\perp \ne 0$$. Choose a non-zero $$f \in M^\perp$$ and consider $$f_0:= f/L(f)$$.