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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.

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


Exercise: Every non-zero functional is surjective!

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