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I have a slight problem with the proof of the Riesz Representation Theorem, particularly the uniqueness part.

Suppose we have a Hilbert space $H$ and a non-zero linear functional $f : H\rightarrow\mathbb F$. The proof I have seen begins as follows. Since $f$ is non-zero, there exists a $z\in(\ker f)^\perp$ such that $f(z) = 1$ and $\|z\| = 1$. It then proceeds to show that $f(x) = \langle x, z\rangle$ for all $x\in H$. This part is fine.

The uniqueness is confusing me though, because to me this seems to suggest that there is precisely one vector $z\in (\ker f)^\perp$ such that $f(z) = 1$ and $\|z\| = 1$. This seems unlikely to me. Is this true? Or am I thinking about this wrong?

IAlreadyHaveAKey
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2 Answers2

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Yes. If there were a second one ($\bar z$, say), you have $$\langle z,x\rangle = \langle \bar z,x\rangle\quad \forall x$$ which implies $$\langle z-\bar z ,x\rangle = 0 \quad \forall x $$ so in particular

$$\langle z-\bar z , z-\bar z\rangle =||z - \bar z ||^2= 0 \quad \Rightarrow z=\bar z $$

Thomas
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  • But doesn't it seem strange that there is only one vector with that value under $f$? That seems quite strange to me. I'm not saying it's not true, I understand the proof, but I guess I'm looking for some intuition as to why there's only one $z$ with $f(z) = 1$ and $|z| = 1$. – IAlreadyHaveAKey Jun 10 '17 at 10:02
  • +1 for the simplicity (oh and thank you!) – Thomas Winckelman Nov 18 '20 at 02:34
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Concerning your question as to why this would be expected, maybe this will help. $$ f : H \to \mathbb{F}$$ If $H$ is finite dimensional (dim = $n$), then by the Rank-Nullity theorem, $$ \text{Null}(f) + \text{Rank}(f)=n$$ but $\mathbb{F}$ has dimension 1 so $\text{Rank}(f) \le 1$. Thus, ''most'' elements of $H$ are in the null space of $f$. In the infinite-dimensional case, all elements except those in the span of $z$ are actually in the null space of $f$.

Walton
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