In Kreyszig's book we have the following version of the Riesz's Representation Theorem:
Theorem 3.8-1 Every bounded linear functional $f$ on a Hilbert space $H$ can be represented in terms of the inner product, namely,
$$f(x)=\langle x,z\rangle $$
where $z$ depends on $f$, is uniquely determined by $f$ and has norm
$$\|z\|=\|f\|.$$
My question: Considering $H$ be a complex Hilbert space, is there any way to obtain $w \in H$ such that
$$f(x)=\langle w,x \rangle$$
for all $x \in H$ and $\|f\|=\|w\|$?
I was trying to apply this theorem to $g(x)=\overline{f(x)}$. Unfortunatelty, $g$ is not linear because $g(\lambda x)=\overline{\lambda}g(x)$.
Best Answer
No. $x \mapsto \langle w, x \rangle$ is not even linear, so how can $f$ be?
Or just take $f:\mathbb{C} \rightarrow \mathbb{C}$, $f(z) := iz$. If there is some $w \neq 0$ such that $f(z) = \langle w, z \rangle_{\mathbb{C}} = w \bar z$, then $$ iz = w\bar{z}. $$ Plug in $i$, then $-1 = -iw \implies w = -i$. Plug in $1$, then $i = w$. Makes no sense.