Let V be the vector space of polynomials over the field of complex numbers, with the inner product $\langle f,g \rangle = \int_{0}^{1} f(t) \overline{g(t)}dt$. Let $L$ be a linear functional defined by $L(f)=f(z)$ where $z$ is a fixed complex number. Prove that there is no polynomial $g$ such that $\langle f,g \rangle=L(f) \forall f \in V$.
This example is given in the Linear Algebra book by Hoffman&Kunze, as a counter example to for the Riesz Representation Theorem when the Inner product space $V$, is not finite dimensional. This is the answer given in the text.
Suppose we have $L(f)=f(z)=\langle f,g \rangle=\int_{0}^{1} f(t) \overline{g(t)}dt \forall f$. Let $h(x)=x-z$.
Then $\color{red}{(hf)(z)=0}$ and so $\color{red}{\int_{0}^{1} f(t)h(t) \overline{g(t)}dt=0} \forall f$.
In particular, choosing $f(t)=\overline{h(t)}g(t)\implies \int_{0}^{1} |h(t)|^2|g(t)|^2dt=0 \implies hg=0$. Since $h \neq 0$, we must have $g=0$. But $L$ is not the zero functional and hence no such $g$ exists.
My doubt is in the red coloured text. How does it follow that $(hf)(z)=h(f(z))=0$ and $\int_{0}^{1} f(t)h(t) \overline{g(t)}dt=0$? It isn't clear to me at all why $h(f(z))=f(z)-z=0$
P.S: I've found similar questions asked before here, however as far I could find, they all give the same counterexample in terms of functional analysis/measure theory and not solely a linear algebraic proof.
The first red bit comes from expanding $$hf(z) = (z-z)f(z)= 0 .$$ The second is just the definition $$hf(z) = \left< hf, g \right> = \int_0^1 h(t) f(t) \overline{g(t)} \mathrm{d} t $$