Are ALL linear functionals on $C[0,1]$ generated by measures? [closed]

Question

Consider derivative of the convolution of a given function $f(\cdot)$ with a fixed $C^\infty$ function $s(\cdot)$, evaluated say at $1/2$. Is there a measure which generates the functional so defined?

Answer 1

For your linear functional (say $L$) and all $f\in C[0,1]$ we have $$L(f)=(f*s)'(1/2)=\int_0^1 f(y)s'(1/2-y)\,dy,$$ so that $L$ is generated by the signed measure $\mu(dy)=s'(1/2-y)\,dy$.

The question in your title, "Are ALL linear functionals on C[0,1] generated by measures?", is different from the one in the body of your post, and the answer to the "title" question is negative, assuming the axiom of choice. Indeed, then $C[0,1]$ has a Hamel basis $(e_i)_{i\in I}$, which is infinite (and even uncountable). Let $J=\{i_1,i_2,\dots\}$, with pairwise distinct $i_1,i_2,\dots$, be a countable subset of $I$. Define the linear functional $K$ on $C[0,1]$ by the formulas $$K(e_{i_k}):=k\|e_{i_k}\|$$ for natural $k$, with $K(e_i):=0$ for $i\in I\setminus J$. Then the linear functional $K$ is not bounded and hence not continuous. So, $K$ is not generated by any signed measure.