Does there exists a measure $\mu$ on $[0,1]$ such that $\int_0^1 p(x) \, d\mu(x) = p'(0)$ for every polynomial $p$

Question

If this polynomial is at most degree $n$, I know this measure exists, but I am not sure that whether there exists $\mu$ for every polynomial?

Answer 1

As every complex measure is of finite total variation, the functional

$$ C([0,1])\to \Bbb{C}, f \mapsto \int f \, d \mu $$

is bounded w.r.t. the sup-norm, hence continuous w.r.t. uniform convergence.

Now let $p_n (x)= (1-x)^n /n$. Then $p_n \to 0$ uniformly on $[0,1]$, hence

$$ \int p_n \, d\mu \to 0, $$

But $p_n '(0)= -1$ for all $n$. Hence there is no such measure.

Answer 2

Let $f(x)=a_0+a_1x\cdots +a_{n}x^{n}, n\ge 1$. Then by linearity we have $$ \int fd\mu=\sum_{i=0}^{n}a_{i}\int x^{i}d\mu=a_{1}\int xd\mu=a_{1} $$ and this is because $$ \frac{d}{dx}|_{0}x^{i}=0, i\not= 1 $$ Therefore the measure $\mu$ can be viewed as a linear functional given by $$ \int_{[0,1]}\sum^{n}_{i=0}a_{i}x^{i}d\mu=a_{1},\forall n $$ I claim that this is not a continuous linear functional on the space of all polynomials at $[0,1]$. If $\mu$ is continuous we would have $$ \exists A, |f'(0)|\le A|f|_{\max} $$ But this is impossible because we can construct sequence of polynomials such that $$ |f_{n}'(0)|\ge 1, |f_{n}|_{\max}\rightarrow 0 $$ To explicitly construct this takes some work, and I am not sure if my construction is optimal. Let $g_{n}$ be a sequence of differentiable functions on $[0,1]$ such that $|g_{n}(0)'|\ge 1+\frac{1}{n}$ for all $n$, and $|g_{n}|_{max}\rightarrow 0$. This can be done geometrically. Then we approximate each $g_{n}$ by a sequence of polynomials $f^{i}_{n}$ globally such that $|f'(0)^{i}_{n}-g'(0)_{n}|\le \frac{1}{2i}$, $|f^{i}_{n}-g_{n}|_{max}\rightarrow 0$. This can be done by Weistrauss approximation theorem by working with $g_{n}$'s derivative one by one. Now taking the diagonal sequence $f^{n}_{n},n\in \mathbb{N}$ would give the desired counter-example.