Find a Borel measure $\mu$ such that $\int_0^1 f(t) d\mu(t)=\Lambda f, \forall f\in C[0,1]$

Question

I'm working through a problem set for my functional analysis course and one of the problems requires me to find a Borel measure such that $\int_0^1 f(t) d\mu(t)=\Lambda f, \forall f\in C[0,1]$, where $\Lambda:C[0,1]\to \mathbb{R}$ is the linear functional $$\Lambda f = \lim_{n\to+\infty} (n+1)\int_0^1 t^n f(t) dt.$$ I've determined that $\Lambda x^k=1, \forall k\in\mathbb{N}$ and, since $\{x^k\}$ is a fundamental set in $C[0,1]$, I'm thinking it would be enough to determine the measure from the condition $\int_0^1 t^n d\mu(t)=1$, but that's where I'm stuck now. I would love an idea of how to continue solving this or another idea of how to approach this problem.

Answer 1

For any smooth $f$, we see \begin{align*}\Lambda f &= \lim_{n\to\infty} (n+1)\int^1_0t^n f(t) dt \\ &= \lim_{n\to\infty} \int^1_0 \frac{d}{dt}(t^{n+1})f(t)dt\\ &= \lim_{n\to\infty} \left(t^{n+1}f(t) \bigg|^1_0 -\int^1_0 t^{n+1} f'(t)dt\right)\\ &= f(1) - \lim_{n\to\infty} \int^1_0 t^{n+1} f'(t)dt. \end{align*} The latter integral goes to zero by the dominated convergence theorem, and thus $\Lambda f = f(1)$ for all smooth $f$. Since every $f$ is a uniform limit of polynomials, this will extend to all continuous functions. Thus the correct Borel measure is the Dirac mass centered at $t = 1$.