TITLE: Find functions family satisfying $ \lim_{n\to\infty} n \int_0^1 x^n f(x) = f(1)$
QUESTION [18 upvotes]: I wonder what kind of functions satisfy 
$$ \lim_{n\to\infty} n \int_0^1 x^n f(x) = f(1)$$
I suppose all functions must be continuous.

REPLY [5 votes]: The equation is true for any integrable function $f$ on $[0,1]$ so that $1$ is a Lebesgue point for $f$, in the sense that $\lim_{y\to1}\bar f(y)=f(1)$, where $$\bar f(y)=\frac1{1-y}\int_y^1 f(y)\,dy.$$
Indeed, using $x^n=(n+1)\int_0^x y^n\,dx$ in the integral and interchanging the order of integration, we find after a bit of calculation $$n\int_0^1x^nf(x)\,dx=n(n+1)\int_0^1(1-y)y^{n+1}\bar f(y)\,dy,$$
in which we notice that $n(n+1)(1-y)y^{n+1}$ is a delta sequence, in this case converging to the delta function at $y=1$.
Noting that $\bar f$ is continuous in $[0,1)$, and can be extended continuously to $[0,1]$ when $1$ is a Lebesgue point for $f$, completes the proof.

REPLY [3 votes]: Certainly all the continuous functions do satisfy this property. But I think we will have to see whether there are classes of functions which are not continuous but satisfy this property. 
To show that continuous functions satisfy this, we shall use the well-known Stone Weierstrass Theorem (SWT) which states that for any continuous function $f:[a,b] \to \mathbb{R} $ , there exist a sequence of polynomials which uniformly converge to $f$.
Define, $$n \int_0^1 x^n f(x)dx=L_n(f)$$ 
Then,$L_n(x^k) = \ n \int_0^1 x^n x^k dx= \frac{n}{n+k+1} $ and $\lim_{n\to\infty} L_n(x^k)=1$ for all non-negative integers $k$.
Now if $P(x)$ is an arbitrary polynomial then from above it follows that $\lim_{n\to\infty} L_n(P)=P(1)$
Let $\epsilon>0.$ Given a continuous function $f$ and a polynomial $P$ we have that $$|L_n(f)-L_n(P)| \leq L_n(|f-P|) \leq L_n( \epsilon/3)=\epsilon/3$$ provided that $d(f,P) \leq \epsilon/3$ and indeed by the SWT we can choose such a polynomial $P$
And also there exist $N \in \mathbb{N}$ such that for all $n > N,  |L_n(P)-P(1)|<\epsilon/3$
Now for such an $n$, 
$$|L_n(f)-f(1)| \leq|L_n(f)-L_n(P)|+|L_n(P)-P(1)|+|P(1)-f(1)|<\epsilon$$ 
This proves that 
$\lim_{n\to\infty}L_n(f)=f(1)$ for all continuous functions.