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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.

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up vote 14 down vote accepted

Your equation can be rewritten as $$\lim_{n \to \infty} (n+1) \int_0^1 x^n (f(x) - f(1))\ dx = 0 $$ (I'd rather use $n+1$ than $n$, because $(n+1) \int_0^1 x^n \ dx = 1$)

Note that as $n \to \infty$, $(n+1) x^n \to 0$ uniformly on $[0,1-\delta]$ for any $\delta > 0$, implying that $(n+1) \int_0^{1-\delta} x^n (f(x) - f(1))\ dx \to 0$ for any $f$ that is integrable on $[0,1]$. On the other hand, if $f$ is continuous from the left at $1$, for any $\epsilon > 0$ there is $\delta > 0$ such that $|f(x) - f(1)| < \epsilon$ for $x \in [1-\delta, 1]$, and then $$\left|(n+1) \int_{1-\delta}^1 x^n (f(x) - f(1))\ dx\right| < \epsilon (n+1) \int_{1-\delta}^1 x^n\ dx < \epsilon$$ So your equation is true for functions that are integrable on $[0,1]$ and continuous from the left at $1$. However, this is only a sufficient condition, which certainly could be weakened. I doubt that there is a simple necessary and sufficient condition.

Note also that $(n+1) x^n < -1/(e x \ln(x))$, so the equation is also true if $(f(x) - f(1))/(x \ln(x))$ is integrable on $[1-\delta,1]$ for some $\delta > 0$. For an example that is not continuous from the left at $1$, take the indicator function of the union of the intervals $J_k = [1-2^{-k}-3^{-k}, 1-2^{-k}]$ for positive integers $k$.

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I certainly concur with the doubt expressed at the end of the penultimate paragraph. – Harald Hanche-Olsen Jul 8 '12 at 8:02
Does $f$ have to be Riemann integrable? Isn't it sufficient that $x^nf$ is Riemann integrable? – Kasun Fernando Jul 8 '12 at 14:29
@KasunFernando: Yes, you can replace "$f$ integrable on $[0,1]$ with "$x^n f$ integrable on $[0,1]$ for some $n$". And "integrable" can be Lebesgue rather than Riemann. – Robert Israel Jul 8 '12 at 16:27

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.

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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.

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From the other answers, we can see that there are other broader classes of functions. For example, the class of all the Riemann integrable functions on $[0,1]$ which are left continuous at $1$. – Kasun Fernando Jul 8 '12 at 8:30

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