Uniform Convergence and limit $(n+1)\int_0^1 x^nf(x) \; dx$

Question

If $f$ is a continuous real-valued function, show that $$ f(1)=\lim_{n\to \infty} \int_0^1 (n+1)\,x^n \,f(x) \; dx $$

I am looking for a general hint or steps to proceed but I want to fill them in. Looking at this I see the '$n+1$' along with the $\int x^n$ makes me very suspicious.

My first thought is that $(n+1)x^nf(x)$ converges to $f(1)$ uniformly on $[0,1]$ but I feel as though this is a fools thought. I then sat and tried the case $f(x)=x$ and $f(x)=x^{n_0}$. This was a bit illuminating. I then tried the case where $f(x)$ is a general polynomial and I can show that case. But of course $f(x)$ need not be a polynomial.

To try to make this method work, I considered using Stone-Weierstrass. I can find a sequence of polynomials $p_n(x)$ such that $p_n(x) \to f(x)$ uniformly. But I am unsure of how to make it fit in with the integral. I assume it is trivial to show $x^np_n(x) \to x^nf(x)$. I know that the above limit works with $x^np_n(x)$. However, does this imply that this -- under the limit -- must be the same as $x^nf(x)$? Or is this not at all how I should go about it? Is there a better or easier way? Or perhaps a more enlightening or 'quaint' way?

Answer 1

Note that since $[0,1]$ is compact $f$ is uniformly continuous and bounded, say $|f| \leq K$ on $[0,1]$. Split the integral into a portion close to $1$ and far from $1$.

$$ \int_0^{1-\epsilon} (n+1) x^n f(x) dx + \int_{1-\epsilon}^1 (n+1)x^n f(x)dx $$

Let's look at the first integral

$$ \left|\int_0^{1-\epsilon} (n+1) x^n f(x) dx\right| \leq (n+1)(1-\epsilon)^n K \to 0 $$ as $n \to \infty$. Now since $f$ is uniformly continuous, we can choose $\epsilon$ small enough that $|f(x)-f(1)| \leq \delta$ on $(1-\epsilon,1]$. Examining the second integral (read from the middle out, not left to right) $$ (f(1)-\delta)[1-(1-\epsilon)^{n+1}] = \int_{1-\epsilon}^1 (n+1)x^n (f(1)-\delta)dx\leq \int_{1-\epsilon}^1 (n+1)x^n f(x)dx \leq\int_{1-\epsilon}^1 (n+1)x^n (f(1)+\delta)dx = (f(1)+\delta)[1-(1-\epsilon)^{n+1}] $$ So $$ \int_0^{1-\epsilon} (n+1) x^n f(x) dx + \int_{1-\epsilon}^1 (n+1)x^n f(x)dx = \text{small} + \text{close to $f(1)$} $$ do you see how to finish/fill in the details?

Answer 2

This is a special case of a much more general phenomenon: Suppose you have a sequence of positive continuous functions $g_n$ on $[0,1]$ such that i) $\int_0^1g_n = 1, n=1,2,\dots $ ii)$\int_0^b g_n \to 0$ for each $b, 0\le b <1.$ Example: $g_n(x) = (n+1)x^n.$ If $f$ is Riemann integrable on $[0,1]$ and $f$ is continuous at $1,$ then

$$\int_0^1fg_n \to f(1).$$

Proof:

$$\tag 1 |\int_0^1fg_n - f(1)| = |\int_0^1(f - f(1))g_n| \le \int_0^1|f - f(1)|g_n.$$

Let $\epsilon>0.$ Let $M= \sup_{[0,1]}|f|.$ Choose $b$ such that $|f(x)-f(1)| <\epsilon$ on $[b,1].$ Then the right side of (1) equals

$$\tag 2 \int_0^b|f - f(1)|g_n + \int_b^1|f - f(1)|g_n \le 2M\int_0^b g_n + \epsilon\int_b^1g_n.$$

Applying the $\limsup$ to both sides of (2), and using our hypotheses, we see the $\limsup$ of the left side of (2) is less than or equal to $2M\cdot 0 + \epsilon.$ Since $\epsilon$ is arbitrary, we're done.