Let $f:[0,1]\rightarrow$ $\mathbb{R}$ be continuous and $f(1)=0$. Show that $\displaystyle\lim_{n \to \infty}\int_0^1 \! x^nf(x) \, \mathrm{d}x=0.$

This is what I've done.

Let $\epsilon>0$. Split the integral into two parts.

$\displaystyle\int_0^1 \! x^nf(x) \, \mathrm{d}x$=$\displaystyle\int_0^{1-\epsilon} \! x^nf(x) \, \mathrm{d}x + \displaystyle\int_{1-\epsilon}^1 \! x^nf(x) \, \mathrm{d}x$.

Since $x^n\rightarrow0$ uniformly on $[0,1-\epsilon]$, the first integral goes to zero when we take the limit.

Now it's the second integral that's giving me trouble. I wanted to say that since f is uniformly continuous on $[0,1]$, we have that

$\displaystyle\int_{1-\epsilon}^1 \! x^nf(x) \, \mathrm{d}x=\displaystyle\int_{1-\epsilon}^1 \! x^n(f(x)-f(1)) \, \mathrm{d}x<\displaystyle\int_{1-\epsilon}^1 \! \epsilon \, \mathrm{d}x$ for any $x\in[1-\epsilon,1]$.

Is this true? Any suggestions would be helpful. Thanks!


Here is a simpler proof. Since $f$ is continuous so is $|f|$; therefore $|f|$ attains its maximum on $[0,1]$. Put $$M = \max\{|f(t)|, t\in [0,1]\}.$$ Then you have $$\int_0^1 |x^n f(x)| dx \le M\int_0^1 x^n\,dx = {M\over n + 1} $$ The last term on the right converges to zero as $n\to\infty$. Your hypothesis that $f(0) = 1$ is not needed.

  • $\begingroup$ Thanks! I was thinking about doing something like this, but since it doesn't use the assumption that f(1)=0, I thought I might be doing something wrong. $\endgroup$
    – Son Goku
    Dec 16 '12 at 0:47

Since $f$ is continuous on the compact interval $[0,1]$ it is bounded, say $|f|<M$. Then $$| \int_0^1 x^nf(x)dx| \leq \int_0^1 | x^n f(x)dx| \leq M \int_0^1x^ndx=\frac{M}{n+1}$$ which tends to zero.

I don't really see why the value $f(1)$ is given because it does not affect the outcome. Uniform convergence isn't really relevant either because you're not trying to interchange any sorts of limits.


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