# Integral x^n *f(x) dx =0 ; f for all n, f in C[0,1], then f(x)=0

1. Jul 18, 2014

### WWGD

I am trying to show that for f in C[0,1] , and $n=0,1,2,...$ we have:

$\int_0^1 x^n f(x)dx =0$ (&&) , then

$f(x)==0$ .

I am using Weirstrass approximation, so that , for any $\epsilon >0$ , there is $P_n(x) = a_0+a_1x +..+x^n$ with : $Sup_{x in [0,1]} | P_n(x)-f(x)| < \epsilon$.

We then sub-in in (&&) to get, for $i=0,1,..,n$. :

$0= \int _0^1 a_i x^i => \int_0^1 P_n(x)f(x)dx =0$ , so :

$\int_0^1 [f(x) \pm \epsilon] f(x)dx = \int_0^1 f(x)^2 + (\epsilon) f(x) =0$

We let $\epsilon →0$ , and then $\int_0^1 f(x)^2 =0 → f(x)=0$.

Last edited: Jul 18, 2014
2. Jul 18, 2014

### micromass

You have substituted $f(x) = P_n(x) \pm \varepsilon$ in the integral. I don't see how this could be correct. The only thing you know is that

$$f(x)-\varepsilon < P_n(x) < f(x)+\varepsilon$$

So you should do something with this to get an inequality of integrals.

But what I would do is make use of the fact that $P_n\rightarrow f$ uniformly and that uniform limits allow us to interchange limit and integral.

3. Jul 18, 2014

### WWGD

Yes, I was a bit fuzzy there, but I think this can be "rigorized" by using an inequality. I will show a cleaned-up version soon; just trying to see if the argument is "spiritually correct" as my prof. says.

4. Jul 18, 2014

### micromass

I would do the following

$$\left|\int_0^1 P_n(x)f(x)dx - \int_0^1 f(x)^2 dx\right|\leq \int_0^1 |P_n(x) - f(x)| |f(x)|dx\leq \textrm{sup}_{x\in [0,1]} |P_n(x) - f(x)| \int_0^1 |f(x)|dx$$

This if $n\rightarrow +\infty$, then

$$\int_0^1 P_n(x)f(x) dx \rightarrow \int_0^1 f(x)^2dx$$

And since each term of the sequence is $0$, we get $\int_0^1 f(x)^2dx = 0$ and thus $f = 0$.

5. Jul 19, 2014

### WWGD

Yes, this follows from the layout pretty straightforward.