If $f$ is continuous on $[0,1]$ and if $\int\limits_{0}^{1} f(x) x^n dx = 0, (n=0,1,2,...)$, prove that $f(x)=0$ on $[0,1]$.

sequences-and-series analysis continuity proof-verification uniform-continuity

Solution 1:

See my comment. This is another way to use Weierstrass's theorem.

If $f(x)$ is a polynomial, the result is clearly true. However, any continuous function can be approximated by a polynomial, the result is true for any continuous function.

As I said in my comments, I like your answer better, but it is here as alternative.

Related

Proving that the natural filtration of Brownian motion (not augmented) is not right-continuous
Fourier transform vs Fourier series
A cycle of size at least $\frac{n}k$ in a graph with at least $3k$ vertices
If $(f_n)\to f$ uniformly and $f_n$ is uniformly continuous for all $n$ then $f$ is uniformly continuous
Green's functions of Stokes flow
How to solve equations with logarithms, like this: $ ax + b\log(x) + c=0$
Nilradical of absolutely flat ring
Equations For Quadratic Regression
In ZF, does there exist an ordinal of provably uncountable cofinality?
Ergodicity of a skew product
Proving $\sum _{k=1}^n \frac{(-1)^{k-1} 16^k (k-1)! k! (k+n-1)!}{((2 k)!)^2 (n-k)!}=\frac{4}{n}\sum _{k=1}^n \frac{1}{2 k-1}$
Having trouble understand the proof of Rouché's Theorem