Let $f, g$ be polynomials of degree n such that $\int_0^1 x^kf(x)dx =\int_0^1x^k g(x)dx$ holds for each $k = 0, 1, . . . , n.$ Show that $f = g.$

I considered the function $ h(x)=f(x)-g(x)$ and got $\int_0^1 x^kh(x)dx=0, k=0,1,...,n$. How do I proceed further?


1 Answer 1


Denote $h(x)= \sum_{k=0}^n a_k x^k.$

By linearity of the integral, you can write that :

$$\int_{0}^1 h(x) h(x) dx = \sum_{k=0}^n a_k \underbrace{\int_0^1 h(x) x^k dx}_{=0} = 0$$

Then you have $\int_0^1 |h(x)|^2 =0$, which implies that $h(x)=0$ for all $x$ in $[0,1]$, which should be enough for you to conclude.


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