How do I prove interval 𝐴⊂[0,3] exists on this integration

Question

Let $f:[0,3] \to \mathbb{R}$ be a continuous function satisfying

$$\int_{0}^{3}x^{k}f(x)dx=0 \quad \text{for each k = 0,1,}\dots,n-1$$ and $$\int_{0}^{3}x^{n}f(x)dx=3.$$

Then prove that there is an interval $A \subset [0,3]$ such that $$|f(x)|\geq \left(\frac{2}{3}\right)^{n}(n+1)$$ for all $x \in A.$

I have tried $3=\int_{0}^{3}(1+x+x^2+......x^{n-1}+x^n)f(x)dx=\int_{0}^{1}\frac{x^{n+1}-1}{x-1}f(x)dx+\int_{1}^{3}\frac{x^{n+1}-1}{x-1}f(x)dx$

but I got stucked on integrating

Answer 1

I'll argue using what @1mdlrjcmed gives as a hint.

Suppose NOT, there is no such interval $A$.
Then for any interval $A,$ $$|f(x)| < \left(\frac{2}{3}\right)^n (n+1),$$ where $x \in A.$
Note that it implies that $M:=\max_{x \in [0,3]} |f(x)| < \left(\frac{2}{3}\right)^n (n+1).$
Also, observe that $$\int_0^3 \left(x-\frac{3}{2}\right)^n f(x) \, dx = \int_0^3 x^n f(x) \, dx= 3,$$ which is true since $\int_0^3 x^k f(x) \, dx =0$ for $k=0, 1,\dots, n-1$ and if we expand $(x-\frac{3}{2})^n$, then all the integrals of lower-degree terms $x^k f(x)$ equal to $0.$ So the translation(I mean the translation only restricted to $x^n$, not whole integrand.) won't make any difference to the integral.

Then \begin{align*} 3 &= \Biggl|\int_0^3 \left(x-\frac{3}{2}\right)^n f(x) \, dx\Biggr| \\& \le \int_0^3 |f(x)|\Biggl|\left(x-\frac{3}{2}\right)^n\Biggr| \, dx \\ &\le \int_0^3 M\Biggl|\left(x-\frac{3}{2}\right)^n\Biggr|\, dx\\ &\le 2 \int_{\frac{3}{2}}^3 M \biggl(x-\frac{3}{2}\biggr)^n \,dx = 3\,\frac{M}{n+1} \biggl(\frac{3}{2}\biggr)^n. \end{align*} Hence,
$${n+1}\biggl(\frac{2}{3}\biggr)^n \le M.$$ This contradicts to the assumption. So the assertion follows.

Answer 2

Here is a hint: if we assume for contradiction that for all intervals $A \subset [0,3],$ $$|f(x)| < \left(\frac{2}{3}\right)^n(n+1)$$ that would imply (let $A = [0,3]$) $$\int_0^3 f(x)x^n dx \leq \int_0^3 |f(x)|x^ndx < \int_0^3 \left(\frac{2}{3}\right)^n(n+1)x^n dx = 3\cdot2^n.$$