3
$\begingroup$

Suppose that $f$ is continuous on $[0,1]$ and $$ \int_0^1 x^kf(x)\ dx=0$$ for $k=0,1,...,n-1$, $$\ \int_0^1x^nf(x) \ dx=1$$ Prove that there exists $\xi\in(0,1)$ such that $|f(\xi)|\geq 2^n(n+1)$.

So my idea is to use proof by contradiction, but nothing comes out

$\endgroup$

2 Answers 2

2
$\begingroup$

You have $\int_0^1 (x-\frac{1}{2})^nf(x)dx =1$ which means

$$1 \leq \int_0^1 \left|(x-\frac{1}{2})^nf(x)\right|dx = |f(\xi)|\int_0^1\left|(x-\frac{1}{2})^n\right|dx = |f(\xi)|\dfrac{1}{n+1}\dfrac{1}{2^n} $$

$\endgroup$
2
  • $\begingroup$ I'm assuming your first equality is an inequality and you are maximizing $f$ on $[0,1]$. Seems strange that the provided information from the original question is not essential except for the last bit. $\endgroup$ Nov 21, 2014 at 19:36
  • $\begingroup$ @JasonKnapp It becomes an inequality if we take $|f(\xi)|$ as the maximum of $|f(x)|$, but we can also find $\xi$ such that the equality holds, then $|f(\xi)|$ is the weighted average of $|f(x)|$ over the interval by the weight $|(x-\frac{1}{2})^n|$ $\endgroup$ Nov 21, 2014 at 19:40
0
$\begingroup$

Let $g(x) = f(x+1/2)$. We then have $$\int_0^1x^k f(x) dx = \int_{-1/2}^{1/2} (t+1/2)^k f(t+1/2) dt = \int_{-1/2}^{1/2} (t+1/2)^k g(t) dt$$ For $k \in \{0,1,2,\ldots,n-1\}$, we have $$\int_{-1/2}^{1/2} (t+1/2)^k g(t) = 0$$ This means we have $$\int_{-1/2}^{1/2} t^k g(t) = 0$$ for $k \in \{0,1,2,\ldots,n-1\}$. Now consider $$1=\int_0^1 x^n f(x) dx = \int_{-1/2}^{1/2} (t+1/2)^n g(t) dt = \int_{-1/2}^{1/2}t^n g(t) dt$$ Now use contradiction to obtain the desired result (you can in fact you a direct proof from here).

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .