If $f$ is continuous on $[0,1]$ and if $\int_0 ^1 f(x) x^n dx = 0$ for $n=0,1,2,3,\cdots$; then prove or disprove $\int _0 ^1 f^2(x) dx = 0$

Question

Attempt: $\int_0 ^1 f(x) x^n dx = 0$ for $n=0,1,2 \cdots \implies \int_0 ^1 f(x) dx = 0 $.

Using Bi Parts rule :

$\int _0 ^1 f^2(x) dx = \int _0 ^1 f(x) f(x) dx = |_0^1f(x) \int f(x) dx - \int [ \int f(x) dx . f'(x) ]dx$

Since, $\int_0 ^1 f(x) dx = 0 \implies $ the above expression reduces to $0$.

Is my attempt correct?

Somehow, I feel that I might be making a blunder as I haven't even used the continuous nature of $f$ given in the problem. Please help me move towards the correct solution.

Thank you.

You are making an assumption that $f$ is differentiable when you only know continuity. — Clayton, Nov 07 '14 at 14:18
It cannot be the solution, since you are not using the fact the the integration is $0$ with any power of $x$. — Martigan, Nov 07 '14 at 14:19
I'll wager you've recently studied the Weierstrass Approximation Theorem. — David Mitra, Nov 07 '14 at 14:20
@SimonS I haven't studied the Weierstrass approximation theorem? Though, I am trying to study this theorem now. I saw this problem in a problem booklet actually — MathMan, Nov 07 '14 at 14:34
closely related to http://math.stackexchange.com/questions/16831/nonzero-f-in-c0-1-for-which-int-01-fxxn-dx-0-for-all-n/16849#16849 — Matthew Towers, Nov 07 '14 at 14:37

score 3 · Answer 1 · answered Nov 07 '14 at 14:26

3

Hint: introduce a sequence of polynomials $P_n$ which uniformly to $f$. Then prove that $$ \int_0^1 P_n f \to \int_0^1 f^2 $$

answered Nov 07 '14 at 14:26

mookid

28,795

Thank you for your answer :-). Do you think a solution exists without using the Weierstrass approximation theorem? – MathMan Nov 07 '14 at 14:45
1

You only need $L^2$ convergence, obviously. But I don't think there is. – mookid Nov 07 '14 at 14:46

score 2 · Accepted Answer · answered Nov 07 '14 at 14:26

2

By the Weierstrass approximation theorem, there exists a polynomial $p(x)$ such that $|f(x)-p(x)|\leq\varepsilon$ for any $x\in[0,1]$. Now we have: $$\int_{0}^{1}f(x)^2\,dx = \int_{0}^{1}f(x)(f(x)-p(x))+f(x)p(x)\,dx $$ but the term $\int_{0}^{1}f(x)p(x)\,dx$ vanishes while: $$\left|\int_{0}^{1}f(x)(f(x)-p(x))\,dx\right|\leq \varepsilon\cdot\max_{x\in[0,1]}|f(x)|.$$ Since $\varepsilon$ was arbitrary, $$\int_{0}^{1}f(x)^2 = 0.$$

answered Nov 07 '14 at 14:26

Jack D'Aurizio

361,689

Thank you very much for the answer. Do you think a solution exists without using the Weierstrass approximation theorem? – MathMan Nov 07 '14 at 14:43
1

If you know that the shifted Legendre polynomials give a complete base of $L^2(0,1)$, from the first condition you know that the integral of $f(x)$ against $\tilde{L}_n(x)$ is zero for any $n$, hence the $L^2$-norm of $f$ is zero, too. – Jack D'Aurizio Nov 07 '14 at 14:46
1

Alright :-) thanks again. – MathMan Nov 07 '14 at 14:47

If $f$ is continuous on $[0,1]$ and if $\int_0 ^1 f(x) x^n dx = 0$ for $n=0,1,2,3,\cdots$; then prove or disprove $\int _0 ^1 f^2(x) dx = 0$

2 Answers2