Is the following assertion true or false?
There exists a nonzero function $f \in C([0,1])$ such that $$\int_0^1f(x)x^ndx=0 (\forall n \in \mathbb N)$$ holds.
(Hint: use the weierstrass approximation theorem)
Asked
Active
Viewed 103 times
0
-
1What have you attempted yet ? – Jean Marie Feb 07 '16 at 14:56
-
The statement in this current form is false, since it states for all n. I think, there should be some restriction on n to make the question interesting at least. – runaround Feb 07 '16 at 15:29
-
Possible duplicate of Nonzero $f \in C([0, 1])$ for which $\int_0^1 f(x)x^n dx = 0$ for all $n$ – Feb 07 '16 at 16:19
1 Answers
4
From weierstrass approximation theorem, for any $\epsilon > 0$, there exists a polynomial $P_{\epsilon}(x)$ such that $$\int_0^1 |f(x) - P_{\epsilon}(x)| dx \le \epsilon$$ Now, from the condition of this question, $$\int_0^1f(x)P_{\epsilon}(x) dx = 0 $$ So, (using fact f is continuous on closed interval) $$\int_0^1f^2(x)dx = \int_0^1f(x)(f(x) - P_{\epsilon}(x))dx \le \int_0^1|f(x)|\epsilon dx = \epsilon \int_0^1|f(x)dx$$
let $\epsilon \to 0$, we have $$\int_0^1(f(x))^2dx = 0$$ since f is continuous, so f(x) = 0 for all $0\le x \le 1$
Therefore, there exists no such function to satisfy conditions of the question.
QED
runaround
- 2,216