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The problem is show that $f=0$ whenever $f\in C[a,b]$ and $$\int_a^bf(x)e^{kx}dx =0, \hspace{1cm}\forall k\in\mathbb{N}.$$

Can someone help me?

Thank you!

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up vote 2 down vote accepted

First, letting $u=e^x$ we note that $\int_{e^a}^{e^b}f(\ln u)u^{k-1}du=0$ for all $k\in\mathbb N$.

Next, see the following old questions:


If $f$ is continuous on $[a , b]$ and $\int_a^b f(x) p(x)dx = 0$ then $f = 0$

problem on definite integral

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Very good @JonasMeyer. But, note that, $f(\log u)$ is approximated by $p(\log u)$ and not by $p(u)$! – Kelson Vieira Jan 29 '13 at 4:14
@KelsonVieira: That is not what I had in mind, else I could have invoked Stone-Weierstrass as julien did. The function $g:[e^a, e^b]\to \mathbb R$ defined by $g(u)=f(\ln u)$ is continuous, hence uniformly approximable by polynomial functions on $[e^a,e^b]$. – Jonas Meyer Jan 29 '13 at 4:16
Really @Jonas. Thank you very much! – Kelson Vieira Jan 29 '13 at 4:18

You can change variables and turn this into an application of the Weierstrass Theorem (density of the real polynomials in $C([e^a,e^b],\mathbb{R})$. That's Jonas' answer.

Or you can use the real version of the Stone-Weierstrass Theorem:

The subalgebra $\{a_o+a_1e^x+\ldots+a_ne^{nx}\;|\;\in\mathbb{N},a_j\in\mathbb{R}\}$ of $C([a,b],\mathbb{R})$ contains a non-zero constant function and separates points. So it is dense in $C([a,b],\mathbb{R})$.

Take $g_n$ a sequence in this subalgebra that converges unformly to $f$ on $[a,b]$. Then $$ 0=\int_a^bfg_n\longrightarrow \int_a^b f^2. $$

Now if $f$ is continuous and $\int_a^bf^2=0$, we find that $f^2=0$, hence $f=0$.

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Thank you for your reply @Julien! – Kelson Vieira Jan 29 '13 at 4:20
You're most welcome. – 1015 Jan 29 '13 at 4:22

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