Show that $f = 0$ if $\int_a^b f(x)e^{kx}dx=0$ for all $k$ 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!
 A: 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: http://en.wikipedia.org/wiki/Stone%E2%80%93Weierstrass_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 $f=0$.
A: 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:
Nonzero $f \in C([0, 1])$ for which $\int_0^1 f(x)x^n dx = 0$ for all $n$
If $f$ is continuous on $[a , b]$ and $\int_a^b f(x) p(x)dx = 0$ then $f = 0$
problem on definite integral
