# Proving that $\int_0^1 f(x)e^{nx}\,{\rm d}x = 0$ for all $n\in\mathbb{N}_0$ implies $f(x) = 0$

I'm trying to show that if $f$ is a continuous function on $[0,1]$ and $\int_0^{1} f(x)e^{nx}\,{\rm d}x = 0$ for all $n = 0, 1, 2, \dots$, then $f(x) = 0$.

I'd like to use Weierstrass approximation theorem to find a sequence of polynomials $p_m$ that converge uniformly to $f(x)$. Then we could say $\lim\limits_{m\to \infty} \int p_m(x)e^{nx}\,{\rm d}x = \int f(x)e^{nx}\,{\rm d}x = 0$, but I'm struggling to deduce that then all the $p_m$ are zero which would give the result.

• Apologies, have corrected now Commented Jan 13, 2015 at 10:27
• What's your assumption on $f$? Continuity as you seem to imply?
– a...
Commented Jan 13, 2015 at 10:28
• $f$ is a continuous function on $[0,1]$, will add that to the question Commented Jan 13, 2015 at 10:29
• This is most likely overkill, but you can note that the set of finite linear combinations of these exponentials is a subalgebra of $C([0,1],\mathbf R)$ which contains a non-zero constant function and separates points, so by the Stone-Weierstrass theorem this subalgebra is dense in $C([0,1],\mathbf R)$. Approximate $f$ with the elements of this subalgebra to find that $\int |f|^2=0$.
– a...
Commented Jan 13, 2015 at 10:41
• @a... That doesn't work because the algebra is not self-adjoint (because the complex conjugate of $e^{int}$ for $n\ge0$ is $e^{int}$ for $n\le0$...) Commented Jul 19, 2022 at 8:11

## 1 Answer

Make a change of variables $u = e^x$

$$\int_0^1 f(x) e^{nx}dx = \int_{1}^e g(u) u^{n}du = 0$$

where $g(u) = \frac{f(\log(u))}{u}$ is a continuous function. You can now apply Weierstrass approximation theorem. Since the above holds for all $n$ we have that

$$\int_{1}^e g(u) P(u)du = 0$$

for any polynomial $P(u)$. Now pick the polynomial to approximate $g$ to within an $\epsilon$ on $[1,e]$ and then take $\epsilon\to 0$ to obtain

$$\int_{1}^e g^2(u)du = 0$$

and it follows that $\frac{f(\log(u))}{u} = 0 \implies f\equiv 0$.