With $f\in C^0([0,1])$ and equal to $0$ for a finite number of values, what is $\lim_{n\to+\infty}|\int_0^1\exp(nx)f(x)\,dx| = ?$ With $f\in C^0([0,1])$ and equal to $0$ for a finite number of values, what is $\lim_{n\to+\infty}|\int_0^1\exp(nx)f(x)\,dx| =\,?$
To add : $f(0)=f(1)=0$
I did get the case $f\ge0$ :
Let $(\alpha_0,\dots,\alpha_n)\in[0,1]^{n+1}$ such that $f(\alpha_i)=0$ and $\alpha_0<\alpha_1<\dots<\alpha_n$.
Let $\varepsilon>0$ such that $\forall i, \alpha_i+\varepsilon<\alpha_{i+1}-\varepsilon$ and $\beta_i = \inf_{]\alpha_i+\varepsilon,-\varepsilon+\alpha_{i+1}[} f(x)$
$$|\int_0^1\exp(nx)f(x)\,dx| = \sum_{i=0}^{n-1} \int_{\alpha_i}^{\alpha_{i+1}}\exp(nx)f(x)\,dx\ge \sum_{i=0}^{n-1} \int_{\alpha_i+\varepsilon}^{\alpha_{i+1}-\varepsilon}\beta_i\exp(nx)\,dx + R_i$$
And $\forall i,  R_i \ge 0$ and $\int_{\alpha_i+\varepsilon}^{\alpha_{i+1}-\varepsilon}\beta_i\exp(nx)\,dx \to +\infty$. So $\lim_{n\to+\infty}|\int_0^1\exp(nx)f(x)\,dx| = +\infty$.
I don't know if there is a better way to explain for $f\ge0$. So If you have a better way, I am ready for it !
The case $\exists x_0\in [0,1],f(x)<0$, I didn't find find anything.
Thanks for the help !
 A: Fix, as you did, $(a_0 = 0, a_1, \dots, a_{n-1}, a_n = 1)$ such that $f(0) = f(1) = f(a_i) = 0$. I claim that if $f(x) > 0$ on $(a_{n-1}, 1)$, then
$$
\lim_{n \to \infty}\int_0 ^ \infty \exp(nx)f(x)dx \to \infty
$$
and if $f(x) < 0$ on $(a_{n-1}, a_n)$, then the limit goes to $-\infty$. I'll only do the first case; the other is very similar.
In particular, we have that there is some $(\alpha, \beta) \subset (a_{n-1}, 1)$ and $\epsilon_n > 0$ such that $f(x) > \epsilon_n$ in $(\alpha, \beta)$, and so $\left| \int_{a_{n-1}} ^ 1 \exp(nx)f(x)dx \right| > \epsilon_n \frac{\exp(n \beta) - \exp(n \alpha)}{n}$.
Similarly, we see that if we pick $\epsilon_{i}$ to be the maximum of $|f(x)|$ on $(a_{i-1}, a_i)$, then $\left| \int_{a_{i-1}} ^ {a_i} \exp(nx)f(x)dx \right| < \epsilon_i \frac{\exp(n a_i) - \exp(n a_{i-1})}{n}$. Then we see that
$$
\lim_{n \to \infty}\int_0 ^ \infty \exp(nx)f(x)dx = \lim_{n \to \infty} \sum_{i=1}^{n}\int_{a_{i-1}}^{a_i}\exp(nx)f(x)dx \\ > \lim_{n \to \infty} \epsilon_n\frac{\exp(n \beta) - \exp(n \alpha)}{n} - \sum_{i=1}^{n-1} \epsilon_i \frac{\exp(n a_i) - \exp(n a_{i-1})}{n}
$$
where the signs come from the assumed positivity on $(a_{n-1}, 1)$, and supposing that the function is negative on all other intervals. But note that if $x < y$, $\exp(nx) = o(\exp(ny))$; so this limit is greater than
$$
\lim_{n \to \infty} \epsilon_n\frac{\exp(n\beta) + o(\exp(n\beta))}{n} = \infty.
$$
Just swap signs everywhere for the other case.
