If $\int_{-1}^1 \frac{|f(t)|}{|t|^n} dt \leq C^n$ for constant $0Let $f$ be a continuous function on $[-1,1]$. Suppose $\exists C$, constant such that $0<C<1$, and  $\int_{-1}^1 \frac{|f(t)|}{|t|^n} dt \leq C^n$, $\forall n$. Then I want to show $f \equiv 0$.
Naively I guess from $\lim_{n \rightarrow \infty} C^n=0$, but how one can prove $f\equiv 0$?

Trial vai comment from @Martin R
Assume $|f(x)| > \epsilon>0$ on some interval $[a,b]$, then
\begin{align}
\int_{-1}^1 \frac{|f(t)|}{|t|^n}dt = 2 \int_{0}^1 \frac{f(t)}{t^n} dt \geq 2 \epsilon \int_{0}^1 t^{-n} dt = \frac{2\epsilon}{n-1}
\end{align}
Hence
\begin{align}
\frac{2\epsilon}{n-1} \leq \int_{-1}^1 \frac{|f(t)|}{|t|^n}dt \leq C^n
\end{align}
For all $n$. Taking $n\rightarrow \infty$ it becomes $0\leq \int_{-1}^1 \frac{|f(t)|}{|t|^n} dt \leq 0$.
But How $|f(t)|=0$?
 A: Assume that $f$ is not identically zero. Then $|f(x)| \ge \epsilon > 0$ on some interval $[a, b]$ of positive length. Without loss of generality we can assume that $0 < a < b < 1$. Then
$$ 
C^n \ge \int_{-1}^1 \frac{|f(t)|}{|t|^n} \, dt \ge \int_a^b \frac{\epsilon}{t^n} \, dt = \frac{\epsilon}{n-1} \left( \frac{1}{a^{n-1}}- \frac{1}{b^{n-1}}\right) = \frac{\epsilon}{(n-1)a^{n-1}}\left( 1 - \left( \frac ab \right)^{n-1}\right)
$$
for all $n \ge 2$, or
$$
 (n-1) C (aC)^{n-1} \ge \epsilon \left( 1 - \left( \frac ab \right)^{n-1}\right) \, .
$$
That is not possible, because the left-hand side converges to zero for $n \to \infty$, whereas the right-hand side converges to $\epsilon > 0$.
A: It is enough to consider the integral over $I=[0,1]$. If $f\neq0$, then, by continuity of $f$,  we may assume with out loss of generality that there is $0<t_0<1$ such that $f(t_0)\neq0$; furthermore, we may assume that for some $\alpha>0$ small enough,  $J=[t_0, a_0+\alpha)\subset [0,1]$,  and  $|f(t)|>\frac12|f(t_0)|>0$ for all $t\in J$.
Then
$$\frac12|f(t_0)|\frac{t_0^{1-n}-(t_0+\alpha)^{1-n}}{n-1}=\frac{1}{2(n-1)}|f(t_0)|t^{1-n}_0\Big(1-\big(\frac{t_0+\alpha}{t_0}\big)^{1-n}\Big)\leq \int^1_{0}\frac{|f(t)|}{t^n}dt\leq C^n$$
Hence
$$0<\frac{1}{2}|f(t_0)|t_0\leq (n-1)C^nt^n_0\frac{1}{\Big(1-\big(\frac{t_0+\alpha}{t_0}\big)^{1-n}\Big)}\xrightarrow{n\rightarrow\infty}0$$
contradiction!
Similar argument works $f$  over $[-1,0]$.
