measure theory f$\in$ $L^1([0,1])$ then $\lim_{n\to \infty}$ (f, $\chi_n$) = 0

Question

Measure theory:

If $f\in$$L^1([0,1])$, then $\lim_{n\to \infty}$ (f, $\chi_n$) = 0, where (f, $\chi_n)$= $\int_{[0,1]}f\space \bar{\chi_n} $
With $\bar{\chi_n}=e^{-nx}$

I am thinking that using simple functions to build a sequence that converges to $f$ in $L^1$[0,1], so that by subtracting it from f in the $L^1$-norm I get 0, and take the absolute value to remove complex coefficient. But after going over it again and again, I'm stuck and I don't seem to get the result I want.

Answer 1

I'm no sure if this is even correct, but my thoughts are the following:

Notice:

$\chi_n \rightarrow 0$ as $n \rightarrow \infty$

and $\forall n\in\mathbb{N}, \lvert\chi_n\rvert \leqslant e^{-x}$

Further

$\int_{[0,1]}e^{-x}d\lambda(x) <\infty$

Now I think you could use Dominated Convergence and take the limit into the integral.