measure theory/inner product/ Fourier question: Suppose f$\in$ $L^1([0,1])$. Show the $\lim_{n\to \infty}$ (f, $\chi_n$) = 0

Question

measure theory question: Suppose f $\in L^1([0,1])$. Show the $\lim_{n\to \infty}$ (f, $ \chi_n$) = 0, where

(f, $ \chi_n)$= $\int_{[0,1]}f\space \bar{\chi_n} $

Note that $\bar{\chi_n}=e^{-nx}$

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 L1-norm I get 0, and take the absolute value to remove complex coefficient. But after going over it again and again, I don't seem to get the result I want.

Answer 1

Observe that $$ \lim_{n \to \infty} e^{-nx} = 0. $$

Set $f_n(x) := f(x) e^{-nx}$ so that for all $ x \in [0,1]$ satisfies $$ |f_n(x)| \leq |f(x)|. $$

Since $f \in L^1$ we may now apply the Lebesgue Dominated Convergence theorem -

$$ \lim_{n \to \infty} \int_0^1 f(x) e^{-nx} dx = \int_0^1 f(x) \lim_{n \to \infty} e^{-nx} dx = 0 .$$