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

Measure theory:

If $$f\inL^1([0,1])$$, 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.

• maybe use Dominated convergence to get the limit into the integral for $e^{-nx}$ to converge to 0? – MathematicalMoose May 2 '20 at 16:14

## 1 Answer

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.

• yeah, I think I have to let fn(x)= f(x)e^(-nx) to have the limit equal 0 – learning_mathematician May 2 '20 at 20:27
• I'm not sure if that's correct, take for example $f=e^{x^{x}}$. Then $\lvert fn \rvert \nless e^{-x}$. Then you have to choose a $g(x)$ that dominates $\forall f$ and that could be difficult. – MathematicalMoose May 2 '20 at 21:18