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.

  • 1
    $\begingroup$ maybe use Dominated convergence to get the limit into the integral for $e^{-nx}$ to converge to 0? $\endgroup$ – MathematicalMoose May 2 '20 at 16:14

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


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

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


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

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

  • $\begingroup$ yeah, I think I have to let fn(x)= f(x)e^(-nx) to have the limit equal 0 $\endgroup$ – learning_mathematician May 2 '20 at 20:27
  • $\begingroup$ 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. $\endgroup$ – MathematicalMoose May 2 '20 at 21:18

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