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so I'm currently tussling with this particular problem. I have $$f(x)={e^{-|x|}-1 \over x} $$ $$f_\varepsilon(x)={e^{-|x|}-{\sin(\varepsilon x) \over \varepsilon x} \over x},\quad x \in \mathbb R, \ \varepsilon > 0$$ I need to determine whether they belong to $L^1(\mathbb R), L^2(\mathbb R)$. The question is, must I necessarily brute-force it via the $\int_\mathbb R|f|^pdx<\infty$ condition or is there a smarter way to go about this? Can I determine the answer by looking for something in the functions?

ViktorStein
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Silence
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  • The bilateral limit doesn't exist? Since $f(x)=-1$ for $x \rightarrow 0^+$ and $f(x)=1$ for $x \rightarrow 0^-$. And yes $\epsilon >0 $ sorry. – Silence Jan 11 '20 at 22:27
  • I'm thinking... could I possibly say that $f(x) \notin L^1(R)$ due to how ${1 \over |x|}$ behaves for $x \rightarrow \infty$? Since I can split the integral in two, the second part will always diverge so might as well leave it? – Silence Jan 11 '20 at 22:33
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    If you're only looking for a way to guess (if you need to justify, then the answer will depend on what tools you are allowed to use), then notice that in both functions the numerator is 0 at 0 so there is no (infinite) singularity at 0, so both functions are locally bounded. The problem lies at infinity; for this, as you note you can split the difference and the term with the exponential is always in any $L^p$ that you want. Therefore the whole difference will be in the same $L^p$s as the remaining term. Use the $p$ test for this. – Jose27 Jan 11 '20 at 23:47

1 Answers1

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Hints:

$1).\ $ If $0\leqslant t \leqslant |x|,\ \left|{e^{-|t|}-1 \over t}\right |={1-e^{-|t|} \over t}=-\sum^{\infty}_{k=1}\frac{(-1)^kt^k}{k!}$ and this series converges uniformly on $[1,x]$ so it may be integrated termwise:

$\displaystyle\int^x_0 {e^{-t}-1 \over t}dt=-\sum^{\infty}_{k=1}\frac{(-1)^kx^{k+1}}{(k+1)!}=-\sum^{\infty}_{k=2}\frac{(-1)^kx^{k}}{k!}=-e^{-x}+1+x$ so the integral diverges.

This shows that $\displaystyle\int^0_{-\infty} {e^{-|t|}-1 \over t}dt$ diverges as well.

Since $\displaystyle\int^1_{-1} {e^{-|t|}-1 \over t}dt$ converges, putting the pieces together, we get that the integral diverges.

$2).\ $ Now, $\displaystyle \int^{\infty}_1\left|{e^{-|t|}-1 \over x}\right|^2dx$ and $\displaystyle \int_{-\infty}^{-1}\left|{e^{-|t|}-1 \over x}\right|^2dx$ clearly converge so all you need consider is $\left|{e^{-|t|}-1 \over x}\right|^2$ on $[-1,1]$.

$3).\ $ Finally, split up $\mathbb R$ and analyze as above to consider the three integrals whose integrand is $f_{\epsilon.}$

Matematleta
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