2
$\begingroup$

Let $f\in H^1(\mathbb{R})$ (Sobolev space). Then we have $\lim_{\lvert x\rvert\to\infty}f(x)=0$. I have to prove that this implies that $\cos(f(x))$ is not Lebesgue-integrable, i.e. $$ \int_{\mathbb{R}}\lvert \cos(f(x)\rvert\, d\lambda=\infty, $$ where $\lambda$ is the Lebesgue-measure.

I do not know exactly how to prove that.

We have $\lim_{\lvert x\rvert\to\infty}\cos(f(x))=1$ by assumption.

My first vague idea was to choose $\lvert a\rvert$ large enough and to write the integral as

$$ \int_R\lvert\cos(f(x))\rvert\, d\lambda=\lim_{z\to\infty}\int_{-a-z}^{-a}\lvert\cos(f(x))\rvert\, d\lambda+\int_{-a}^a\lvert\cos(f(x))\rvert\, d\lambda+\lim_{z\to\infty}\int_a^{a+z}\lvert\cos(f(x))\rvert\, d\lambda $$

Now, can't we compute the first and the last integral as $\approx z$ and then taking the limit $z\to\infty$?

$\endgroup$
2
$\begingroup$

Let us prove that if $\lim \limits _{|x| \to \infty} g(x) = 1$ then $g \notin L^1 (\Bbb R)$ (any other non-zero number could be used instead of $1$).

If $\lim \limits _{|x| \to \infty} g(x) = 1$, then there exist $R>0$ such that for $|x| > R$ we have $|g(x) - 1| < \frac 1 2$ (this comes from the $\varepsilon-\delta$ definition of the concept of limit, with $\varepsilon = \frac 1 2$ and $\delta = R$). This means that for $|x| > R$ we have $-\frac 1 2 < g(x) - 1 < \frac 1 2$, i.e. $\frac 1 2 < g(x) < \frac 3 2$.

Then

$$\int \limits _{\Bbb R} |g(x)| \ \Bbb d x = \int \limits _{|x| \le R} |g(x)| \ \Bbb d x + \int \limits _{|x| > R} |g(x)| \ \Bbb d x \ge \int \limits _{|x| \le R} |g(x)| \ \Bbb d x + \int \limits _{|x| > R} \frac 1 2 \ \Bbb d x = \text{finite} + \infty = \infty ,$$

which shows that $g \notin L^1 (\Bbb R)$. Notice that the only thing used here is that $g$ is locally-integrable (because I have integrated it on $[-R,R]$) - which is clearly true for functions in $H^1 (\Bbb R)$. This means that your statement remains true even for functions in $L^1_{loc} (\Bbb R)$.

Now, take $g = \cos \circ f$ and you are done.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.