Let $f \in L^p(\Bbb{R},m)$, then prove $\lim_{x \to \infty} \inf \vert f(x) \vert=0$. Here, $m$ denotes Lebesgue measure, and $p \in [1,\infty)$.
So I thought of either using Chebychev's Inequality to get a bound or Fatou's lemma but the only sequence of functions I can think of which are not even necessarily nonnegative would be something like
$$f_n(x)=f(x)+\frac{1}{n}.$$
Then I thought to prove if $\lim_{x \to \infty} \inf \vert f(x) \vert>0$, then $f \not\in L^p(\Bbb{R},m)$. So if thats the case, there exists some $x_0 \in \Bbb{R}$ beyond which $f(x)>\epsilon>0.$ But I get stuck here, any tips or hints appreciated.
Am I safe to say, for any $\epsilon >0$, there exists some $x_0 \in \Bbb{R}$ such that $\vert f(x_0) \vert >\epsilon$? that is, if I go by contrapositive.
And sorry if my initial guesses of the two inequalities (Fatou, Chebychev) are way off, a man wonders a lot when thinking of a solution to a math proof lol. But if one of them works please let me know!!
Furthermore, give an example of an $L^p$ function which is absolutely continuous on bounded intervals but $\lim_{x \to \infty} \sup \vert f(x) \vert>0$.
Im REALLY stuck on the furthermore part by the way.
Suppose $a=\liminf_{x\rightarrow\infty}|f(x)|>0$. Then there is $x_0$ such that $|f(x)|>a/2$ for all $x> x_0$. Then $$\int|f|^p\,dm\geq \int_{\{x>x_0\}}|f|^p\,dm>\frac{a^p}{2^p}\int_{\{x>x_0\}}\,dm=\infty$$ contradicting the assumption that $f\in L_p(\mathbb{R},m)$.
As for an example of a funciton $f\in L_p$ that is absolutely continuous in bounded sets but $\limsup_{x\rightarrow\infty}|f(x)|>0$, suffices to consider $p=1$. Consider the function $$\phi(x)=(1-|x|)_+$$ and define $$f(x)=\sum_n\phi(n^2(x-n))$$ Observe that $\int_{\mathbb{R}}\phi(n^2(x-n))\,dx=\frac{1}{n^2}\int^1_{-1}\phi(x)\,dx=\frac1{n^2}$ and so $f\in L_1(\mathbb{R})$. Observe that $$\limsup_{x\rightarrow\infty}f(x)=1$$ I leave all details to the OP.