I have a fairly elementary question regarding Lebesgue spaces that I can't seem to figure out.
As the title suggests, I wish to know what additional condition(s) do we need to impose on a function $f\in L^p(\mathbb{R})$ so that it is bounded everywhere (as oppose to almost everywhere)? Would continuity suffice?
Smooth counterexample: Let me do an example of a smooth function that is in $L^p(\mathbb{R})$, but not bounded. Pick your favourite smooth function $\varphi$ supported in $[0;1/2]$ and such that $\Vert \varphi \Vert_p=1$. Then we define $$ f(x) = \sum_{n\geq 1} n^{1/2}\varphi(n^{2p}(x-n)). $$ Then we have $$ \Vert f \Vert_p \leq \sum_{n\geq 1} n^{1/2}\Vert \varphi(n^{2p}(\cdot - n) \Vert_p = \sum_{n\geq 1} \frac{1}{n^{3/2}}. $$ Here we used a couple of substitutions $$ \Vert \varphi(n^{2p}(\cdot - n) \Vert_p =\left( \int_{\mathbb{R}} \vert \varphi(n^{2p}(x-n)) \vert^p dx \right)^{1/p} = \left( \int_\mathbb{R} \vert \varphi(n^{2p}y) dy \right)^{1/p} = \left(\int_{\mathbb{R}} n^{-2p} \vert \varphi(z) dz \right)^{1/p} = n^{-2} \Vert \varphi \Vert_p = n^{-2}. $$ Thus, $f$ is in $L^p(\mathbb{R})$. As the $\varphi(n^{2p}(x-n))$ have disjoint support and $\varphi$ is smooth, so is $f$. Furthermore, again as the $\varphi(n^{2p}(x-n))$ have disjoint support, we get that $\Vert f \Vert_{L^\infty([0;n+1))}= n^{1/2}\Vert \varphi \Vert_\infty$ and hence $\Vert f \Vert_\infty = \infty$.
Of course, this is just a complicated version of the example of copper.hat from the comments. I just wanted to show that continuity or differentiability is not quite enough.
Lipschitz continuity is enough: As I wrote in the comments, lipschitz continuity would be sufficient. As $f$ is lipschitz continous, is is enough to show that $\lim_{\vert x \vert \rightarrow \infty} f(x) = 0$.
Assume by contradiction that $\lim_{x\rightarrow \infty} f(x)=0$ does not hold. Then there exists $C >0$ and a sequence $(x_n)_{n\geq 1}$ such that $\lim_{n\rightarrow\infty} x_n =\infty$, such that $f(x_n)\geq 2C$ (I assumed wlog that those values are positive and that $\vert x_n - x_m \vert \geq 1$ for $n\neq m$). But our function is lipschitz continuous with, say, lipschitz constant $L>0$. Then we have for $y\in [x_n; x_n +a]$ with $a:=\min \{1/2, C/L \}$ $$ f(y)\geq f(x_n)- \vert f(y)-f(x_n)\vert \geq 2C - L \vert x_n - y\vert \geq C.$$ Thus, we get $$\Vert f \Vert_p^p \geq \sum_{n\geq 1} \int_{x_n}^{x_n+a} \vert f(y) \vert^p dy \geq \infty,$$ which contradicts our assumption that $f\in L^p(\mathbb{R})$.
Alternatively, you could assume that $f$ is $C^1(\mathbb{R})$ and impose some integrability for $f'$.
As PhoemueX pointed out in the comments below, uniform continuity works as well (would be essentially the same proof).