Consider $f\in \mathcal{C}^1(\mathbb{R})$, and suppose moreover $f\in L^p(\mathbb{R})$. Prove (or give a counterexample) that $f'\in L^p(\mathbb{R})$.
I bump into this question with a friend, and we have no idea if this is true or not. If we replace $\mathbb{R}$ with an open subset of $\mathbb{R}$ of finite Lebesgue measure then it should be immediate, since then we can bound the integral with the $\infty$-norm of the derivative times the length of the interval, obtaining the thesis (I think).
But the case of the real line is still misterious: my gut feeling is this is true, because the hypothesis' are quite strong; I also thought that if $f\in L^p(\mathbb{R})$ then the integral of the two tails of $f^p$ goes to $0$, so there exists a $K\subset \mathbb{R}$ compact in which $f^p$ has major impact in the integral; unfortunately I cannot say anything about the behaviour of the derivative, it may explode outside $K$.
I also think this is a pretty simple problem, but I'm missing the right way to see it. Any hint or answer would be much appreciate, thanks in advance.
What about something like $$ f(x) = \frac{\sin(x^2)} x $$ with $$ f'(x) = 2 \cos(x^2) - \frac{\sin(x^2)}{x^2}. $$ This function is in $L^p(\mathbb R)$ for all $p>1$, and $f'$ is bounded but not in $L^q(\mathbb R)$ for all $q<\infty$.