If $f\in L^p(\mathbb{R})$ and $f'\in L^1(\mathbb{R})$, where $p\in[1,\infty]$, must it follow that $\lim_{x\to\infty}f(x) = 0$?
My proof thus far is:
Suppose, for contradiction, $\lim_{x\to\infty}f(x) = L > 0$. Then there exists $M>0$ such that $$ x > M \quad\Rightarrow\quad \left|f(x)\right| >\frac{L}{2} $$ Then \begin{align*} \int_{-\infty}^\infty \left|f(x)\right|^p\,dx &= \int_{-\infty}^M \left|f(x)\right|^p\,dx + \int_{M}^\infty \left|f(x)\right|^p\,dx \\ &> \int_{-\infty}^M \left|f(x)\right|^p\,dx + \left(\frac{L}{2}\right)^p(\infty-M) \\ &> \infty \end{align*} Since $f\in L^p(\mathbb{R})$, this leads to a contradiction.
Two things about this proof attempt worries me: (1) I did not use the condition that $f'\in L^1(\mathbb{R})$, and (2) I began with an (implicit) assumption that the limit exists.
Is the condition $f'\in L^1(\mathbb{R})$ necessary to establish this fact? Also, how can we rigorously prove the existence of this limit?