$C^\infty$ functions with all derivatives in $L^p$ Let $1\le p<\infty$ and $f\in C^{\infty}(\mathbb{R})$ be such that its all derivatives $f^{(n)} \in L^p(\mathbb{R})$ for every $n\in \mathbb{N} \cup \{0\}$. 
Is it true that $f\in L^\infty(\mathbb{R})$?
If it is true, do we have a generalization in $\mathbb{R}^n$? 
My partial answer for $p=1$:
For $p=1$ and  $t\in \mathbb{R}$, we have
$$
|g(t)|=
|\int_{0}^{t} g'(x) \ dx+g(0)|
\le
\int_{0}^{t}
|g'(x)| \ dx
+|g(0)|
\le 
\|g'\|_{L^1}
+|g(0)|.
$$ 
Thus, $g\in L^\infty(\mathbb{R})$ with 
$\|g\|_{L^\infty} \le \|g'\|_{L^1}+|g(0)|$.
Thanks for help and hint.  
 A: That $f'\in L^1(\mathbb R)$ implies already that $f \in L^\infty$. To see this, first we observe that 
$$\lim_{x\to\infty} f(x) = 0.$$
Indeed like you did, 
$$f(x) - f(y) = \int_x^y f'(s) ds\Rightarrow |f(x) - f(y)| \le \int_x^y |f'(s)|ds \to 0$$
as $x, y\to \infty$ (this is where we use $f' \in L^1$). Thus $\lim_{x\to \infty} f(x)$ exists and must be zero since $f\in L^1$. Then 
$$f(x) = \int_\infty^x f'(s)ds\Rightarrow |f(x)| \le \|f'\|_{L^1}.$$
When $p>1$, we consider $g = f^p$. Then $g' = pf^{p-1} f'$ and so by Holder's inequality
$$\begin{split} \int |g'| &= p \int |f^{p-1} f'|  \\
& \le p \left(\int |f^{p-1}|^{\frac{p}{p-1}}\right)^{1-1/p} \left(\int |f'|^p\right)^{1/p}\\
&= p \| f\|_{L^p}^{p-1} \| f'\|_{L^p}
\end{split}$$
Thus $g$ satisfies $g,g' \in L^1$ and so $g\in L^\infty$ by the first argument. Hence $f\in L^\infty$. 
The generalization in $\mathbb R^n$ is the general Sobolev embedding, which shows that your $f$ has pointwise bounds on all derivatives
$$ \sup_{x\in \mathbb R^n} |D_I f(x)| \le C(I,f) <\infty.$$
A: If $f\in C^1$ and $f' \in L^1,$ then
$$|f(x)-f(0)| = |\int_0^x f'| \le \|f'\|_1,$$
giving $\|f\|_\infty \le |f(0)| + \|f'\|_1.$
Suppose $p>1.$ If $f\in C^1,$ then $|f|^p$ is the composition of $|x|^p \in C^1$ with $f;$ hence $|f|^p\in C^1.$ Check that the formula $(|f|^p)'(x) = p|f|^{p-1}f'(x)\text { sgn }(f(x))$ holds. If we assume now that $f, f'\in L^p,$ then we know $|f|^{p-1}f' \in L^1$ as in John Ma's answer. By the first paragraph, we have $|f|^p,$ and hence $f,$ in $L^\infty.$
