Convergence of the $p$-norm of $f+f(x-h)$ I'm reading a book in harmonic analysis, and there is one step in a proof that the author thought should be trivial: 
Suppose $f\in L^p(\mathbb{R})$, $p\geq1$, then $\lim_{h\to \infty} \| f+f(x-h)\|_{p}=2^{\frac{1}{p}}\|f\|_{p}$.
I have never seen this before, neither can I show it's correct. Could someone give some help?
 A: As noted by @copper.hat the statement is wrong for $p=\infty$.
Proof for $p\in[1,\infty)$:
For convenience, I will drop the $p$ in $\|\,\cdot\,\|_p$ and denote $ g(\,\cdot\, - h)$ by $\tau_h g$.
Let $f_n = f\cdot 1_{[-n, n]}$. Notice that $f_n\to f$ in $p$-norm by dominated convergence theorem.
Let $\epsilon > 0$. Then, there exists some $n\in\mathbb N$ such that $\|f - f_n\| < \epsilon$. Now, for every $h > h_0 := 2n$ we have
$$ \| f_n - \tau_h f_n \| = 2^{1/p} \|f_n\|, $$
as their supports are disjoint.
Further, by the reversed triangle inequality and substitution, we have
$$ |\| f - \tau_h f \| - \|f_n - \tau_h f_n\|| \le \| f- \tau_h f - f_n + \tau_h f_n \| \le \| f - f_n\| + \underbrace{\|\tau_h f - \tau_h f_n \|}_{=\|f - f_n\|} \le 2\epsilon $$
In particular, we have
$$ \begin{align*}  
|\| f - \tau_h f \| - 2^{1/p} \| f\||
&\le \underbrace{|\| f - \tau_h f \| - \|f_n - \tau_h f_n\||}_{\le 2\epsilon} \\ 
&+ \underbrace{|\|f_n - \tau_h f_n\| - 2^{1/p} \|f_n\||}_{=0} \\
&+ \underbrace{|2^{1/p} \|f_n\| - 2^{1/p} \| f\||}_{\le 2^{1/p}\epsilon}.
\end{align*}$$
