Finding limit of $p$-norm of succession of functions Let $\,f,g \in L^{p} ( \mathbb R^{n} ,\mathcal{L}, m)$, $\mathcal{L}$ being the $\sigma$-algebra of Lebesgue-measurable sets and $m$ being the Lebesgue measure on $\mathbb R^{n}$. Now, for $1\le p < \infty$ find the value of $\lim\limits_{|x|_{n} \to \infty} \|f(x+x_{n}) + g(x-x_{n})\|_{p}$.
My thoughts:
Since the space of all simple measurable functions with zero space of finite measure is dense in $L^{p}(\mathbb R^{n},\mathcal{L},m)$, if we take two such functions $f, g$ then as $|x_{n}| \to \infty$ $f(x+x_{n}) $ and $g(x-x_{n})$ will have disjoint compact support, and by the translation invariance of Lebesgue Measure in $\mathbb R^{n}$ the limit value should be $\|f\|_{p} + \|g\|_{p}$.
Now, please check my idea and rectify if necessary. But what I need mostly, because I think my thought out solution is correct, is a formal proof! Beacuse I am confused about how to write this proof! Thank you!
 A: $\newcommand{\xint}{\int\limits}
\newcommand{\dd}{\mathrm{d}}$
As the comments state, if $f,g$ are two Lebesgue-$L^p$ functions, surely their supports $A,B\subseteq\mathbb{R}^n$ are of finite measure. If we take $f_n(x):= f(x+x_n)$ for some $x_n\in\mathbb{R}^n$, we can conclude the support $A_n$ of $f_n$ is $A-x_n$. Indeed take a point $x\in A$. That means $f(x)\neq0$. If we take $x-x_n$, then $f_n(x-x_n)=f(x-x_n+x_n)=f(x)\neq0$, so $x-x_n\in A_n$. Therefore $A-x_n\subseteq A_n$. Viceversa, let $x\in A_n$. Then $f_n(x)\neq0$ But $f_n(x)=f(x+x_n)$, so $x+x_n\in A\implies x\in A-x_n$. Thus, $A_n\subseteq A-x_n$, which combined with the other inclusion gives an equality. A similar argument shows that letting $g_n(x):=g(x-x_n)$ we obtain that the support $B_n$ of $g_n$ is $B+x_n$. Summing up, we have:
$$\left\lbrace\begin{array}{@{}l@{}}
A_n=A-x_n \\
B_n=B+x_n
\end{array}\right. \quad\quad\ast$$
for all $n\in\mathbb{N}$. Let $x\in A_n,\,y\in B_n$ for some $n\in\mathbb{N}$. Then $x=a-x_n$ for $a\in A$ and $y=b+x_n$ for $b\in B$, by $\ast$. So their distance is $|x-y|=|(a-x_n)-(b+x_n)|=|a-b-2x_n|$. Now as said in the comments from the triangular equality $|x|\leq|x-y|+|y|$, and carrying $|y|$ to the left and applying an argument of symmetry yields $|\|x\|-\|y\||\leq\|x-y\|$, where I had to use $\|\|$ to avoid confusing the norm with the absolute value. If we apply this above we get:
$$d(x,y)=|x-y|=|a-b-2x_n|\geq2|x_n|-d(A,B),$$
where $d(A,B)$ is the infimum of $d(a,b)$. So fixing any $(a,b)\in A\times B$, the shifted points are further apart than $|x_n|-d(A,B)$, which tends to infinity for $n\to\infty$. This means that for all $(a,b)\in A\times B$ there exists an $n\in\mathbb{N}$ such that $d(a-x_n,b+x_n)>0$ strictly. Since $|x_n|\uparrow+\infty$, we can find an $n$ as above described such that the strict positivity holds for all $m\geq n$. Note that the inequality above holds for all couples, meaning the supremum of such $n$ is necessarily finite. This implies that, for $n$ sufficiently big, $A_n\cap B_n=\varnothing$, where by "sufficiently big" I clearly mean "greater than or equal to that supremum". Since we are taking the limit, we are not interested in what happens before the supremum. So we can, without any loss of generality, assume that supremum is 1. This is merely to write what follows without repeating "for $n$ etc" every time. Let's see the norm. Inside it we have an integral. Obviously, integrating over $\mathbb{R}^n$ is the same as integrating over the support of the integrand. But that support is $A_n\cup B_n$, which is a disjoint union, and the integral is additive over the integration domain. So we can write:
$$\|f(x+x_n)+g(x-x_n)\|_p=\sqrt[p]{\xint_{A_n\cup B_n}|f(x+x_n)+g(x-x_n)|^p\mathrm{d}m}=$$ $$=\sqrt[p]{\xint_{A_n}|f(x+x_n)+g(x-x_n)|^p\mathrm{d}m+\xint_{B_n}|f(x+x_n)+g(x-x_n)|^p\mathrm{d}m}.$$
Now the disjointness of the two supports also implies the first integral reduces to that of $|f(x+x_n)|^p$, since $g(x-x_n)$ is zero, and the second one analogously reduces to that of $|g(x-x_n)|^p$. By translation invariance of $m$, we get that those two integrals are the $p$-th powers of the $p$-norms:
$$\|f(x+x_n)+g(x-x_n)\|_p=\sqrt[p]{\|f\|_p^p+\|g\|_p^p}.$$
Now that very rarely reduces to the sum of the $p$-norms, except for the trivial cases when one of the norms (or both) is 0 or $p=1$. So your arguments were correct, but you conclusion wasn't careful enough. Of course, the limit value you gave is greater than or equal to this one: just take the $p$-th powers and see the extra terms on the side of your limit, which are all positive since they are positive-coefficient products of norms, which are non-negative. Note that this is a limit, since the disjointness and all that follows only holds for $n$ above the supremum above mentioned.
Edit: I wasn't too explicit on this, but the fact that the $p$-norms of the shifted functions are all equal to those of the unshifted ones is due to the translation invariance of the Lebesgue measure, which grants the equality for the integrals, as can be seen by the definition with the repartition function. You can easily add details to this, I take it.
Edit 2: In fact, I just realized there are $L^p$ functions with infinite support, meaning it is not compact, as you assumed, but most relevantly that the supremum above is in fact $+\infty$. If you opportunely connect $\frac{1}{x^2+y^2}$ and $x^2+y^2$, for example, you can easily see by switching to polar coordinates that the resulting function is $L^p$ for any $p$. However, it is never zero. The above limit holds only if the supremum is finite, because otherwise the supports of the shifted functions are never disjoint. I guess, however, it can still be a good approximation of the actual limit, since for very big $n$'s the intersection of the supports is a place where both shifted functions are really close to zero. That is required by the $L^p$ condition, so in general the limit wil be a little greater than the $p$-th root of the sum of the $p$-th powers of the $p$ norms. When it is not equal, you can give a range of values for the limit, since it is less than the sum of the $p$ norms as I showed in a comment somewhere.
