Limits in matrix norm if convergence of integration is guaranteed the answer of this question probably is very obvious, but I want to make sure this is correct. I have a function $F: \mathbb{R}\rightarrow \mathbb{R}^{n\times n}$ that is continuous and $\int\limits_{-\infty}^\infty \|F(s)\|ds=C<\infty$, where $\|.\|$ denotes any matrix norm. Then, for me, since $\|F(s)\|\ge 0$ for all $s$,   $\|F(s)\|$ must satisfy
$\mathop {\lim }\limits_{s\rightarrow\infty} \|F(s)\|=0$, and
$\mathop {\lim }\limits_{s\rightarrow \ \ -\infty} \|F(s)\|=0$
However, I'm not sure if this is true or not, or if there should be additional conditions on $F(s)$ such as uniform continuity so the previous statement is true.  Any insights will be appreciated. 
 A: There is a little caveat you should consider. Except for that, what you say is true, independently of the norm you put on $F$. More generally, every integrable function $g$ from the real line to itself (notice this: by considering the norm of $F$ you're just considering a function from $R$ to $R$) has most of its mass in a compact. 
In fact, you can prove that for all $\epsilon$ there exists a compact set $K$ such that 
\begin{equation}
\int_{K^c}g(s)ds<\epsilon
\end{equation}
(integrating on the complement gives a small number). 
This leads to the conclusion that you state, the limit as $s\to\infty$ has to be zero, ONLY IF said limit exists. (How to prove this? Assume the limit exists and it's not 0, and you'll get a contradiction). 
But it's possible to cook up some examples of functions which are perfectly integrable on the real line, but do not admit limit at the extreme values of the variable. And the continuity hypothesis won't save you: it is possible to build up a counterexample which is smooth (infinitely many times differentiable).
A: (Note that the norm is not relevant here, this is really a question about continuous non-negative integrable functions.)
It is not true. Let $t(x) = \max(0,1-|x|)$, and note that $t$ is continuous and $\int t = 1$. Then let $f(x) = \sum_{n \in \mathbb{Z}} t(n^2(x-n))$. It is easy to verify that $f$ is continuous and $\int f = \sum_{n \in \mathbb{Z}} {1 \over n^2}$, but $f(n) = 1$ for all $n \in \mathbb{Z}$.
If $f$ is uniformly continuous, the statement is true. To see this, suppose $f$ does not converge to zero as $x \to \infty$. Then there is some $\epsilon>0$ and a sequence of points $x_n$ such that $x_n \to \infty$ and $\|f(x_n)\| \ge \epsilon$. Since $f$ is uniformly continuous, there is some $\delta>0$ such that if $\|x-x_n\| < \delta$, then $\|f(x)\| \ge {1 \over 2} \epsilon$. Since
$\int_{B(x_n,\delta)} \|f\| \ge {1 \over 2} \epsilon \  m B(0,\delta)>0$, this contradicts $\|f\|$ being integrable.
