Do smooth and $L^1$ functions vanish at infinity?

Question

Sometimes when proving some estimates on a smooth solution of a PDE over the whole space domain, one integrates over space terms like $\partial_x u$, and those terms vanish for some reason. Since we are working on the whole space and no boundary conditions are prescribed, my guess was that the reason was that $u \in L^1 \cap C^1$, hence $\lim_{|x| \to +\infty} u(x) = 0$ and then $\int_{-\infty}^{+\infty} \partial_x u \, dx = \left[u(x)\right]_{-\infty}^{+\infty} = 0$ basically.

However, I failed to prove such a statement and don't even know if it is true. I know that there are $L^1$ functions that do not vanish at infinity, and that aren't even bounded, but I couldn't think of any function that would be smooth as well, let's say $C^1$. I thought that if $u$ is $C^1$ then you can't have peaks as stiff as you want, so you can't imagine peaks which mass would tend to zero ... but I failed to make this rigorous.

Does anybody have a clue on that?

Answer 1

Not true. Here is a hint: Take a sequence of intervals $(a_n,b_n)$ separated by a fixed positive distance moving off to $\infty$. Consider the function $c_n(x-a_n)^{2}((x-b_n)^{2}$. This is a $C^{1}$ function on the interval with the function and the derivative both vanishing at the end points. So Let $f$ have this value on $(a_n,b_n)$ for each $n$ and $0$ outside these intervals . If $\sum (b_n-a_n) <\infty$ and $c_n=\frac 1 {(b_n-a_n)^{4}}$ you get a $C^{1}$ function which is integrable but does not have limit $0$ at $\infty$.

Answer 2

This is not addressing your question, but rather the underlying problem, but let me mention it.

In most places, one chooses to work with $u \in C^1_c$, i.e., a $C^1$-function with compact support (which is trivially in $L^1$). Then integration by parts gets very easy - one can restrict the integration to some large ball $B_R(0)$, and since both $u$ and its derivatives vanish on the boundary of $B_R(0)$, the boundary term vanishes as well.

Sometimes more derivatives are needed, so one can even work with $u \in C_c^\infty$. Of course, one always needs to check that this restriction is justified - this usually amount to showing that $C_c^\infty$-functions are a dense subset of your function space.