TITLE: Control over bump function's second derivative
QUESTION [19 upvotes]: The usual constructions of bump functions supported on an interval $[a,b]$ are based on the non-analytic function $f(x) = e^{\frac{1}{(x-a)(x-b)}}$. However, the $n$-th order derivatives of these functions are very high (in fact, they grow exponentially with $n$).
I wonder if one can control some of the derivatives (at the expense, for example, of the other ones). In my application, I'd like to construct the following function:

Question: Is it possible to construct a bump function $\phi: \mathbb{R}
 \rightarrow \mathbb{R}$ which satisfies:
$$ \phi(x)  = \begin{cases} x^2 \,\,\,, |x| \leq 1\\ 0\,\,\,, |x|\geq
 2\\ \end{cases} $$
and  $$ \phi''(x) \leq 2 \,\,\, \forall x \in [1,2] ? $$

To give some context, this function is useful in the so-called virial-like identities in Nonlinear Dispersive PDE's. If $u_0 \in H^1\left(\mathbb{R}^N\right)$ is such that $|\cdot |u_0 \in L^2\left(\mathbb{R}^N\right)$, the solution $u(x,t)$ of the equation:
$$
\begin{cases}
\partial_t u + \Delta u + |u|^{p-1}u = 0\\
u(x,0) = u_0(x)
\end{cases}
$$
Also satisfies $u(\cdot,t) \in H^1\left(\mathbb{R}^N\right)$ and $|\cdot|u(\cdot,t) \in L^2\left(\mathbb{R}^N\right)$ for all t in the existence time of the solution. Moreover, we have the inequality:
$$
\frac{d^2}{dt^2} \int |x|^2 |u(x,t)|^2 dx \leq C E[u_0]
$$
which allows us to prove blow-up in some cases.
However, if one drops the decay assumption, the term $|x|^2$ in the integral must be replaced for some compactly supported function. I'd like to use the function $\phi$ above, and the bound on the second derivative would be useful on controlling some error terms. The obvious try, that is multipliyng $|x|^2$ by a function constant (and equal to 1) in some neighborhood of the origin and constant (and equal to 0) away from the origin doesn't seem to work.

REPLY [2 votes]: I think unfortunately not. If the second derivative is to be always less than 2, then moving from $x=2$ to $x=1$, the function could at most describe the parabola $y=(x-2)^2$, in which case it would not be smooth at $x=1$. 
Put more sophisticatedly, the function $\phi$  must be greater than 1 on some interval $(1, 1+\epsilon)$. Then if its second derivative is to be smaller than 2 on the interval $(1+\epsilon, 2)$, then it satisfies $\phi(2)>0$, at which point it cannot be continuous at 2. 
Proof: we’ll show that if u is smooth at $-2$ and satisfies $u’’\leq 2$, then it cannot be smooth at $-1$. 
If $u$ is as above, then upon integrating $u’’$ we see that $u’ \leq 2(x+2)$ on [-2, -1]. Integrating again, we see that $u$ satisfies $u\leq (x+2)^2$ on the same interval. However, in order for $u$ to be smooth at 2, there must be an interval $[-1-\epsilon, -1]$ on which $u$ exceeds 1. This is a contradiction.