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.

  • 1
    $\begingroup$ Please note that I have not chosen the values of $\phi$ in the interval $(1,2)$. I only ask it to be smooth, and would like to control its second derivative $\endgroup$ – Luccas Campos Apr 19 '18 at 2:16
  • $\begingroup$ Do you want the second derivative bounded in absolute value by $2$? If so, I don't think its possible. If the second derivative were as small as it could possibly be, that is $-2$ throughout the interval (1,2), the graph ``wouldn't have time'' to make it back down to 0 (it would look like an upside down parabola, so that the graph on (-1, 2) would look sin-esque). $\endgroup$ – Van Latimer Apr 19 '18 at 18:25
  • $\begingroup$ No, I just need it bounded from above. I'd like to get rid of a term $\phi''-2$ that appears in the calculations, $\endgroup$ – Luccas Campos Apr 20 '18 at 0:31

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 it’s 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.

  • $\begingroup$ Could you please include a proof that $ \ \phi(2) >0 \ $? $\endgroup$ – Gustavo Apr 26 '18 at 18:15

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