An argument similar to the one outlined here shows that the sup-norm of $f''$ on $[0,1]$ can be bounded in terms of the sup-norms of $f$, $f'$, and $f'''$. Can the sup-norm of $f''$ be bounded in terms of sup-norms of $f$ and $f'''$ alone, without assuming a bound on $f'$? Are there counterexamples?


Setting $p,q,r = \infty$, $n=1$, $j=2$, $m=3$ in the Gagliardo-Nirenberg interpolation inequality yields $$\sup |f''| \le C \sup |f'''|^{2/3} \sup |f|^{1/3}.$$

You can establish this pretty easily starting from the easier case $$\|f'\| \le c \|f''\|^{1/2} \|f\|^{1/2},$$

where I'm using $\| \cdot \|$ for the sup norm. This is a quantification of the following fact: if the velocity is large at some time, then the only way to make the displacement stay small is to impose a large acceleration. This holds without any condition on the domain (other than one-dimensionality).

Applying this twice (to the functions $f'$ and $f$) yields $$\|f''\| \le c \|f'''\|^{1/2} \|f'\|^{1/2}\le c \|f'''\|^{1/2} (c \|f''\|^{1/2} \|f\|^{1/2})^{1/2}= c^{3/2} \|f''\|^{1/4} \|f'''\|^{1/2}\|f\|^{1/4}.$$

Now just divide both sides by $\| f''\|^{1/4}$ to get $$\|f''\|^{3/4} \le c^{3/2} \|f'''\|^{2/4} \|f\|^{1/4},$$

which is the G-N inequality I stated with $C = c^2$.

  • $\begingroup$ Anthony, thanks a lot! For users not intimately familiar with the G-N inequality could you possibly provide some motivation and/or explanation? This is a site for students, not professors. Also, is there hope of an explicit form for $C$? What does $C$ depend on? $\endgroup$ – Mikhail Katz Jul 9 '17 at 8:10
  • $\begingroup$ @MikhailKatz: I've added a derivation of this from a simpler inequality similar to the one in your link. $\endgroup$ – Anthony Carapetis Jul 9 '17 at 8:32

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