We have a function for $x_i, t_i >0$

$$|f(x_1,t_1)-f(x_0,t_0)| \leq C (|t_1-t_0|^{1/2} + |x_1-x_0|)$$

Why does this mean $f_t$ is locally bounded?

$f$ is non-increasing and convex in $x$. $f$ is non decreasing in $t$. Also $f$ is bounded below by 0 and above by a constant.

We have $$f_t(x,s) = \lim\limits_{u\rightarrow s} \dfrac{|f(x,u) - f(x,s)|}{u-s}$$

The bound only tells us

$$\lim\limits_{u\rightarrow s} \dfrac{|f(x,u) - f(x,s)|}{|u-s|}\leq C|u-s|^{-1/2}$$

Which can blow up.

  • $\begingroup$ Is that $x_2$ up there meant to be an $x_0$? $\endgroup$ – Patrick Da Silva Aug 14 '13 at 16:55
  • 1
    $\begingroup$ @PatrickDaSilva yes. thanks $\endgroup$ – Lost1 Aug 14 '13 at 17:00
  • $\begingroup$ Why are you sure $f_t$ is locally bounded? I am not convinced either. $\endgroup$ – Patrick Da Silva Aug 14 '13 at 17:03
  • $\begingroup$ @PatrickDaSilva because a paper says so. I will try to see if I missed any conditions. $\endgroup$ – Lost1 Aug 14 '13 at 17:03
  • $\begingroup$ Papers can seem wrong if you read them wrong, or sometimes papers can be wrong. You should figure out if something weird is happening in there, I'm trying to think of a counter example ; I don't believe your conditions are sufficient. $\endgroup$ – Patrick Da Silva Aug 14 '13 at 17:05

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