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Assume that the function $f(t,x) \in C^{3,4}(\Omega)$, i.e. three times continuously differentiable as a function of $t$ and four times continuously differentiable as a function of $x$. Why does the partial derivative $f_{xxt}$ exist?


Background:

I am reading a proof which uses

  1. the Taylor expansion of $f(t,x)$ w.r.t. $t$ up to second partial derivative $f_{tt}$.
  2. the Taylor expansion of $f(t,x)$ w.r.t. $x$ up to the third partial derivative $f_{xxx}$.
  3. the Taylor expansion of $f_{xx}(t,x)$ w.r.t. $t$ up to the first partial derivative $f_{xxt}$.

So how does one actually make use of of the fact that $f \in C^{3,4}(\Omega)$?

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  • $\begingroup$ If $f\epsilon C^{3,4}$ then $f\epsilon C^{3,2}$? $\endgroup$ – herb steinberg Aug 7 '18 at 15:42
  • $\begingroup$ Being four times continuously differentiable w.r.t. $x$ necessarily implies being two times continuously differentiable w.r.t. $x$. $\endgroup$ – Holden Aug 7 '18 at 15:47
  • $\begingroup$ My comment was in reference to the need for 4 times diff. I know that 4 times diff. implies 2 times diff. I was just commenting on your original question. I also wonder why 4 times is a condition. That was the only answer I could think of and that was the reason for ?. $\endgroup$ – herb steinberg Aug 7 '18 at 16:00
  • $\begingroup$ Still, why does $f_{xxt}$ exist? $\endgroup$ – Holden Aug 7 '18 at 16:03
  • $\begingroup$ That the function is partially differentiable in coordinate directions does not ensure that the function is differentiable, there exist standard counter examples. The same holds for the higher derivatives, there is no reason that the mixed derivatives exist under the stated assumptions. $\endgroup$ – LutzL Aug 7 '18 at 16:17

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