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We all know that if the partial derivatives exist, then f is not necessarily differentiable. My question is: what if f has got EVERY partial derivative of nth order, for each n in N? This appears to me as a very strong condition; can we argue the analiticity, or the smoothness, of f if all its derivatives of whatever order exist?

(It that is true, when I can find a proof? If it is false, can I have a counterexample of a function having all its partial derivatives,and not being smooth?

Rosario
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    If $f$ has continuous 1st order partial derivatives then $f$ is differentiable. So the existence of partial derivatives of any order implies that $f$ is indeed smooth. This does not imply analyticity as some smooth functions are not analytic: https://en.wikipedia.org/wiki/Non-analytic_smooth_function – Nicolas Jan 27 '22 at 20:12
  • How can i find a proof of this? I have by hypotesis the existence of all partial derivstives, not the continuity of them – Rosario Jan 27 '22 at 21:05
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    How about $f(x,y)=\frac{xy}{x^2+y^2}$ for $(x,y) \neq (0,0)$ and $f(0,0)=0$? It is smooth on $\mathbb{R}^2 \backslash {0}$, and its restriction to the two coordinate axes is zero so every partial derivative at $(0,0)$ should exist and vanish, no? – Aphelli Jan 27 '22 at 21:55
  • There are real two variable functions with mixed partial derivatives of all orders which are discontinuous. See https://math.stackexchange.com/questions/456581/is-there-a-discontinuous-function-on-the-plane-having-partial-derivatives-of-all – subrosar Aug 19 '23 at 19:47

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