Sunday, 14 August 2016

calculus - One-variable continuity of one partial derivative implies differentiability?



I am interested in a particular instacne of the phenomena




"Partial derivatives + (A certain degree of) continuity" implies differentiablilty. My case assumes less regularity than usual:



Let f:R2R,(x0,y0)R2.



Assume that the partial derivatives of f with respect to x and y exist at the point (x0,y0) and one of them exists and continuous w.r.t to the other variable (for instance the function yfx(x0,y) is continuous at the point y0).



Is it true that f is differentiable at (x0,y0)?



Note: It is known that if both partial derivatives exists, and one of them is continuous (as a functions of two variables) then f is differentiable. (For a proof see here).




However, the proof uses:



1) The existence of fx on some ball around (x0,y0)



(I assume only fx exists on {x0}×(y0ϵ,y0+ϵ)).



2) The continuity of (x,y)fx(x,y) at (0,0).



(I assume only continuity of fx(x0,y)).


Answer




This is false.



Counterexample, at the point (x0,y0)=(0,0):
f(x,y)={x3/y,y0,0,y=0.


This function satisfies the hypotheses:




  • fy(0,0)=0 (since f is constant along the y axis: f(0,y)=0 for all y),
    so in particular fy(0,0) exists.

  • fx(0,y)=0 for all y (since the restriction of f to a line y=c is either identically zero or a constant times x3, which has zero derivative at the origin), so the one-variable function yfx(0,y)=0 is continuous.



But f isn't even continuous at the origin, hence not differentiable there.



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