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:R2→R,(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 y↦∂f∂x(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 ∂f∂x on some ball around (x0,y0)
(I assume only ∂f∂x exists on {x0}×(y0−ϵ,y0+ϵ)).
2) The continuity of (x,y)↦∂f∂x(x,y) at (0,0).
(I assume only continuity of ∂f∂x(x0,y)).
Answer
This is false.
Counterexample, at the point (x0,y0)=(0,0):
f(x,y)={x3/y,y≠0,0,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 y↦fx(0,y)=0 is continuous.
But f isn't even continuous at the origin, hence not differentiable there.
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