does existence of gradient for every point $(x,y)\in \Bbb{R}^2$ for some real function $f$ mean that function is continuous?
Asked
Active
Viewed 67 times
0
-
1It's clearly false, you just need an example – Peanut Jul 16 '20 at 17:27
-
Try with something like $f(x,y) = xy/(x^2+y^2)$ outside the origin, $f(0,0) = 0$. – Rigel Jul 16 '20 at 17:43
1 Answers
0
Imho, most simple is to consider $f(x,y)=|x|$ in $\mathbb{R}^3$.In any point $(0,y)$ we have continuity but not differentiability, as $f_x$ do not exist.
As more difficult case we can consider $f(x,y)=\sin\left(\frac{y^2}{x}\right)\sqrt{x^2+y^2}$ for $x \ne 0$ and $f(x,y)=0$ for $x=0$. It can be shown, that partial derivatives exists and function is continuous in $(0, 0)$, but we have not differentiability in $(0, 0)$.
zkutch
- 13,410
-
Actually we suppose that that $\nabla f(x, y) $ exists right, in your exemple $\nabla f(0, y) $doesn't exits – EDX Jul 16 '20 at 21:00
-