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June 9

Differentiable multivariate function with discontinuous partial derivatives

As far as I know, having continuous partial derivatives is strictly a stronger condition than differentiability. Does anyone have a simple example of a function of more than one real variable that is differentiable, but whose partial derivatives are discontinuous somewhere?--Jasper Deng (talk) 10:07, 9 June 2016 (UTC)

Not at all sure of myself on this, but I start from the second (univariate) example in Smoothness#Examples:

The function

$f(x)=\{\begin{array}{cc}{x}^2\sin (\frac{1}{x})& \text{if }x\ne 0,\\ 0& \text{if }x=0\end{array}$

is differentiable, with derivative

$f^{\prime }(x)=\{\begin{array}{cc}-\cos (\frac{1}{x})+2x\sin (\frac{1}{x})& \text{if }x\ne 0,\\ 0& \text{if }x=0.\end{array}$

Because cos(1/x) oscillates as x → 0, f ’(x) is not continuous at zero. Therefore, this function is differentiable but not of class C¹.

If you multiply the function f by y to get g(x, y) and treat y times the f ′(x) expression as the partial of g with respect to x, does that give you what you're looking for? Here the partial of g wrt y equals the above expression for f (x).Loraof (talk) 14:27, 9 June 2016 (UTC)

The question is if that resulting function is differentiable, which is stronger than the existence of the partial derivatives.--Jasper Deng (talk) 19:48, 9 June 2016 (UTC)

No, the resulting function, the partial derivative, is not. As the quote from the article says, f ′(x) (hence partial g partial x) is not continuous at zero, so it is not differentiable there. Loraof (talk) 20:38, 9 June 2016 (UTC)

But like I said, that is not strictly necessary for differentiability of the multivariable function (in general; to be clear I'm talking about g(x, y) = yf(x))... so the example as-is still may or may not be differentiable.--Jasper Deng (talk) 20:42, 9 June 2016 (UTC)

f′, and thus g_x, is not even defined at x=0, since as the quote says cos(1/x) oscillates as x→0. If it's not defined there, it can't be differentiable there. Loraof (talk) 21:52, 9 June 2016 (UTC)

The quote says that f' is 0 at x = 0. It has no limit as it approaches 0, but is certainly defined there.--Jasper Deng (talk) 22:06, 9 June 2016 (UTC)

Right, sorry, it is defined at 0. But since it doesn't approach that value as x→0, it's not continuous at 0 (which is what your original question asked about). So it can't be differentiable at 0, since a derivative of g_x at 0 is defined as the limit of [g_x(0 + h) – g_x(0)] / h, as h goes to 0, and that limit does not exist because the numerator oscillates. Loraof (talk) 03:01, 10 June 2016 (UTC)

Huh? I sense you're talking about a different difference quotient. You surely mean $\underset{h\to 0}{\lim }\frac{g(0+h,y)-g(0,y)}{h}$, right? That certainly exists with value 0 by what the quote says. The original function $f(x)=x^2\sin (1/x)$ has limit 0 as x approaches zero (since it is bounded above in absolute value by $x^2$, which has limit 0), so defining it to be equal to 0 at 0 makes it continuous. The derivative $f^{\prime }(x)=\frac{\partial g}{\partial x}/y$ might not be continuous but it certainly is well defined there. Therefore, f is differentiable (everywhere) as a function of a single variable. But is g? It's not continuously differentiable because the partial derivative with respect to x isn't continuous. But it nonetheless could still be differentiable in the sense of having a total derivative.--Jasper Deng (talk) 03:16, 10 June 2016 (UTC)