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I am asked to determine whether or not a function can have all of its partial derivatives exist at a point but not be continuous at that point. I have tried to construct a counterexample but am unsure whether or not I have succeeded.

Consider the following function:

$$ f(x,y) = \left\{ \begin{array}{cc} x & y=0 \\ 0 & \text{otherwise} \end{array}\right. $$

Let $(x_0,y_0) \not= 0$ then we can compute the partials of $f$ at this point. Doing so is not difficult we just have to make sure we are careful about the partials with respect to $y$.

$$ \begin{align*} f_x(x_0,y_0) &= \lim_{h\rightarrow 0} \frac{f(x_0+h,y_0)-f(x_0,y_0)}{h} \\ \text{if $y_0=0$} & \implies \lim_{h\rightarrow 0} \frac{x_0+h-x_0}{h} = 1 \\ \text{if $y_0\not=0$} &\implies \lim_{h\rightarrow 0} \frac{0}{h} = 0 \\ f_y(x_0,y_0) &= \lim_{h\rightarrow 0} \frac{f(x_0,y_0+h)-f(x_0,y_0)}{h} \\ \text{if $y_0=0$} &\implies \lim_{h\rightarrow 0}\frac{x_0-x_0}{h} = 0 \\ \text{if $y_0\not=0$} &\implies \lim_{h\rightarrow 0} \frac{0}h = 0 \end{align*} $$

In all of these cases the partials of $f$ exist, however it is clear that for $x_0\not=0$ we will have that $f$ is not continuous at $(x_0,0)$. Becuase along any path where $y\not=0$ we will have that the limit is $0$, but along the path $y=0$ we obtain the limit being $x_0$. So $f$ is not continuous, but its partials exist.

Is this a valid construction? It felt kind of fishy, any advice would be appreciated.

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  • $\begingroup$ $f$ may not be continuous even if partial derivatives exist. $\endgroup$
    – daulomb
    Dec 16, 2017 at 21:14
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    $\begingroup$ No, $f_x=0$ except on the $x$ axis, furthermore $f_y$ exists nowhere on the $x$ axis if $x\ne 0.$ $\endgroup$
    – zhw.
    Dec 16, 2017 at 22:05
  • $\begingroup$ What is stopping you from taking a function in $\mathbb{R}^1?$ Its partial derivative would then just be its regular derivative. $\endgroup$
    – Jbag1212
    Dec 21, 2017 at 5:29

4 Answers 4

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You may also consider this function: $$ f(x,y) = \left\{ \begin{array}{cc} \displaystyle\frac{xy}{x^2+y^2} & (x, y)\neq (0,0) \\ 0 & \text{otherwise.} \end{array}\right. $$ If you check, you see that this function is not continuous at $(0,0)$ but both partials exist and equal zero, that is, $f_x(0,0)=f_y(0,0)=0$.

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  • $\begingroup$ Sorry to bring revive this post, but what do you mean to say that the p.d. exist and equal zero at the origin? Clearly they are not defined there, so if you want to assign a value at the origin you must assign their limit (which doesn't exist in this case), or else you could suppose any other value. $\endgroup$
    – Bcpicao
    Apr 4 at 8:32
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On $\mathbb R^2,$ define $f=1$ except on the axes, where we define $f=0.$ Then $f_x(0,0)= f_y(0,0) = 0,$ but $f$ is not continuous at $(0,0).$

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Hint: this is the graph of $f(x,y)=\frac{(2x^2-3y^2)^2}{(x^2+y^2)^2}$ over $[-1,1]^2$:

enter image description here

To consider functions which are constant over lines through the origin is a very good idea.

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You might also consider the following (fishier) example. Let $f(x,y)=0$ except on points of the form $(x,x^2)$, with $x>0$, where $f$ takes the value $1$. Then the directional derivatives of $f$ along any ray exist and vanish everywhere, but the function is not continuous at every point of the form $(x,x^2)$ with $x\geq 0$.

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