# about a not continuous function and its derivatives

I'll start with a function $f: \mathbb{R}^2 \rightarrow \mathbb{R}$ whose partial derivatives exists at a point, but is not continuous at that point.

Let $f: \mathbb{R}^2 \rightarrow \mathbb{R}$ be defined by $f(x)=\begin{cases}1, &\mbox{if}& x=0 &\mbox{or if}& y=0\\ 0, &\mbox{otherwise}& \end{cases}$

because $f$ is constant on the $x$ and $y$ axes where it equals $1$, $$\frac{\partial f}{\partial x}(0,0)=0$$ $$\frac{\partial f}{\partial y}(0,0)=0$$ But $f$ is not continuous at $(0,0)$ because $\displaystyle \lim_{(x,y)\rightarrow (0,0)}f(x,y)$ does not exist.

Now my issue:

if $f:\mathbb{R} \rightarrow \mathbb{R}$ it is defined by $f(x)=\begin{cases}1, \mbox{if}& x=0 \\ 0, &\mbox{otherwise} \end{cases}$

this function is not continuous but my question is: is there a derivative for this function?- I say no. For the function $f :\mathbb{R}^2\rightarrow \mathbb{R}$ it is possible because $x=y=0$ not in same time? Am I wrong?

Thanks:)

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## 2 Answers

Differentiability (when all variables are allowed to vary simultaneously) $\implies$ continuity. However, partial differentiability $\not\implies$ continuity. A multivariate function $\mathbf{f}:\mathbb{R}^m\rightarrow\mathbb{R}^n$ is differentiable if there exists a linear function $\mathbf{J}:\mathbb{R}^m\rightarrow\mathbb{R}^n$ such that:

\begin{align} \lim_{\mathbf{h}\rightarrow \mathbf{0}} \frac{\mathbf{f}(\mathbf{x}+\mathbf{h}) - \mathbf{f}(\mathbf{x}) - \mathbf{J}(\mathbf{x})\mathbf{h}}{||\mathbf{h}||_2} &= 0 \end{align}

Note that the direction of approach $\mathbf{h}\rightarrow \mathbf{0}$ has not been specified, and the above limit should exist for all possible directions. In case $m = 1$, we have two directions in which $h\rightarrow 0$, corresponding to the left and right hand limits.

If a multi-variate function is differentiable, then the partial derivatives exist $\mathbf{J}$ is the Jacobian matrix. However, if the partial derivatives exist, the above limit may still not exist, and the function may not be differentiable. Your example illustrates this fact. However, in case of a uni-variate function, partial derivative and the derivative are the same, because you can talk about change in only one variable's direction.

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Can you tell me , please what means "total derivative"? Thanks. –  Iuli Aug 25 '12 at 17:21
I have clarified the answer a lot more now. The term "total derivative" may have another meaning, so I am simply calling it "derivative" now. –  Kartik Audhkhasi Aug 25 '12 at 17:34
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In $x=0$, the derivative of this last function does not exist. In two dimensions, the function is smooth along certain curves (the coordinate axes) passing through $0$, this is why certain directional derivates exist (those you wrote down). If you rotate the coordinate system by say, $45$ degree, then in the new coordinate system the ordinary partial derivative will not exist at the origin.

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