Show That F is Not Differentiable at 0 0 by Showing That F is Not Continuous at 0 0

mathmari

Hey!! :giggle:

Let's consider the function \begin{align*}f:\mathbb{R}^2&\rightarrow \mathbb{R} \\ (x,y)&\mapsto \sqrt{|x|\cdot |y|}\end{align*}
Show that $f$ is partially differentiable in $(0,0)$ but not total differentiable.

I have done the following:
We prove that $f$ is partially differentiable in each direction:
Let $0 \neq v =(r, s) \in \mathbb{R}^2$ be the direction and $0 \neq h \in \mathbb{R}$ :
\begin{align*}\frac{1}{h}\cdot \left (f\left ((0,0)+hv\right )-f(0,0)\right )&=\frac{1}{h}\cdot \left (f(hr,hs)-0\right )=\frac{1}{h}\cdot f(hr,hs)=\frac{1}{h}\cdot \sqrt{|hr|\cdot |hs|}=\frac{1}{h}\cdot \sqrt{|h|^2\cdot |r|\cdot |s|}\\ &=\frac{1}{h}\cdot |h|\cdot \sqrt{|r|\cdot |s|} =\begin{cases}\frac{1}{h}\cdot h\cdot \sqrt{|r|\cdot |s|} & \text{ if } h\geq 0\\ \frac{1}{h}\cdot \left (-h\right )\cdot \sqrt{|r|\cdot |s|} & \text{ if } h< 0\end{cases}\\ &=\begin{cases}\sqrt{|r|\cdot |s|} & \text{ if } h\geq 0\\ - \sqrt{|r|\cdot |s|} & \text{ if } h< 0\end{cases}\end{align*}
Since this term is independent from $h$, the limit for $h \rightarrow 0$ is equal to $\pm \sqrt{|r|\cdot |s|}$,which means that the limit exists.
That means that $f$ is partially differentiable in every direction, since for each $v$ there is a directional derivative.

Is that correct? Or do we show that $f$ partially differentiable by calculating the partial derivatives $f_x$ and $f_y$ ? :unsure:

To show that that $f$ is not totally differentiable do we show that that $f$ is not continuous in $(0,0)$ ? :unsure:

HOI

So a function of two or more variables is "partially differentiable" at a point if its partial derivatives exist there?
(I don't believe that is stanard terminology. I would just say "has partial derivatives".)

There is a theorem that says that a function, f, of two or more variables is "differentiable" at a point if and only if its partial derivative exist and are continuous there (just "f is continuous" is not sufficient).

Here, $f(x,y)= \sqrt{|x||y|}= (|x||y|)^{1/2}$.

The partial deriatives, at (0,0), are particularly easy. To find the partial derivative with respect to x, at (0,0) we take the derivative with respect to x while holding y= 0. Then $f(x, 0)= (|x|(0))^{1/2}= 0$ for all x. That is constant so the partial derivative with respect to x is 0. The same is true of the partial derivative with respec to y.

To find the total derivative at (0,0) we would have to have a neighborhood of (0,0) in which we could calculate $\frac{f(x, y)- f(0, 0)}{\sqrt{x^2+ y^2}}$. But any such neighborood woud include points in the second and fourth quadrants where xy is negative and the function value does not even exist!

I like Serena

Partial differentiable just means that the partial derivatives exist. That is different from differentiable in a specific direction, which is again different from (totally) differentiable.

Using your argument, we can substitute r=1 and s=0 to see that $f_x$ exists and is 0 at (0,0).

Using your argument, we can also pick a direction where both r and s are non-zero and see that we get a different limit depending on the sign of h. Therefore there is a direction in which f is not differentiable, which implies that f is not totally differentiable either.

Btw, isn't f continuous?
Then we can't use that as an argument.

mathmari

So to show that $f$ is partially differentiable we just have to calculate the partial derivatives?

For that do we do the following?

Let $h \in \mathbb{R}\setminus \{0\}$ and $j \in \{1, 2\}$. Then we get \begin{equation*}\frac{f\left ((0,0)+he_j\right )-f(0,0)}{h}=\frac{f\left (he_j\right )-0}{h}=\frac{f\left (he_j\right )}{h}\end{equation*}
For $j=1$ :
\begin{equation*}\frac{f\left (he_1\right )}{h}=\frac{f\left (h(1,0)\right )}{h}=\frac{f(h,0)}{h}=0\rightarrow 0\end{equation*}
For $j=2$ :
\begin{equation*}\frac{f\left (he_2\right )}{h}=\frac{f\left (h(0,1)\right )}{h}=\frac{f(0,h)}{h}=0\rightarrow 0\end{equation*}
Therefore $f$ is in $(0, 0)$ partially differentiable with \begin{equation*}\frac{\partial{f}}{\partial{x}}(0,0)=\frac{\partial{f}}{\partial{y}}(0,0)=0\end{equation*}

Is that correct? Or have I understand that wrongly?

:unsure:

I like Serena

Therefore $f$ is in $(0, 0)$ partially differentiable with \begin{equation*}\frac{\partial{f}}{\partial{x}}(0,0)=\frac{\partial{f}}{\partial{y}}(0,0)=0\end{equation*}

Correct. (Nod)

mathmari

So to show that $f$ partially differentiable in $(0,0)$ do we need to do only what I have done in post #4 ? Or do we have to do something further? Do we need the part I wrote in post #1 ? :unsure:

I like Serena

Yes. Post 4 covers partial differentiability.
That leaves total differentiability for which we can use post 1 to show f is not differentiable..

mathmari

That leaves total differentiability for which we can use post 1 to show f is not differentiable..

For that we have to pick a specific direction $(r,s)$ with $r,s\neq 0$, or not?
For example $r=s=1$ then we have that
\begin{align*}\frac{1}{h}\cdot \left (f\left ((0,0)+h(1,1)\right )-f(0,0)\right )&=\frac{1}{h}\cdot \left (f(h,h)-0\right )=\frac{1}{h}\cdot f(h,h)=\frac{1}{h}\cdot \sqrt{|h|^2}\\ &=\frac{1}{h}\cdot |h| =\begin{cases}\frac{1}{h}\cdot h & \text{ if } h\geq 0\\ \frac{1}{h}\cdot \left (-h\right )& \text{ if } h< 0\end{cases}\\ &=\begin{cases}1 & \text{ if } h\geq 0\\ - 1 & \text{ if } h< 0\end{cases}\end{align*}
Since for $h\rightarrow 0^+$ we get $1$ and for $h\rightarrow 0^-$ we get $-1$, this means that we don't have continuity there and so $f$ cannot be differentiable.

Is that correct? :unsure:

I like Serena

It's not about continuity.
Instead we have that the limit does not exist to find a directional derivative.
And if a directional derivative does not exist, then that implies that the function is not differentiable. (Nerd)

mathmari

It's not about continuity.
Instead we have that the limit does not exist to find a directional derivative.
And if a directional derivative does not exist, then that implies that the function is not differentiable. (Nerd)

Ahh I see!! But do we do that as I did it in my previous post with $(1,1)$ ? :unsure:

I like Serena

Ahh I see!! But do we do that as I did it in my previous post with $(1,1)$ ?

For instance yes. (Nod)

mathmari

Great!! Thanks a lot!! (Sun)

Show That F is Not Differentiable at 0 0 by Showing That F is Not Continuous at 0 0

F is partially differentiable in (0,0) but not total differentiable

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