Is Differentiability at the Origin Determined by Partial Derivatives?

[mathmari]

Hey!

Let $g:\mathbb{R}\rightarrow \mathbb{R}$ be arbitrary and $f:\mathbb{R}^2\rightarrow \mathbb{R}$ be defined by $f(x,y)=yg(x)$.
I want to prove that $f$ is differentiable in the origin if and only if $g$ is continuous in $x=0$.

So that $f$ is differentiable in $(0,0)$ does the following has to hold?

• $f_x(0,0)$ exists
• $f_y(0,0)$ exists
• $\displaystyle{\lim_{(x,y)\rightarrow (0,0)}\frac{f(x,y)-f(0,0)-f_x(0,0)(x-0)-f_y(0,0)(y-0)}{\|(x,y)-(0,0)\|}=0}$

Or do we have to check in an other way the differentiability? (Wondering)

 

[I like Serena]

Hey mathmari! (Wave)

Yes, that is the definition that we have to verify. (Nod)

 

[mathmari]

Yes, that is the defintion that we have to verify. (Nod)

So, we have the following:

• The partial derivative is $f_x(x,y)=yg'(x)$. But $g$ is an arbitrary function. Can we just take its derivative? Or do we have to take this as a condition?
  
  Then $f_x(0,0)$ is equal to $0\cdot g'(0)$. Here it has to hold that $g'(0)$ exists, or not? (Wondering)
  
• The partial derivative is $f_y(x,y)=g(x)$. At the point $(0,0)$ the partial derivative is equal to $f_y(0,0)=g(0)$. So $g(0)$ has to exist.
  
• We have the following limit
  \begin{align*}\lim_{(x,y)\rightarrow (0,0)}&\frac{f(x,y)-f(0,0)-f_x(0,0)(x-0)-f_y(0,0)(y-0)}{\|(x,y)-(0,0)\|}\\ & =\lim_{(x,y)\rightarrow (0,0)}\frac{y\cdot g(x)-0\cdot g(0)-0\cdot g'(0)\cdot x-g(0)\cdot y}{\|(x,y)\|}\\ & =\lim_{(x,y)\rightarrow (0,0)}\frac{y\cdot \left [g(x)-g(0)\right ]}{\sqrt{x^2+y^2}}\end{align*} right? How could we continue? (Wondering)

 

[I like Serena]

So, we have the following:

• The partial derivative is $f_x(x,y)=yg'(x)$. But $g$ is an arbitrary function. Can we just take its derivative? Or do we have to take this as a condition?
  
  Then $f_x(0,0)$ is equal to $0\cdot g'(0)$. Here it has to hold that $g'(0)$ exists, or not? (Wondering)

We can't make assumptions about $g$ can we?
As yet we don't even know if $g$ is continuous in $0$, which is what we want to verify.
Instead we should deduce the value of $f_x(0,0)$ from the limit.

Suppose we take the path given by $(x(t),y(t))=(t,0)$.
Then the limit becomes:
$$\lim_{t\to 0} \frac{0\cdot g(t) - f_x(0,0)\cdot t - f_y(0,0)\cdot 0}{\sqrt{t^2+0^2}} = \lim_{t\to 0} \frac{-f_x(0,0)\cdot t}{|t|} = 0$$
Therefore we must have that $f_x(0,0)=0$ don't we? (Wondering)

• The partial derivative is $f_y(x,y)=g(x)$. At the point $(0,0)$ the partial derivative is equal to $f_y(0,0)=g(0)$. So $g(0)$ has to exist.

It's already given that $g(0)$ exists, isn't it?
And yes, we must have $f_y(0,0)=g(0)$.

• We have the following limit
  \begin{align*}\lim_{(x,y)\rightarrow (0,0)}&\frac{f(x,y)-f(0,0)-f_x(0,0)(x-0)-f_y(0,0)(y-0)}{\|(x,y)-(0,0)\|}\\ & =\lim_{(x,y)\rightarrow (0,0)}\frac{y\cdot g(x)-0\cdot g(0)-0\cdot g'(0)\cdot x-g(0)\cdot y}{\|(x,y)\|}\\ & =\lim_{(x,y)\rightarrow (0,0)}\frac{y\cdot \left [g(x)-g(0)\right ]}{\sqrt{x^2+y^2}}\end{align*} right? How could we continue? (Wondering)

We can't use $g'(0)$.
However, if instead we substitute the $f_x(0,0)=0$ that we found before, we'll find the same result.

Suppose we pick the path given by $(x(t),y(t))=(t,t)$, what will we get? (Wondering)

 

[mathmari]

We can't make assumptions about $g$ can we?
As yet we don't even know if $g$ is continuous in $0$, which is what we want to verify.
Instead we should deduce the value of $f_x(0,0)$ from the limit.

Suppose we take the path given by $(x(t),y(t))=(t,0)$.

Do we consider that path because then the term $f_y(0,0)$ would vanish at the limit? (Wondering)

Then the limit becomes:
$$\lim_{t\to 0} \frac{0\cdot g(t) - f_x(0,0)\cdot t - f_y(0,0)\cdot 0}{\sqrt{t^2+0^2}} = \lim_{t\to 0} \frac{-f_x(0,0)\cdot t}{|t|} = 0$$
Therefore we must have that $f_x(0,0)=0$ don't we? (Wondering)

The third condition of the definition of differentiabilty must hold, i.e. the limit must be equal to $0$ and so since $\displaystyle{\lim_{t\to 0} \frac{-f_x(0,0)\cdot t}{|t|}=\lim_{t\to 0} -f_x(0,0)\cdot \text{sgn}(t)}$ it follows that $f_x(0,0)$ must be equal to $0$.
Is this correct? (Wondering)

It's already given that $g(0)$ exists, isn't it?

How do we know that this exists? I got stuck right now. (Wondering)

Suppose we pick the path given by $(x(t),y(t))=(t,t)$, what will we get? (Wondering)

We have the following:
\begin{align*}\lim_{(x,y)\rightarrow (0,0)}&\frac{f(x,y)-f(0,0)-f_x(0,0)(x-0)-f_y(0,0)(y-0)}{\|(x,y)-(0,0)\|}\\ & =\lim_{(x,y)\rightarrow (0,0)}\frac{y\cdot g(x)-0\cdot g(0)-0\cdot x-g(0)\cdot y}{\|(x,y)\|}\\ & =\lim_{(x,y)\rightarrow (0,0)}\frac{y\cdot g(x)-g(0)\cdot y}{\sqrt{x^2+y^2}}\\ & =\lim_{t\rightarrow 0}\frac{t\cdot g(t)-g(0)\cdot t}{\sqrt{t^2+t^2}} \\ & = \lim_{t\rightarrow 0}\frac{t\cdot g(t)-g(0)\cdot t}{\sqrt{2t^2}} \\ & = \lim_{t\rightarrow 0}\frac{t\cdot [g(t)-g(0)]}{\sqrt{2}\cdot |t|}\\ & = \frac{\text{sgn}(t)}{\sqrt{2}}\cdot \lim_{t\rightarrow 0}(g(t)-g(0))\end{align*} Since the limit must be equal to $0$ it must holt that $\displaystyle{\lim_{t\rightarrow 0}g(t)=g(0)}$, which means that $g$ is continuous at $0$.

Is everything correct? (Wondering)

 

[I like Serena]

Do we consider that path because then the term $f_y(0,0)$ would vanish at the limit?

Yep. (Nod)

The third condition of the definition of differentiabilty must hold, i.e. the limit must be equal to $0$ and so since $\displaystyle{\lim_{t\to 0} \frac{-f_x(0,0)\cdot t}{|t|}=\lim_{t\to 0} -f_x(0,0)\cdot \text{sgn}(t)}$ it follows that $f_x(0,0)$ must be equal to $0$.
Is this correct?

Indeed.

How do we know that this exists? I got stuck right now.

It's given that $g$ is a function $\mathbb R \to \mathbb R$.
That means that $g$ is defined for every real number isn't it? (Wondering)

We have the following:
\begin{align*}\lim_{(x,y)\rightarrow (0,0)}&\frac{f(x,y)-f(0,0)-f_x(0,0)(x-0)-f_y(0,0)(y-0)}{\|(x,y)-(0,0)\|}\\ & =\lim_{(x,y)\rightarrow (0,0)}\frac{y\cdot g(x)-0\cdot g(0)-0\cdot x-g(0)\cdot y}{\|(x,y)\|}\\ & =\lim_{(x,y)\rightarrow (0,0)}\frac{y\cdot g(x)-g(0)\cdot y}{\sqrt{x^2+y^2}}\\ & =\lim_{t\rightarrow 0}\frac{t\cdot g(t)-g(0)\cdot t}{\sqrt{t^2+t^2}} \\ & = \lim_{t\rightarrow 0}\frac{t\cdot g(t)-g(0)\cdot t}{\sqrt{2t^2}} \\ & = \lim_{t\rightarrow 0}\frac{t\cdot [g(t)-g(0)]}{\sqrt{2}\cdot |t|}\\ & = \frac{\text{sgn}(t)}{\sqrt{2}}\cdot \lim_{t\rightarrow 0}(g(t)-g(0))\end{align*} Since the limit must be equal to $0$ it must holt that $\displaystyle{\lim_{t\rightarrow 0}g(t)=g(0)}$, which means that $g$ is continuous at $0$.

Is everything correct?

All correct (for evaluating the limit on the specified path). (Nod)

So we have proven that if $f$ is differentiable at $0$, that then $g$ must be continuous at $0$.
That leaves the converse.
How about switching to polar coordinates to prove the converse? (Wondering)

 

[mathmari]

It's given that $g$ is a function $\mathbb R \to \mathbb R$.
That means that $g$ is defined for every real number isn't it? (Wondering)

Ahh ok! (Nerd)

So we have proven that if $f$ is differentiable at $0$, that then $g$ must be continuous at $0$.
That leaves the converse.
How about switching to polar coordinates to prove the converse? (Wondering)

For the converse we suppose that $g$ is continuous at $x=0$ and we want to show that $f$ is differentiable at the origin, so we want to show that the three conditions of the first thread are satisfied, right?

Do you mean to use the polar coordinates at the third condition, at the limit?

(Wondering)

 

[I like Serena]

For the converse we suppose that $g$ is continuous at $x=0$ and we want to show that $f$ is differentiable at the origin, so we want to show that the three conditions of the first thread are satisfied, right?

Do you mean to use the polar coordinates at the third condition, at the limit?

Yes. (Thinking)

 

[mathmari]

Yes. (Thinking)

• How can we check if $f_x(0,0)$ exists?
  
• $f_y(0,0)$ exists
  
  We have that $f_y(x,y)=g(x)$. So $f_y(0,0)=g(0)$. Since $g(0)$ exists, it follows that $f_y(0,0)$ exists.
  
• \begin{equation*} \\ \end{equation*}
  \begin{align*}\lim_{(x,y)\rightarrow (0,0)}&\frac{f(x,y)-f(0,0)-f_x(0,0)(x-0)-f_y(0,0)(y-0)}{\|(x,y)-(0,0)\|}\\ & =\lim_{(x,y)\rightarrow (0,0)}\frac{yg(x)-0\cdot g(0)-f_x(0,0)\cdot x-g(0)\cdot y}{\|(x,y)\|} \\ & = \lim_{(x,y)\rightarrow (0,0)}\frac{yg(x)-f_x(0,0)\cdot x-g(0)\cdot y}{\sqrt{x^2+y^2}} \\ & = \lim_{(x,y)\rightarrow (0,0)}\frac{y[g(x)-g(0)]-f_x(0,0)\cdot x}{\sqrt{x^2+y^2}}\end{align*} Using polar coordinates we have $x=r\cos\theta$ and $y=r\sin\theta$. Then we consider the limit that $r$ goes to $0$, don't we? (Wondering)

 

[I like Serena]

• How can we check if $f_x(0,0)$ exists?

By treating $f_x(0,0)$ as an unknown in condition 3 and solving for it, as we did before. (Thinking)

• $f_y(0,0)$ exists
  
  We have that $f_y(x,y)=g(x)$. So $f_y(0,0)=g(0)$. Since $g(0)$ exists, it follows that $f_y(0,0)$ exists.

Indeed.
As a possible alternative, we could also solve $f_y(0,0)$ from condition 3.

• \begin{equation*} \\ \end{equation*}
  \begin{align*}\lim_{(x,y)\rightarrow (0,0)}&\frac{f(x,y)-f(0,0)-f_x(0,0)(x-0)-f_y(0,0)(y-0)}{\|(x,y)-(0,0)\|}\\ & =\lim_{(x,y)\rightarrow (0,0)}\frac{yg(x)-0\cdot g(0)-f_x(0,0)\cdot x-g(0)\cdot y}{\|(x,y)\|} \\ & = \lim_{(x,y)\rightarrow (0,0)}\frac{yg(x)-f_x(0,0)\cdot x-g(0)\cdot y}{\sqrt{x^2+y^2}} \\ & = \lim_{(x,y)\rightarrow (0,0)}\frac{y[g(x)-g(0)]-f_x(0,0)\cdot x}{\sqrt{x^2+y^2}}\end{align*} Using polar coordinates we have $x=r\cos\theta$ and $y=r\sin\theta$. Then we consider the limit that $r$ goes to $0$, don't we?

Yes. (Thinking)

 

[mathmari]

By treating $f_x(0,0)$ as an unknown in condition 3 and solving for it, as we did before. (Thinking)

But before why assumed that the condition 3 holds, i.e. that the limit is equal to 0. What do we have to do here? (Wondering)

Indeed.
As a possible alternative, we could also solve $f_y(0,0)$ from condition 3.

Ah ok!

 

[I like Serena]

But before why assumed that the condition 3 holds, i.e. that the limit is equal to 0. What do we have to do here?

Condition 3 is the definition of the derivative for multi-variable functions.
$f_x(0,0)$ and $f_y(0,0)$ are defined as the solutions of condition 3 - if we can find them.
If we can find them (they exist), it means that the function is differentiable.
And if we cannot find them, it means that the function is not differentiable.

Note that if we evaluate the limit following e.g. the path $(x,t)=(0,t)$, the limit in condition 3 collapses to the usual definition of the partial derivative. (Thinking)

 

[mathmari]

I haven't really understood that. (Thinking)

$f_x(0,0)$ and $f_y(0,0)$ are defined as the solutions of condition 3 - if we can find them.

Does does this mean? (Wondering)

Note that if we evaluate the limit following e.g. the path $(x,t)=(0,t)$, the limit in condition 3 collapses to the usual definition of the partial derivative. (Thinking)

Why? (Wondering)

 

[I like Serena]

We already know that whatever g is, the dervative at (0,0) must be (0, g(0)), don't we?
That is, if f is differentiable at (0,0).

What is left is to verify that the limit of condition 3 exists, isn't it? (Wondering)
And we can use that g is continuous at 0.

 

[mathmari]

We already know that whatever g is, the dervative at (0,0) must be (0, g(0)), don't we?
That is, if f is differentiable at (0,0).

What is left is to verify that the limit of condition 3 exists, isn't it? (Wondering)
And we can use that g is continuous at 0.

\begin{align*}\lim_{(x,y)\rightarrow (0,0)}&\frac{f(x,y)-f(0,0)-f_x(0,0)(x-0)-f_y(0,0)(y-0)}{\|(x,y)-(0,0)\|}\\ & =\lim_{(x,y)\rightarrow (0,0)}\frac{y\cdot g(x)-0\cdot g(0)-f_x(0,0)\cdot x-g(0)\cdot y}{\|(x,y)\|} \\ & = \lim_{(x,y)\rightarrow (0,0)}\frac{y\cdot [g(x)-g(0)]-f_x(0,0)\cdot x}{\sqrt{x^2+y^2}} \\ & = \lim_{(x,y)\rightarrow (0,0)}\frac{y\cdot [g(x)-g(0)]}{\sqrt{x^2+y^2}}-\lim_{(x,y)\rightarrow (0,0)}\frac{f_x(0,0)\cdot x}{\sqrt{x^2+y^2}}\end{align*} The first limit is zero because $g$ is continuous at $0$, or can't we just say that?
What can we do for the second limit? I got stuck right now.

(Wondering)

 

[I like Serena]

For the first part, no, we can't just say that it is zero.
We have to prove it for any path.
To do so, I suggest to switch to polar coordinates and see what happens. (Thinking)

For the second part, we know that the complete expressiion must be 0.
So if the first part is 0, then the second part must also be 0.
What does it take for it to be 0? (Wondering)

 

[mathmari]

For the first part, no, we can't just say that it is zero.
We have to prove it.
To do so, I suggest to switch to polar coordinates and see what happens. (Thinking)

Using polar coordinates $x=r\cos\theta, y=r\sin\theta$ we get: \begin{align*}\lim_{(x,y)\rightarrow (0,0)}\frac{y\cdot [g(x)-g(0)]}{\sqrt{x^2+y^2}}&=\lim_{r\rightarrow 0}\frac{r\sin\theta\cdot [g(r\cos\theta)-g(0)]}{r}=\lim_{r\rightarrow 0}\left (\sin\theta\cdot [g(r\cos\theta)-g(0)]\right )\\ &=\sin\theta\cdot\lim_{r\rightarrow 0} [g(r\cos\theta)-g(0)]\end{align*} If $r\rightarrow 0$ then $w:r\cos\theta\rightarrow 0$. So $\lim_{r\rightarrow 0} [g(r\cos\theta)-g(0)]=\lim_{w\rightarrow 0} [g(w)-g(0)]=0$, since $g$ is continuous at $0$.

Is everything correct?

For the second part, we know that the complete expressiion must be 0.
So if the first part is 0, then the second part must also be 0.
What does it take for it to be 0? (Wondering)

We have that \begin{equation*}\lim_{(x,y)\rightarrow (0,0)}\frac{f_x(0,0)\cdot x}{\sqrt{x^2+y^2}}=f_x(0,0)\cdot \lim_{(x,y)\rightarrow (0,0)}\frac{x}{\sqrt{x^2+y^2}}\end{equation*} Using again polar coordinates we get \begin{equation*}f_x(0,0)\cdot \lim_{(x,y)\rightarrow (0,0)}\frac{x}{\sqrt{x^2+y^2}}=f_x(0,0)\cdot \lim_{r\rightarrow 0}\frac{r\cos\theta}{r}=f_x(0,0)\cdot \lim_{r\rightarrow 0}\cos\theta=f_x(0,0)\cdot\cos\theta\end{equation*} right? Does this mean that $f_x(0,0)$ must be equal to $0$ ? (Wondering)

 

[I like Serena]

All correct.
And yes, $f_x(0,0)$ must be equal to $0$, meaning it exists, and condition 3 holds. (Nod)

 

[mathmari]

All correct.
And yes, $f_x(0,0)$ must be equal to $0$, meaning it exists, and condition 3 holds. (Nod)

So, at this direction we do the following:
We suppose that $g$ is continuous at $x=0$.
Then we have $f_y(x,y)=g(x)$. At $(0,0)$ we get $f_y(0,0)=g(0)$ which exists.

The third condition holds, i.e. the limit is equal to $0$ if and only if $f_x(0,0)$ is equal to $0$ (so it exists). That means either all condition hold, or just the second one.
Have I understood that correctly? (Wondering)

 

[I like Serena]

So, at this direction we do the following:
We suppose that $g$ is continuous at $x=0$.
Then we have $f_y(x,y)=g(x)$. At $(0,0)$ we get $f_y(0,0)=g(0)$ which exists.

The third condition holds, i.e. the limit is equal to $0$ if and only if $f_x(0,0)$ is equal to $0$ (so it exists). That means either all condition hold, or just the second one.
Have I understood that correctly? (Wondering)

Yes. (Nod)
And for differentiability it suffices if we can find a $f_x(0,0)$ and $f_y(0,0)$ such that all 3 conditions hold.
The 3rd condition effectively determines $f_x(0,0)$ and $f_y(0,0)$.

There is a theorem that says that $f_x(0,0)$ and $f_y(0,0)$ will be equal to the partial derivative functions at (0,0) if the partial derivative functions are well-defined on a neighborhood. This is what we used for $f_y(0,0)$, but what we couldn't use for $f_x(0,0)$. (Nerd)

 

[mathmari]

Yes. (Nod)
And for differentiability it suffices if we can find a $f_x(0,0)$ and $f_y(0,0)$ such that all 3 conditions hold.
The 3rd condition effectively determines $f_x(0,0)$ and $f_y(0,0)$.

There is a theorem that says that $f_x(0,0)$ and $f_y(0,0)$ will be equal to the partial derivative functions at (0,0) if the partial derivative functions are well-defined on a neighborhood. This is what we used for $f_y(0,0)$, but what we couldn't use for $f_x(0,0)$. (Nerd)

Ah ok! Thanks a lot! (Yes)