Question about the differentiability of a function of more than one variable

[00Donut]

I've been thinking about this for a while... sorta.

If a function of two or more variables is differentiable at some point, does this imply that all its partial derivatives are continuous at that point?

 

[alexfloo]

It's probably not really correct to talk about differentiability of a function in that sense. Instead, we'd say that it is differentiable with respect to a particular variable. For instance, the function:

f(x, y) = y*sgn(x)

where "sgn" is the sign function (+1 for positive values, -1 for negative values, and sgn(0)=0). This function is differentiable at 0 with respect to y but not with respect to x. To say that it's just "differentiable" might be a useful shorthand for saying that it's "differentiable in both variables," but I've actually never heard that, and it's a bit ambiguous.

One other note: a function can be "differentiable," and not have a continuous derivative. A function that is "continuously differentiable" has a continuous derivative.

 

[TylerH]

It sounds like you may have the implication backwards. In general, differentiable implies continuous, not the other way around. For example, since f(x, y) = y*sgn(x) is differentiable with respect to y along the slice x=0, we can say that slice is continuous, or that f(y)=f(0,y)=y*0 is continuous.

 

[alexfloo]

I'm not sure who you're talking to, but I don't believe I or the original poster ever assumed that continuous implies differentiable. What are you referring to exactly?

In hindsight, however, I picked a crappy example, since when either argument is 0, the function is constant and identically 0.

f(x,y) = y + sgn(x)

illustrates the property I intended better. At the point (0,0), the function is differentiable with respect to y, but not with respect to x.

 

[TylerH]

I'm not sure who you're talking to, but I don't believe I or the original poster ever assumed that continuous implies differentiable. What are you referring to exactly?

If a function of two or more variables is differentiable at some point, does this imply that all its partial derivatives are/is continuous at that point?

It sounds like it could have been incorrectly generalized from the converse, differentiable -> continuous. Of course, I could be completely wrong.

 

[alexfloo]

Oh I see what you're saying, and I think it's a different mistake from that. It doesn't appear that he's going from f differentiable to f continuous, but from f differentiable to f' continuous, which is also incorrect, but slightly different.

 

[micromass]

It's probably not really correct to talk about differentiability of a function in that sense. Instead, we'd say that it is differentiable with respect to a particular variable. For instance, the function:

f(x, y) = y*sgn(x)

where "sgn" is the sign function (+1 for positive values, -1 for negative values, and sgn(0)=0). This function is differentiable at 0 with respect to y but not with respect to x. To say that it's just "differentiable" might be a useful shorthand for saying that it's "differentiable in both variables," but I've actually never heard that, and it's a bit ambiguous.

One other note: a function can be "differentiable," and not have a continuous derivative. A function that is "continuously differentiable" has a continuous derivative.

Saying that a function is differentiable is standard terminology really and has little to do with the partial derivatives. In general a function $f:\mathbb{R}^n\rightarrow \mathbb{R}^m$ is differentiable in a if there exists a linear function $L:\mathbb{R}^n\rightarrow \mathbb{R}^m$ such that

$$\lim_{h\rightarrow 0}{\frac{f(a+h)-f(h)-L(h)}{\|h\|}}=0$$

 

[00Donut]

Im confused now... According to my book, a function of two or more variables is differentiable at a point P if a linear function can approximate it "Well enough" at other points "Close to P". I believe there was also another one that involved delta z. (Where z=f(x,y)) If z is differentiable at some point, then $\Delta$z=f$_{x}$$\Delta$x+f$_{y}$$\Delta$y+$\epsilon$1$\Delta$x+$\epsilon$2$\Delta$y

Where both $\epsilon$1 and $\epsilon$2 approach zero as delta x and delta y approach zero.

If $\Delta$z can indeed be expressed in that way at some point (a,b), then the function is differentiable at the point.

Now, there is an easier way of finding out if a function of two or more variables is differentiable at some point. If all the partial derivatives of a function are continuos at some point, then f(x,y) is differentiable at that same point.

This is where my question comes in... does the differentiability of a function (The way I defined it above) imply that its partial derivatives are all continuos at that point? or does it merely imply that they exist?

Im sorry if my initial post was a bit unclear... also, I apologize if my equation seems messy, its my first shot at using latex

 

[TylerH]

Im confused now... According to my book, a function of two or more variables is differentiable at a point P if a linear function can approximate it "Well enough" at other points "Close to P". I believe there was also another one that involved delta z. (Where z=f(x,y)) If z is differentiable at some point, then $\Delta$z=f$_{x}$$\Delta$x+f$_{y}$$\Delta$y+$\epsilon$1$\Delta$x+$\epsilon$2$\Delta$y

Where both $\epsilon$1 and $\epsilon$2 approach zero as delta x and delta y approach zero.

If $\Delta$z can indeed be expressed in that way at some point (a,b), then the function is differentiable at the point.

Now, there is an easier way of finding out if a function of two or more variables is differentiable at some point. If all the partial derivatives of a function are continuos at some point, then f(x,y) is differentiable at that same point.

This is where my question comes in... does the differentiability of a function (The way I defined it above) imply that its partial derivatives are all continuos at that point? or does it merely imply that they exist?

Im sorry if my initial post was a bit unclear... also, I apologize if my equation seems messy, its my first shot at using latex

To determine if f(x,y) is differentiable at (x₀, y₀), you see if both the limits, $$\lim_{h \to 0} \frac{f(x_0+h,y_0)-f(x_0,y_0)}{h}$$ and $$\lim_{h \to 0} \frac{f(x_0,y_0+h)-f(x_0,y_0)}{h}$$
exist.

Hopefully, knowing the definition in 1D and 2D, you can see how it generalizes to higher dimensions, because I don't feel like trying to make a general formula. :)

 

[HallsofIvy]

No, that is not sufficient. That only shows that the two partial derivatives exist and, as has been said before in this thread, partial derivatives existing at a point is not sufficient to say that the function is differentiable there.

As an example, take f(x,y)= 0 is xy= 0, 1 otherwise.

For all h, f(h, 0)= 0 so f(h, 0)- f(0, 0)= 0 and f(0, h)- (0, 0)= h so the limits you give exist and are 0 but the function is not even continuous at (0, 0).

 

[micromass]

Indeed, only seeing that the partial derivatives exist is not enough. But if the partial derivatives exist AND they are continuous, then the function is differentiable. Maybe use that?

 

[HallsofIvy]

Yes, if a function has continuous partial derivatives in some neighborhood of a point (not just continuous at the point) then the function is differentiable at that point.

 

[TylerH]

Yes, if a function has continuous partial derivatives in some neighborhood of a point (not just continuous at the point) then the function is differentiable at that point.

Can a function be continuous at a point? Wouldn't the fact that $\lim_{x \to c}f(x)=f(c)$ imply that it is continuous in some neighborhood of c?

 

[micromass]

Can a function be continuous at a point? Wouldn't the fact that $\lim_{x \to c}f(x)=f(c)$ imply that it is continuous in some neighborhood of c?

Yes, a function can be continuous in one point only. The example are perhaps a little farfetched, but they exist anyways. If you're interested:

$$f:[0,1]\rightarrow \mathbb{R}:x\rightarrow \left\{\begin{array}{c} x~\text{if x is rational}\\ 0~\text{if x is irrational}\end{array}\right.$$

is continuous only in 0.

 

[TylerH]

Yes, a function can be continuous in one point only. The example are perhaps a little farfetched, but they exist anyways. If you're interested:

$$f:[0,1]\rightarrow \mathbb{R}:x\rightarrow \left\{\begin{array}{c} x~\text{if x is rational}\\ 0~\text{if x is irrational}\end{array}\right.$$

is continuous only in 0.

Does $\lim_{x \to 0} f(x)$ even exist? (Out of curiosity, I would like to see a proof of that.)

I'm not disagreeing that it does. I don't have the knowledge to know for sure either way.

 

[micromass]

Does $\lim_{x \to 0} f(x)$ even exist? (Out of curiosity, I would like to see a proof of that.)

I'm not disagreeing that it does. I don't have the knowledge to know for sure either way.

Yes, the limit is 0. The easiest way by seeing this is by using the squeeze theorem:

$$0\leq f(x)\leq x$$

so in 0, the limit of 0 and x is 0. So the limit of f(x) is 0.We can also do it by epsilon-delta. Indeed, if $|x|<\delta$, then $|f(x)|\leq |x|<\epsilon$. So by the epsilon-delta definition of limits, the limit is 0.

Analogously, every sequence $(x_n)_n$ which converges to 0 will have that $f(x_n)\rightarrow 0. this is very easily seen. So the limit indeed exists!$