Prove a statement regarding differentiable mv-function

[lep11]

Homework Statement

Let $f$: $G\subset\mathbb{R}^2\rightarrow\mathbb{R}$ be differentiable at $(x_0,y_0)\in{G}$ and $\lim_{(x, y) \to (x_0, y_0)} \frac{f(x,y) -a -b(x-x_0) -c(y-y_0)}{\sqrt{(x-x_0)^{2} + (y-y_0)^{2}}} = 0.$ The task is to prove that then $a=f(x_0,y_0),b=f_x(x_0,y_0)$ and $c=f_y(x_0,y_0)$.

Homework Equations

The Attempt at a Solution

[/B]
Since $f$ is differentiable at $(x_0,y_0)$, there exists a linear function $L$ such that

$$\lim_{(h_1, h_2) \to (0, 0)} \frac{f(x_0 + h_1, y_0 + h_2) - f(x_0, y_0) - L(h_1,
h_2)}{\sqrt{h_1^{2} + h_2^{2}}} = 0,$$where

$$
L(h_1, h_2) = \frac{\partial f}{\partial x}(x, y)\, h_1 + \frac{\partial f}{\partial y}(x, y)\, h_2.
$$Now we can substitute $L(h_1,h_2)$ in and get

$$\lim_{(h_1, h_2) \to (0, 0)} \frac{f(x_0 + h_1, y_0 + h_2) - f(x_0, y_0) - \frac{\partial f}{\partial x}(x, y)\, h_1 - \frac{\partial f}{\partial y}(x, y)h_2}{\sqrt{h_1^{2} + h_2^{2}}} = 0,$$
which is equivalent to $$ \lim_{(x, y) \to (x_0, y_0)} \frac{f(x,y) -f(x_0,y_0) -\frac{\partial f}{\partial x}(x, y)(x-x_0) -\frac{\partial f}{\partial y}(x, y)(y-y_0)}{\sqrt{(x-x_0)^{2} + (y-y_0)^{2}}} = 0,$$because $(x-x_0)\rightarrow{0}$ and $(y-y_0)\rightarrow{0}$ as $(x,y)\rightarrow(x_0,y_0)$.

Therefore it must (?) hold that $f(x,y) -a -b(x-x_0) -c(y-y_0)=f(x,y) -f(x_0,y_0) -\frac{\partial f}{\partial x}(x, y)(x-x_0) -\frac{\partial f}{\partial y}(x, y)(y-y_0)$

which implies (?) that $a=f(x_0,y_0),b=f_x(x_0,y_0)$ and $c=f_y(x_0,y_0)$. I am thinking that maybe my proof is wrong and based too much on intuition rather than formal mathematical reasoning. I am asking for comments and tips on how to possibly modify or rewrite it in case it's not correct. There may be other approaches to this problem as well. I will appreciate your answers.

 

[fresh_42]

The equation has to hold in an entire neighborhood of $(x_0,y_0)$. So you may vary $(x,y)$ and regard limits $\lim_{x \rightarrow x_0}\, , \, \lim_{y \rightarrow y_0}$.

 

[lep11]

The equation has to hold in an entire neighborhood of $(x_0,y_0)$. So you may vary $(x,y)$ and regard limits $\lim_{x \rightarrow x_0}\, , \, \lim_{y \rightarrow y_0}$.

Could you please elaborate? Is there something to correct?

 

[fresh_42]

Could you please elaborate? Is there something to correct?

No, it looks o.k. to me. If you take the equation after "Therefore it must (?) hold that" and after subtracting $f(x,y)$ first apply the limit $\lim_{x \rightarrow x_0}$, then $\lim_{y \rightarrow y_0}$ you get two conditions, for $b$ and then for $c$ and finally for $a$.

 

[Ray Vickson]

Homework Statement

Let $f$: $G\subset\mathbb{R}^2\rightarrow\mathbb{R}$ be differentiable at $(x_0,y_0)\in{G}$ and $\lim_{(x, y) \to (x_0, y_0)} \frac{f(x,y) -a -b(x-x_0) -c(y-y_0)}{\sqrt{(x-x_0)^{2} + (y-y_0)^{2}}} = 0.$ The task is to prove that then $a=f(x_0,y_0),b=f_x(x_0,y_0)$ and $c=f_y(x_0,y_0)$.

Homework Equations

The Attempt at a Solution

[/B]
Since $f$ is differentiable at $(x_0,y_0)$, there exists a linear function $L$ such that

$$\lim_{(h_1, h_2) \to (0, 0)} \frac{f(x_0 + h_1, y_0 + h_2) - f(x_0, y_0) - L(h_1,
h_2)}{\sqrt{h_1^{2} + h_2^{2}}} = 0,$$where

$$
L(h_1, h_2) = \frac{\partial f}{\partial x}(x, y)\, h_1 + \frac{\partial f}{\partial y}(x, y)\, h_2.
$$Now we can substitute $L(h_1,h_2)$ in and get

$$\lim_{(h_1, h_2) \to (0, 0)} \frac{f(x_0 + h_1, y_0 + h_2) - f(x_0, y_0) - \frac{\partial f}{\partial x}(x, y)\, h_1 - \frac{\partial f}{\partial y}(x, y)h_2}{\sqrt{h_1^{2} + h_2^{2}}} = 0,$$
which is equivalent to $$ \lim_{(x, y) \to (x_0, y_0)} \frac{f(x,y) -f(x_0,y_0) -\frac{\partial f}{\partial x}(x, y)(x-x_0) -\frac{\partial f}{\partial y}(x, y)(y-y_0)}{\sqrt{(x-x_0)^{2} + (y-y_0)^{2}}} = 0,$$because $(x-x_0)\rightarrow{0}$ and $(y-y_0)\rightarrow{0}$ as $(x,y)\rightarrow(x_0,y_0)$.

Therefore it must (?) hold that $f(x,y) -a -b(x-x_0) -c(y-y_0)=f(x,y) -f(x_0,y_0) -\frac{\partial f}{\partial x}(x, y)(x-x_0) -\frac{\partial f}{\partial y}(x, y)(y-y_0)$

which implies (?) that $a=f(x_0,y_0),b=f_x(x_0,y_0)$ and $c=f_y(x_0,y_0)$. I am thinking that maybe my proof is wrong and based too much on intuition rather than formal mathematical reasoning. I am asking for comments and tips on how to possibly modify or rewrite it in case it's not correct. There may be other approaches to this problem as well. I will appreciate your answers.

Your agrument is OK, but really assumes more than you need. Assuming ONLY that
$$\lim_{(x, y) \to (x_0, y_0)} \frac{f(x,y) -a -b(x-x_0) -c(y-y_0)}{\sqrt{(x-x_0)^{2} + (y-y_0)^{2}}} = 0$$
for some finite constants $a,b,c$ (but not assuming differentiability, or even continuity), you can prove that $f$ is continuous at $(x_0,y_0)$, that $a = f(x_0,y_0)$, that partial derivatives $f_x, f_y$ exist at $(x_0,y_0)$, and that they are, respectively, equal to $b, c$. The condition you assumed to start is essentially one of the definitions of differentiability at $(x_0,y_0)$, so the differentiability assumption is redundant, at least in certain approaches to the topic.

My preferred type of argument would be to write
$$f(x,y) = a + b (x-x_0) + c(y - y_0) + e(x,y),$$
where $e(x,y) = o(\|(x-x_0,y-y_0)\|)$, meaning that $e(x,y)/\|(x=x_0,y-y_0)\| \to 0$ as $(x,y) \to (x_0,y_0)$. So, taking $(x,y) \to (x_0,y_0)$ we get $f(x,y) \to a = f(x_0,y_0)$, hence $f$ is continuous at $(x_0,y_0)$. Then we have $[f(x_0+h,y_0)-f(x_0,y_0)]/h \to b$ as $x-x_0 = h \to 0$, so $f_x(x_0,y_0)$ exists and equals $b$, etc.

 

[lep11]

Your agrument is OK, but really assumes more than you need. Assuming ONLY that
$$\lim_{(x, y) \to (x_0, y_0)} \frac{f(x,y) -a -b(x-x_0) -c(y-y_0)}{\sqrt{(x-x_0)^{2} + (y-y_0)^{2}}} = 0$$
for some finite constants $a,b,c$ (but not assuming differentiability, or even continuity), you can prove that $f$ is continuous at $(x_0,y_0)$, that $a = f(x_0,y_0)$, that partial derivatives $f_x, f_y$ exist at $(x_0,y_0)$, and that they are, respectively, equal to $b, c$. The condition you assumed to start is essentially one of the definitions of differentiability at $(x_0,y_0)$, so the differentiability assumption is redundant, at least in certain approaches to the topic.

My preferred type of argument would be to write
$$f(x,y) = a + b (x-x_0) + c(y - y_0) + e(x,y)$$
where $e(x,y) = o(\|(x-x_0,y-y_0)\|)$, meaning that $e(x,y)/\|(x=x_0,y-y_0)\| \to 0$ as $(x,y) \to (x_0,y_0)$. So, taking $(x,y) \to (x_0,y_0)$ we get $f(x,y) \to a = f(x_0,y_0)$, hence $f$ is continuous at $(x_0,y_0)$. Then we have $[f(x_0+h,y_0)-f(x_0,y_0)]/h \to b$ as $x-x_0 = h \to 0$, so $f_x(x_0,y_0)$ exists and equals $b$, etc.

Why can you write $$f(x,y) = a + b (x-x_0) + c(y - y_0) + e(x,y)?$$ And I assume I'm supposed to use the fact that $f$ is differentiable since it's given in the problem statement. Thank you for the alternative approach,though. I will try that one also.

 

[Ray Vickson]

Why can you write $$f(x,y) = a + b (x-x_0) + c(y - y_0) + e(x,y)?$$ And I assume I'm supposed to use the fact that $f$ is differentiable since it's given in the problem statement. Thank you for the alternative approach,though. I will try that one also.

Because the numerator $f(x,y) -a - bx - cy$ is some function of $(x,y)$, and I can call it $e(x,y)$. The letter "e" is reminiscent of "error"---as in "error in the linear approximation".

I already said that what you did was OK, but the fact that somebody told you to use differentiablility was unnecessary. However, since they told you to do it, you have no real choice.