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I struggle with a proof in multivariable calculus:

Let $f:\mathbb{R}^2 \rightarrow \mathbb{R}$ be a continuous function that is differentiable at $\mathbb{R}^2\backslash \{a\}$ with $a\in\mathbb{R}^2$. The limit $\lambda:=\lim_{x\rightarrow a}\nabla f(x)$ exists. Show that $f$ is differentiable at the point $a$, also show that $\nabla f(a)=\lambda$.

I think this statement can be proven by using the Mean Value Theorem for multiple variables, but I don't know how to do it. I would appreciate your help.

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  • $\begingroup$ Have you tried to show it in one dimension? $\endgroup$
    – md2perpe
    Jun 15, 2017 at 20:06

1 Answer 1

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Note that \begin{eqnarray} f(a+h)-f(a) - \lambda h &=& f((a_1+h_1,a_2+h_2))-f((a_1,a_2+h_2)) \\ & & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +f((a_1,a_2+h_2)) - f(a) - \lambda h \\ &=& {\partial f(a_1+\xi_1,a_2+h_2) \over \partial x_1}h_1 + {\partial f(a_1,a_2+\xi_2) \over \partial x_2}h_2 - \lambda h \end{eqnarray} for some $\xi_1 \in (a_1,a_1+h_1), \xi_2 \in (a_2,a_2+h_2)$.

Since ${\partial f(x) \over \partial x} \to \lambda$, for any $\epsilon>0$ we can find some $\delta>0$ such that $\|{\partial f(a+h) \over \partial x} - \lambda\| <\epsilon$ for all $\|h\| < \delta$. Then the above shows that $\|f(a+h)-f(a) - \lambda h \| \le \epsilon \|h\|$. Hence $f$ is differentiable at $a$ with derivative $\lambda$.

Alternative:

Note that ${\partial f(x) \over \partial x}$ is bounded in a punctured neighbourhood of $a$, and so is bounded on any segment of the form $(a,x)$ (that is, the line segment not including end points) as long as $x$ is in this punctured neighbourhood.

Hence for any sufficiently small $h$, the function $\phi(t)=f(a+th)$ is absolutely continuous and so $\phi(1) = \phi(0) + \int_0^1 \phi'(t) dt$.

Hence $f(a+h) - f(a) - \lambda h = \int_0^1 ({\partial f(a+th) \over \partial x}-\lambda)dt h$.

Let $\epsilon >0$ and choose $\delta>0$ such that $\|{\partial f(x) \over \partial x}-\lambda\| < \epsilon$ whenever $\|h\| < \delta$. Then $\|f(a+h)-f(a) - \lambda h \| \le \epsilon \|h\|$.

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  • $\begingroup$ Can we be certain that $\frac{\partial f(x)}{\partial x}$ is bounded at $a$? $\endgroup$
    – md2perpe
    Jun 15, 2017 at 20:37
  • $\begingroup$ @md2perpe: If the given limit exists, then it must be bounded by $\|\lambda\| + \epsilon$. $\endgroup$
    – copper.hat
    Jun 15, 2017 at 20:56
  • $\begingroup$ The derivative is clearly bounded on a punctured neighbourhood of $a$, but we don't even know that the derivative exist at $a$ so can we conclude that $f$ is Lipschitz at $a$? Isn't that needed for $\phi$ to be absolutely continuous? $\endgroup$
    – md2perpe
    Jun 15, 2017 at 21:40
  • $\begingroup$ @md2perpe: Choose a convex neighbourhood of $a$ such that the derivative is bounded. Pick two points $x_0,x_1$ in this neighbourhood. If $a \notin (x_0,x_1)$ then the derivative bound applies. If $a \in (x_0,x_1)$ then split in interval into two parts $[x_0,a]$ and $[a,x_1]$ and now use the mean value theorem to show that the same bound applies. $\endgroup$
    – copper.hat
    Jun 15, 2017 at 21:45
  • $\begingroup$ @md2perpe: While the assertion above is correct, from the proof perspective one only needs to bound the derivative on segments of the form $(a,x)$ to obtain the formula for $\phi$ above. $\endgroup$
    – copper.hat
    Jun 15, 2017 at 21:48

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