Prove the following theorem, using the definition of total differentiability:

Let $u: \mathbb{R}^m \rightarrow \mathbb{R}$ be totally differentiable in $x \in \mathbb{R^m}$. then $u$ is continuous in $x$.

i need to use: a function $u$ is continuous in $x_0 \in \mathbb{R^m}$ if $\lim_{x \rightarrow x_0} u(x) = u(x_0)$

Some one please help - i have no idea how to set this out but i know the theorem.. thanks

  • $\begingroup$ Camilla! I did some editing which is currently being reviewed. Once that edit (or another) comes through, could you please check if we got what you are asking right? Thanks. $\endgroup$ – gnometorule Feb 19 '13 at 2:28
  • $\begingroup$ yes that is exactly what i was asking! sorry i am new to this and was unsure how to do the symbols properly! thank you :) $\endgroup$ – camilla Feb 19 '13 at 2:30
  • $\begingroup$ thank you i will use that as a reference for future questions $\endgroup$ – camilla Feb 19 '13 at 2:35
  • $\begingroup$ Can someone please help with this i dont understand how to set the answer out!!! thanks $\endgroup$ – camilla Feb 19 '13 at 3:00

Recall that total differentiability of $u: \mathbb{R}^m \to \mathbb{R}$ at $x_0$ means that there exists a vector $v \in \mathbb{R}^m$ (which we identify after the fact as $\nabla u(x_0)$) such that $$\lim_{x \to x_0} \frac{\|u(x) - u(x_0) - v \cdot (x-x_0)\|}{\|x-x_0\|} = 0.$$

Suppose that $u$ is indeed differentiable at $x_0$, so that $v = \nabla u(x_0) \in \mathbb{R}^m$ exists. By the triangle inequality, one has that $$ \| u(x) - u(x_0) \| \leq \|u(x) - u(x_0) - v \cdot (x-x_0)\| + \|v \cdot (x-x_0)\|.$$ Now, when $x \neq x_0$, how does the right hand of this inequality relate to the quantities $\|x-x_0\|$ and $\tfrac{\|u(x) - u(x_0) - v \cdot (x-x_0)\|}{\|x-x_0\|},$ whose behaviour as $x \to x_0$ you do understand?

  • $\begingroup$ okay, so do i just conclude that it is totally differentiable from the triangle inequaility?? im a bit lost, i have done all of this before but in Analysis, not in a differential equation module!! $\endgroup$ – camilla Feb 19 '13 at 13:28
  • $\begingroup$ Wait, aren't you assuming total differentiability at a point $x_0$, and then using this to prove continuity at $x_0$? $\endgroup$ – Branimir Ćaćić Feb 19 '13 at 19:02

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