0
$\begingroup$

I had an idea for how to prove that if $f$ has continuous partial derivatives, then it's differentiable. To make things simpler, take a two variable function $f(x, y):\mathbb R^2 \to \mathbb R$. Let's call its partial derivatives $\partial_xf$ and $\partial_yf$, and suppose they're continuous at $(x_0, y_0)$. Let $dx$ and $dy$ denote two small real numbers.

$$ \begin{align} f(x_0, y_0) + \partial_xf(x_0, y_0)dx &= f(x_0 + dx, y_0) + o(dx) \\ f(x_0, y_0) + \partial_xf(x_0, y_0)dx + \partial_y(x_0 + dx, y_0)dy &= f(x_0, + dx, y_0 + dy) + o(dx) + o(dy) \\ \end{align} $$

The idea is that since that partial derivative with respect to $y$ is continuous, we can replace that term with $\partial_y(x_0, y_0)dy$ without too big of an error, but I can't think of a way to make this rigorous. All I can think of is to write

$$\partial_y(x_0 + dx, y_0)dy = \partial_y(x_0, y_0)dy + \epsilon(dx)dy$$

Where $\epsilon(dx)$ is some function which converges to $0$ as $dx\to 0$. So when I plug that in, I get a remainder term of

$$o(dx) + o(dy) + \epsilon(dx)dy$$

But I can't see how to show this is $o(||(dx, dy)||)$ Also, if this approach works, where am I using the assumption that all partial derivatives (rather than just $\partial_y$) are continuous?

$\endgroup$
  • $\begingroup$ Using this "dx" notation is not such a great idea. $\endgroup$ – zhw. May 20 '16 at 23:33
0
$\begingroup$

Of course both $o(dx)$ and $o(dy)$ are $o(||(dx, dy)||)$. Then you have only to prove that $ \epsilon(dx)dy$ is $o(||(dx, dy)||)$. Now $$ \frac{|\epsilon(dx)dy|}{||(dx, dy)||}\leq\frac{|\epsilon(dx)||dy|}{|dy|}=|\epsilon(dx)|\rightarrow 0 $$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ So do we always have $||(a, b)|| \geq |a|$ for any norm? $\endgroup$ – Jack M May 20 '16 at 17:01
  • $\begingroup$ There are different but equivalent norms (in topological sense). Think for concreteness to the euclidean norm: $ \|(a,b)\|=(a^2+b^2)^{1/2}\geq a$. Of course also $ \|(a,b)\|=\frac12 (a^2+b^2)^{1/2}\geq \frac12 a$ is a norm... You see that the change is not important for this problem. $\endgroup$ – guestDiego May 20 '16 at 17:11
  • $\begingroup$ Okay, so where did we use the assumption that $\partial_x$ is continuous? It seems like we've only assumed $\partial_y$ is continuous. $\endgroup$ – Jack M May 20 '16 at 17:47
  • $\begingroup$ The argument provided in the next answer is complete rigorous. In your framework the problem is (very) hidden in your $o(dy)$. For example, in the first line you write $o(dx)$ which is fine, because, even if you have a different function $o(dx)$ for different $(x_0,y_0)$, here $x_0,y_0$ are fixed. In the second line however, you have a $o(dy)$ which depends on $(x_0+dx,y_0)$ which changes in the limit process. $\endgroup$ – guestDiego May 20 '16 at 18:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.