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?

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

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 $$ \frac{|\epsilon(dx)dy|}{||(dx, dy)||}\leq\frac{|\epsilon(dx)||dy|}{|dy|}=|\epsilon(dx)|\rightarrow 0 $$

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  • $\begingroup$ 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

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