Let $E$ be an open set in $\mathbb{R}^n$ and $\mathbf{f}:E\to \mathbb{R}^m$. Let $\mathbf{f}\in C^1(E)$ where $C^1$ - the space of all continuously differentiable functions.

How to prove that $C^1(E)\subset C(E)$.

Here's my thought: Let $f\in C^1(E)$ then all partial derivatives $D_jf$ exists and continuous on $E$. How to prove that $f$ is also continuous?

  • $\begingroup$ If all partial derivatives are continuous, then the function is continuous $\endgroup$ – zhw. Mar 2 '16 at 19:06
  • $\begingroup$ @zhw., How to prove it? $\endgroup$ – Raheem Najib Mar 2 '16 at 19:07
  • $\begingroup$ @zhw., I know the following theorem: $f\in C'$ if and only if all partial derivatives exists and continuous. So if all partial derivatives exists and continuous then $f\in C'$ how to conclude that $f$ is continuous? $\endgroup$ – Raheem Najib Mar 2 '16 at 19:27
  • $\begingroup$ That's not a theorem is it? Isn't that the definition? $\endgroup$ – zhw. Mar 2 '16 at 19:46
  • $\begingroup$ @zhw., In Rudin's book it's a theorem $\endgroup$ – Raheem Najib Mar 2 '16 at 20:03

Differentiability at $x_0 \in E$ implies continuity at $x_0$. We don't need continuity of the derivative or partial derivatives.

Indeed differentiability means that there exists a linear map $Df_{x_0}:\mathbb{R}^n \to \mathbb{R}^m$ such that for all $\epsilon > 0$ we have $$\|f(x) - f(x_0) - Df_{x_0}(x-x_0)\| < \epsilon \| x - x_0 \|$$ whenever $x$ is close enough to $x_0$. Fix any $\epsilon > 0$.

By the reverse triangle inequality in particular this means that $$\|f(x) - f(x_0)\| - \|Df_{x_0}(x-x_0)\| < \epsilon \| x - x_0 \|,$$ or

$$\|f(x) - f(x_0)\| < \epsilon\|x-x_0\| + \|Df_{x_0}(x-x_0)\|.$$

But as $Df_{x_0}$ is linear and has the finite-dimensional domain $\mathbb{R}^n$, it is bounded, i.e., there is some constant $L > 0$ such that $\|Df_{x_0}(x-x_0)\| \leq L\|x-x_0\|$ for all $x$. Thus $$\|f(x) - f(x_0)\| < (\epsilon+L)\|x-x_0\|.$$

Since we can make the right-hand side arbitrarily small by taking $x$ close enough to $x_0$, $f$ is continuous at $x_0$.

  • $\begingroup$ Thanks but how it related with my problem? $\endgroup$ – Raheem Najib Mar 2 '16 at 19:28
  • $\begingroup$ @RaheemNajib Didn't you want to prove that all continuously differentiable functions on $E$ are continuous? $\endgroup$ – Alex Provost Mar 2 '16 at 19:29
  • $\begingroup$ Yes. Where you used that $f$ is continuously differentiable? $\endgroup$ – Raheem Najib Mar 2 '16 at 19:30
  • $\begingroup$ @RaheemNajib As I've said, we don't need this hypothesis. We have inclusions $C^1(E) \subset D(E) \subset C^0(E)$, where $D(E)$ is the space of differentiable functions on $E$. The first inclusion is obvious, and I've shown the second one. $\endgroup$ – Alex Provost Mar 2 '16 at 19:33

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.