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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?

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  • $\begingroup$ 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
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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 $x_0$.

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  • $\begingroup$ 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

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