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Let $U\subset R^m$ be an open set and $f:U\to\mathbb R^n$ a constant function. My question is really simple. I know that this function is $C^{\infty}$, but I'm having troubles to prove it formally. What I know is its derivative is zero and $f \in C^k$ iff $f'\in C^{k-1}$. I don't know how to manage all this information to prove $f\in C^k$ for every $k=0,1\ldots$

I've already proved $f\in C^0$ and $f\in C^1$. My problem is to prove the induction part: $f\in C^k\implies f\in C^{k+1}$

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    Formally, all its partial derivatives of *any* order exist and are continuous (They all equal zero) , and thus it is differentiable as much as one as can expect from a function... – 2017-01-06
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    Did you try to prove this for $k=0$ or $k=1$ for example? – 2017-01-06
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    @md5 of course. In fact I'm having problems to prove the induction part: $f\in C^k\implies f\in C^{k+1}$ – 2017-01-06
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    @user42912 prove by induction that $f^{(n)}$ exists and is constant zero. Then, the derivative of $f$ of arbitrary order exists and is continuous. That is, $f\in C^\infty$. – 2017-01-06
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    Obviously, you can't prove $f\in C^n \implies f\in C^{n+1}$ in general... – 2017-01-06

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Let's strengthen your induction hypothesis. We prove $f \in \mathcal{C}^{k}$ and $d^kf = 0$ for all $k \ge 1$.

  • $k = 1$: you should prove $df=0$, using the definition of differential.
  • If $f \in \mathcal{C^{k-1}}$ and $d^{k-1}f = 0$, you should prove $f \in \mathcal{C^{k}}$ and $d^{k}f = 0$. Once again apply the definition to the hypothesis.