$f\colon\mathbb{R}^n \to\mathbb{R}^m$ has derivative $0$ everywhere implies $Df=0$?

Question

I was trying to prove that $f\colon\mathbb{R}^n \to\mathbb{R}^m$ has derivative $0$ everywhere implies $f=const.$ and I thought about using the multivariable MVT $f_i(b)-f_i(a)=(b-a)\cdot\nabla f_i(x_0)$ (for some suitable $x_0$) on each of the $m$ $f_i$ but to use this I have to know first that all of the $m$ gradients are zero and so the derivative matrix $D f$ is zero but as far as I know I can conclude this from the hypothesis only if I know beforehand that the partial derivatives of $f$ are continuous.

So, is it possible to amend the sketch of my proof above somehow or to prove this theorem I have to know also that the partial derivatives (and thus the gradients and the derivative matrix) are zero?

Thanks.

Answer 1

If the derivative is $0$ everywhere, then the partial derivatives are $0$ everywhere and, in particular, they are continuous.

Answer 2

If $f$ has "derivative $0$ everywhere" it is everywhere differentiable by assumption. The chain rule then implies that $\phi(t):=f\bigl((1-t)a+tb\bigr)$ has derivative $\equiv0$, hence is constant. It follows that $f(a)=\phi(0)=\phi(1)=f(b)$, whatever $a$ and $b$.