Rank of $DH(x,x^{\prime})$ where $H$ does not depend on $x^\prime$

Question

Suppose $f:\mathbb{R}^m\rightarrow\mathbb{R}^n$, $\text{rank}(Df(x))=k.$ Now Define a function $H: \mathbb{R}^m\times \mathbb{R}^n \rightarrow \mathbb{R}^k$, $H(x,x')=f(x)$.

Does $\text{rank}(DH(x,x^{\prime}))= \text{rank}(Df(x))$? Can I conclude that $\text{nullity}DH(x,x') = m+n-\text{rank}(Df(x))=m+n-k$?

Answer 1

I would argue in a few steps. First, look that \begin{align} H(x + h, x') &= f(x + h)\\ &= f(x) + Df(x)h + \varepsilon(h)\\ &= H(x, x') + Df(x)h + \varepsilon(h) \end{align} and \begin{align} H(x, x' + k) = f(x) = H(x, x'). \end{align} From this follows that \begin{align} H(x + h, x' + k) &= H(x, x' + k) + Df(x)h + \varepsilon(h)\\ &= H(x, x') + Df(x)h + \varepsilon(h). \end{align} Therefore, by the uniqueness of the derivative, $DH(x,x')(h,k) = Df(x)h$. From this one have $\text{rank}(DH(x,x')) = k$ and $\text{null}(DH(x,x')) = m + n - k$.