Continuity of Taylor remainder for a multivariate $C^1$ function

Question

Suppose I have a function jointly $C^1$ in two variables.

By Taylor's theorem I can expand in one of the variables, say the second, and get a remainder term

$R(x,y,h) = \frac{f(x,y+h) - f(x,y)}{h} - \frac{\partial}{\partial y}f(x,y)$,

taking $R(x,y,0) = 0$. Certainly, $R$ is separately continuous in each variable. What I would like to know is, is $R(x,y,h)$ jointly continuous in $x$ and $h$?

(Or, relatedly, is the limit $\displaystyle\lim_{h \to 0} R(x,y,h) = 0$ uniform in $x$ near a given point $(x_0, y_0)$?)

Answer 1

Let $f:U\subset \mathbb{R}^n\to \mathbb{R}^m$ being a $\mathcal{C}^1$ function. Then $$f(u+h)=f(u)+f'(u)(h)+R(u,h),\qquad h\in\mathbb{R}^n, \quad u,u+h\in U,$$ in which $f'(u)$ is viewed as a linear transform in $\mathcal{L}(\mathbb{R}^n,\mathbb{R}^m)$, $f':U\to \mathcal{L}(\mathbb{R}^n,\mathbb{R}^m)$ is continuous, and $$\lim_{h\to 0}\frac{R(u,h)}{\|h\|}=0.$$

This means that $$R(u,h)=f(u+h)-f(u)-f'(u)(h),\qquad h\in\mathbb{R}^n, \quad u,u+h\in U,$$ is continuous.

If we choose $u_0\in U$, then there exists $r>0$, such that the closed ball $B_r[u_0]\subset U$. Since $B_r[u_0]$ is a compact set then $R_{\vert B_r[u_0]}$ is uniformly continuous.

A similar argument applyies when $R(u,h)$ comes from partial derivatives, if you note that $f'(u)$ can be viewed as the Jacobian matrix of $f(u)$.

You can find related discussions searching for "\(R(h)=f(u+h)-f(u)-f'(u)(h)\)" on SearchOnMath, for instance.