Proof of $\lim_{i\to\infty}\lambda_i^{-1}\left|f(\hat{x}+\lambda_ix,u_i) - f(\hat{x},u_i) - D_xf(\hat{x},u_i)(\lambda_ix)\right| = 0$

Question

I am trying to prove or disprove the next statement that seems necessary for the proof of Proposition 2.9 of this book.

Let $U\subset R^k$ be compact and $f:R^n\times U \to R^m$ be twice differentiable with respect to $x\in R^n$. Assume that $f$ and the derivatives are continuous with respect to $(x,u) \in R^n\times U$. Let $\lambda_i > 0$ and $u_i \in U$, $i=1,2,3,\dots$, such that $\lambda_i \to 0$ and $u_i \to u_0\in U$ as $i\to\infty$. Then, \begin{equation} \tag{1}\label{1} \lim_{i\to\infty}\lambda_i^{-1}\left|f(\hat{x}+\lambda_ix,u_i) - f(\hat{x},u_i) - D_xf(\hat{x},u_i)(\lambda_ix)\right| = 0 \end{equation} where $D_xf(\hat{x},u_i)\in L(R^n,R^m)$ is the derivative of $f$ with respect to $x$ at $(\hat{x},u_i)$.

\eqref{1} is obvious if $u_i = u_0$, $i = 1,2,\dots$, by the definition of the derivative but seems not trivial for me in other case. Would you give me any hint or a reference?

Answer 1

By Taylor's theorem with remainder,

$$ | f(\hat{x} + \lambda_i x, u_i - f(\hat{x},u_i) - D_x f(\hat{x},u_i)(\lambda_i x) | \leq C \lambda_i^2 |x|^2 \sup_{y\in B_{\lambda_i |x|} (\hat{x})} |D_x^2 f(y,u_i)| $$

for some universal constant $C$. For sufficiently large $i$, the right hand side is bounded by

$$ \sup_{y\in B_1(x), u\in B_1(u_0)} |D^2_xf(y,u)| $$

the claim then follows as $\lambda_i \searrow 0$.