Consider functions $f_{\epsilon}: K_x \times K_u \rightarrow \mathbb{R}$ for all $\epsilon >0 $ and $f: K_x \times K_u \rightarrow \mathbb{R}$. $K_x \subset \mathbb{R}^n$ and $K_u \subset \mathbb{R}^m$ are compact spaces. Suppose that given $x \in K_x$, $f_{\epsilon}(x, \cdot)$ and $f(x, \cdot)$ are continuous (on $K_u$), and $f_{\epsilon}(\cdot, u)$ and $f(\cdot, u)$ are continuous (on $K_x$).
If $f_{\epsilon}(x, \cdot)$ converges uniformly to $f(x, \cdot)$ as $\epsilon \rightarrow 0^+$, then does it imply that $f_{\epsilon} \rightarrow f$ uniformly? I guess that from the uniform convergence of $f_{\epsilon}(x, \cdot)$ to $f(x, \cdot)$, we may choose $\overline{\epsilon}(x) > 0 $ for any $\epsilon' >0$ such that $|f_{\epsilon(x)}(x, u) - f(x, u)| < \epsilon', \forall \epsilon(x) \leq \overline{\epsilon}(x), \forall (x, u) \in K_x \times K_u$. Then, $\epsilon := \min_{x \in K_x}\overline{\epsilon}(x)$ implies the desired result.
Actually, I'm not sure that such $\epsilon$ exists. Is it correct? If not, are there any sufficient conditions to derive similar conclusions?
EDIT: with an additional condition: $f_{\epsilon}$ converges pointwise to $f$ as $\epsilon \rightarrow 0^+$.
For a counterexample, take $f_\epsilon : [0,1] \times [0,1] \to \mathbb{R}$ where
$$f_\epsilon(x,u) = \begin{cases}\frac{\epsilon xu}{x^3+u^3}, &(x,u) \neq (0,0)\\0, & (x,u) = (0,0)\end{cases}$$
The function $f_\epsilon$ is separately continuous in $x$ and $u$ and converges pointwise to the (continuous) zero function as $\epsilon \to 0+$. For fixed $x$ and all $u \in [0,1]$ we have
$$\left|\frac{\epsilon xu}{x^3+u^3} \right|\leqslant \begin{cases}\frac{\epsilon}{x^2}, & 0 < x \leqslant 1\\0, & x= 0 \end{cases}\underset{\epsilon \to 0+}\longrightarrow 0,$$
and, thus, $f_\epsilon(x,\cdot) \to 0$ uniformly for $u \in [0,1]$.
However, for $\epsilon \in (0,1)$
$$\sup_{(x,u) \in [0,1]^2}\left|\frac{\epsilon xu}{x^3+u^3} \right| \geqslant \frac{\epsilon \cdot \epsilon \cdot \epsilon}{\epsilon^3 + \epsilon^3 } = \frac{1}{2} \underset{\epsilon \to 0+}\longrightarrow \frac{1}{2} \neq0$$
and the convergence is not uniform on $[0,1]\times[0,1]$.