Does $f_n(s,\varphi_n(s)) \to f(s,\varphi(s))$ pointwise under convergence of $f_n,\varphi_n$?

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Let $f_n:\mathbb{R} \times \mathbb{R}^d \to \mathbb{R}$ such that in certain compact set $M$, we have that $f_n \to f$.

Assume that $\varphi_n:\mathbb{R} \to \mathbb{R}^d$ is such that $\varphi_n(s) \to \varphi(s)$ uniformly and that $f_n(s,\varphi(s))$ remains in $M$.

Is it true that:

$f_n(s,\varphi_n(s)) \to f(s,\varphi(s))$ pointwise

when:

  1. $f_n \to f$ pointwise?
  2. $f_n \to f$ uniformly?

My question from an intent of proving Cauchy-Peano theorem on the existence of solutions for a differential equations.

My approach

I don't think pointwise convergence suffices, see: Pointwise convergence does not imply $f_n(x_n)$ converges to $f(x)$.

If we consider uniform convergence:

$\forall \epsilon > 0.\exists n_1 \in \mathbb{N}.\forall n \ge n_1.\|f_n-f\|_{\infty} < \epsilon$

the goal to prove is:

$\forall s, \epsilon > 0.\exists n_2 \in \mathbb{N}.\forall n \ge n_2.\|f_n(s,\varphi_n(s)) - f(s,\varphi(s))\| < \epsilon$

I think that it is evident from here is it so simple?