Let $I=[0,1] \subseteq \mathbb{R}$ and let $(u_n),(v_n)$ be sequences in $C(I)$ such that $$|u_n(0)|+|v_n(0)| \le 1,~~~~ |u'_n(t)|+|v'_n(t)| \le t+e^t ~~~~ \forall t \in I, ~\forall n.$$ I would like to prove that the sequence $(f_n)$ in $C(I \times I)$ given by $$f_n(x,y)=(1+|u_n(x)v_n(y)|)^2$$ is relatively compact in $L^p(I \times I)$ for every $p \in [1,\infty].$
My idea is to prove that $(f_n)$ is bounded (with respect to $||\cdot||_{\infty}$) and equicontinuous, and then use Ascoli Arzelà in order to conclude that $(f_n)$ is relatively compact in $C(I\times I)$ (this implies relative compactness in every $L^p$, since uniform convergence implies convergence in every $L^p$).
My attempt so far:
I show that $(u_n),(v_n)$ are bounded and equicontinuous: let $x,y \in I$, then $$|u_n(x)-u_n(y)| = |u'_n(\xi)|\cdot|x-y| \le |\xi+e^{\xi}||x-y| \le (1+e)|x-y|$$ so $(u_n)$ is equicontinuous. Moreover, for any $x \in I$, $$|u_n(x)-u_n(0)| = |\int_0^x u'_n(t)dt| \le \int_0^1 |u'_n(t)|dt \le \int_0^1 (t+e^t)dt \le M,$$ hence $|u_n(x)| \le |u_n(x)-u_n(0)|+|u_n(0)| \le M+1. $ Analogous results hold for $(v_n)$, so we are done.
Now, I'm trying to prove that the equicontinuity of $(u_n),(v_n)$ implies therelative compactness of $(f_n)$. My idea was to prove the relative compactness of $(g_n)$, where $g_n(x,y)=u_n(x)v_n(y)$, and the use the fact that $g(s)=(1+|s|)^2$ is continuous to conclude that $(f_n=g \circ g_n)$ is realtively compact.
Is there another way to reach the conclusion? Is my argument correct so far?