We are given a positive function $f(x,y)$ on the domain $\mathbb{N} \times [0,1]$. We know all the limits $g(y) = \lim_{n \to \infty} f(n,y)$ for $y\in[0,1]$. $g(y)$ is finite for every $y \in [0,1]$.
Now, consider any sequence $\{y_k\}_{k\geq 1}$ such that $0 \leq y_k \leq 1 \forall k$. What can we say about $$\liminf_{n \to \infty} f(n,y_n)$$ Can we say $$\liminf_{n \to \infty} f(n,y_n) \in [\min g(y),\max g(y)]$$
My approach: I can prove this if I assume 'uniform convergence' of $f(n,y)$ over the index variable $y$.
For any $y \in [0,1]$, for any $\epsilon > 0 $, there exists $N(\epsilon,y)$ such that $\forall n \geq N(\epsilon,y)$, $|f(n,y) -g(y)| \leq \epsilon$. Define $N(\epsilon) = \max_y N(\epsilon,y)$ then for any $n \geq N(\epsilon)$ and any $\{y_n\}$, $|f(n,y_n) - g(y_n)| \leq \epsilon$. Therefore, $\liminf_{n \to \infty} f(n,y_n) \in [\min g -\epsilon,\max g + \epsilon] $.
However, in the above proof, I have assumed $N(\epsilon)$ is finite. I don't know if this can be assumed. Is there any other way to proceed?
This is not true. Let $f(n,x)=n$ for $0 <x<\frac 1 n$ and $0$ for $x=0$ and for $x \geq \frac 1 n$. Then $f(n,y) \to 0$ for all $y$. If you take $f_n=\frac 1 {2n}$ the $f(n,y_n)=n$ so $\lim \inf f(n,y_n)=\infty$.
Actually we can even make the functions smooth . A standard construction using bump functions shows that there exists functions $f(n,y)$ such that $f(x)= \begin{array} {c} 1 \, \,\text {if} \,\,\frac1 {3n} <x<\frac 1 n \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \\0 \,\, \text{if} \, \,x \in [0,\frac 1 {4n}] \cup [\frac 2 n, 1] \end{array}$ with $0 \leq f(n,y) \leq 1$ for all $n$ and $y$.