If a family of sequences converge to the same limit $g$ and uniformly approach each other, do they uniformly converge to $g$?

Question

Let $I$ be an arbitrary index set and for each $i \in I$ let $(g^i_n)$ be a sequence of real-valued functions on some arbitrary set $X$.

Suppose that there exists a function $g: X \to \mathbb{R}$ such that for all $i \in I$ and all $x \in X$, $g_n^i(x) \to g(x)$ as $n \to \infty$. That is, each sequence $(g^i_n)$ has the same pointwise limit $g$.

Suppose also that $\sup_{i,j \in I}|g^i_n(x) - g^j_n(x)| \to 0$ as $n \to \infty$ for all $x \in X$. That is, the family of sequences get closer to each other uniformly as $n$ increases.

Does it follow that $\sup_{i \in I}|g^i_n(x) - g(x)| \to 0$ as $n \to \infty$ for all $x \in X$?

In other words, is the pointwise convergence to $g$ uniform in $I$? It seems very intuitive to me that the answer is affirmative: each sequence in the family is approaching the same limit and the sequences are approaching each other uniformly, so they must be approaching their limit uniformly. For some reason, I am struggling to find a formal proof.

It seems like something simple, like the triangle inequality, should work. For example,

$$|g^i_n(x) - g(x)| \leq |g^i_n(x) - g^j_n(x)| + |g^j_n(x) - g(x)|,$$

and then taking a supremum over $i,j \in I$ in order to use our assumption about uniform convergence on the right-hand side gets

$$\sup_{i \in I}|g^i_n(x) - g(x)| \leq \sup_{i,j \in I}|g^i_n(x) - g^j_n(x)| + \sup_{j \in J}|g^j_n(x) - g(x)|.$$

But this is just a dead end: $\sup_{j \in J}|g^j_n(x) - g(x)|$ on the right-hand side is the same quantity I'm trying to estimate on the left-hand side.

Answer 1

For any $j$, there is $N_j$ such that for all $n > N_j$, $$ |g_n^j(x) - g(x)| \le \epsilon/2$$ Then we have \begin{align} |g^i_n(x) - g(x)| &\leq |g^i_n(x) - g^j_n(x)| + |g^j_n(x) - g(x)|\\ &\leq |g^i_n(x) - g^j_n(x)| + \epsilon/2. \end{align} By taking sup only w.r.t. $i$, we have \begin{align} \sup_{i\in I}|g^i_n(x) - g(x)| \leq \sup_{i\in I} |g^i_n(x) - g^j_n(x)| + \epsilon/2. \end{align} Since $\sup_{i\in I} |g^i_n(x) - g^j_n(x)| \le \sup_{i,j\in I}|g^i_n(x) - g^j_n(x)|$, we obtain \begin{align} \sup_{i\in I}|g^i_n(x) - g(x)| \leq \sup_{i,j \in I} |g^i_n(x) - g^j_n(x)| + \epsilon/2. \end{align} Since $\sup_{i,j \in I} |g^i_n(x) - g^j_n(x)| \to 0$ as $n \to \infty$, by choose $N$ appropriately, we conclude that $$\sup_{i\in I}|g^i_n(x) - g(x)| \le \epsilon$$

Answer 2

The claim you're trying to prove is intended to hold for each $x$, so you can suppress $x$. Pick any $j$, say $j=J$. Then for each $i$ and $n$ we have $$ |g_n^i-g|\le |g_n^i-g_n^J| + |g_n^J-g|\le \sup_{i,j}|g_n^i-g_n^j|+|g_n^J-g|\tag1 $$ The RHS of (1) depends on $n$ only, which implies the sup over all $i$ of the LHS is bounded by the same thing: $$ \sup_i |g_n^i-g|\le \sup_{i,j}|g_n^i-g_n^j|+|g_n^J-g| $$ Now let $n\to\infty$, and you're done.