I Uniform convergence of family of functions with continuous index

[psie]

TL;DR Summary

I'm reading Rudin's PMA. In particular, his proof of the Stirling formula (see below). I'm stuck trying to figure out how a certain family of functions parameterized by a continuous variable converges uniformly to its limit.

In Rudin's PMA, when he proves the Stirling formula, he defines a continuous, decreasing function $h:(-1,\infty)\to\mathbb R$ such that $h(u)\to\infty$ as $u\to -1$ and $h(u)\to 0$ as $u\to\infty$. Then he derives an integral expression for $\Gamma(x+1)$, where $\Gamma$ is the Gamma function, in terms of the function $\psi_x(s)$ defined in the screenshot below. He claims this function converges uniformly on $[-A,A]$ for every $A<\infty$.

I think I know which theorem in the book he uses to justify this claim (Theorem 7.13), however, Theorem 7.13 deals with sequences of functions, whereas here it seems we have a function that is indexed by a continuous variable $x\in (0,\infty)$. I struggle with understanding the definition of uniform convergence when the index is continuous like this. What is the correct definition in this case?

Here's an attempt. This is really just a guess, since I don't know the proper definition of uniform convergence when the index variable is continuous. I'd like to somehow phrase it in terms of sequences, so I'd say $\psi_{x}\to f$ uniformly iff given $\epsilon>0$ and any sequence $(x_n)\subset(0,\infty)$ diverging to infinity, there exists an $N\in\mathbb N$ such that for all $s\in[-A,A]$, $$n>N\implies |\psi_{x_n}(s)-f(s)|<\epsilon.$$Again, I have no clue if this is correct, since I don't know the non-sequential definition. Grateful for any help.

 

[mathwonk]

if you know the definition of point wise convergence for a continuous index, then you get the definition of uniform convergence by interchanging the quantifiers of the index bound N, and of the variable s, just as for sequential convergence.

I.e. (Uniform) for any epsilon e>0, there is a bound N such that for all s, x >N implies |gx(s) - f(s)| < e.

(Pointwise) for any epsilon e>0, and any s, there is a bound N such that x>N implies |gx(s) - f(s)| < e.


your clever sequential version may work too.

 

[fresh_42]

I like to write the $N$ as $N=N(\varepsilon)$ in the uniform case and $N=N(s,\varepsilon)$ in the pointwise case to reflect the dependencies in the notation.

 

[pasmith]

I prefer to regard uniform convergence as being convergence with respect to the sup norm/distance in the appropriate function space. A family of functions f_\alpha : X \to Y converges uniformly with respect to the metric d to f : X \to Y as \alpha \to \infty if and only if \lim_{\alpha \to \infty} \sup_{x \in X} d(f_\alpha(x), f(x)) = 0. It does not matter when taking \alpha \to \infty whether \alpha \in \mathbb{N} or \alpha \in \mathbb{R}. If instead we wanted uniform convergence as \alpha tends to a finite value then it would make a difference.