Pointwise convergence lets each $x$ converge at its own speed. Uniform convergence demands one speed for all of them, and the cleanest way to measure it is a single number, the sup norm. Computing it is a maximum problem, so the extreme value theorem and calculus do the work.
Key idea
The sup norm is $\|g\| = \sup_{x \in S} |g(x)|$. $f_n \to f$ uniformly on $S$ when $\|f_n - f\| \to 0$; equivalently, for every $\varepsilon > 0$ there is an $N$ with $|f_n(x) - f(x)| < \varepsilon$ for every $x \in S$ and every $n \ge N$. Uniform implies pointwise; not conversely.
Why it works
$\|f_n - f\| < \varepsilon$ says the whole graph of $f_n$ lies inside the band between $f - \varepsilon$ and $f + \varepsilon$. So uniform convergence means: for every band width, from some $N$ on, every $f_n$ fits entirely inside the band. To decide it, compute $\|f_n - f\|$ (find the largest gap, often by setting a derivative to $0$) and see whether it goes to $0$. For $x^{n}$ on $[0, 1)$ the sup is $1$ for every $n$: never inside a band of width $\frac{1}{2}$, so the convergence is not uniform. On $[0, \frac{1}{2}]$ the sup is $\left(\frac{1}{2}\right)^{n} \to 0$: uniform.
Worked examples
Worked example: Calculus finds the sup
Find $\|f_n - 0\|$ on $\mathbb{R}$ for $f_n(x) = \frac{x}{1 + nx^{2}}$.
- The derivative $\frac{1 - nx^{2}}{(1 + nx^{2})^{2}}$ is $0$ at $x = \pm\frac{1}{\sqrt{n}}$.
- The value there is $\frac{1/\sqrt{n}}{2} = \frac{1}{2\sqrt{n}}$, and $f_n \to 0$ at $\pm\infty$.
Answer: $\|f_n\| = \frac{1}{2\sqrt{n}} \to 0$: uniform convergence to $0$.
Worked example: A bump that does not shrink
Is $f_n(x) = \frac{nx}{1 + n^{2}x^{2}} \to 0$ uniform on $[0, 1]$?
- At $x = \frac{1}{n}$ the value is $\frac{1}{2}$, the maximum, for every $n$.
- So $\|f_n\| = \frac{1}{2}$, which does not go to $0$, although $f_n(x) \to 0$ at each $x$.
Answer: No: pointwise but not uniform.
Worked example: Finding $N$ for every $x$ at once
For $f_n(x) = x^{n}$ on $\left[0, \frac{1}{2}\right]$ and $\varepsilon = 0.01$, find the smallest $N$ that works for all $x$.
- $\|f_n\| = \left(\frac{1}{2}\right)^{n}$, the value at the right end.
- $\frac{1}{2^{n}} < 0.01$ means $2^{n} > 100$, so $n \ge 7$.
Answer: $N = 7$, one $N$ for every $x$ in the interval.
Common mistake
Fixing $x$ first. "For each $x$, $|f_n(x) - f(x)| \to 0$, so the convergence is uniform" is the definition of pointwise convergence, not uniform. The test for uniform is the supremum over all $x$: it must go to $0$. A moving bump like $\frac{nx}{1 + n^{2}x^{2}}$ passes the pointwise test at every $x$ and fails the uniform one.