Swap the order of the quantifiers
Pointwise convergence puts the for all x outside the there exists N: for each x, some N. Uniform convergence pulls the *for all x* inside, after the N is chosen. The definition reads: for every epsilon > 0 there exists N such that for all n >= N and all x in E, |f_n(x) - f(x)| < epsilon. One N must service the entire domain simultaneously.
Repackage with the sup norm
The clause *for all x, |f_n(x) - f(x)| < epsilon* is exactly a bound on the worst point. Define the sup norm ||g|| = sup over x in E of |g(x)|, the largest height the function reaches (using the supremum so it is defined even when no maximum is attained). Then uniform convergence is the crisp statement ||f_n - f|| -> 0: a single number sequence whose convergence to zero we already understand.
- Find the candidate limit f by computing the pointwise limit lim f_n(x) at each x.
- Form the error function e_n(x) = f_n(x) - f(x).
- Compute M_n = sup over x of |e_n(x)| — often by calculus, locating where the error is largest.
- Convergence is uniform iff M_n -> 0. If M_n stays bounded away from 0, convergence is merely pointwise.
Worked: a bump that fails to vanish
Compare two sequences with the same pointwise limit 0 on [0,1]. The recipe gives a one-line verdict. The first is x^n from the previous guide; the second is a moving spike. The estimate M_n decides each case.
Sequence A: f_n(x) = x^n on [0,1). Pointwise limit f = 0.
M_n = sup_{0<=x<1} |x^n - 0| = sup x^n = 1 (approached as x -> 1).
M_n = 1 does NOT go to 0 => NOT uniform. (matches Guide 1)
Sequence B: g_n(x) = n x (1 - x)^n on [0,1]. Pointwise limit g = 0
(for fixed x in (0,1], (1-x)^n -> 0 beats the factor n; g_n(0)=0).
Maximize: g_n'(x) = 0 gives the peak near x = 1/(n+1),
height g_n(1/(n+1)) = n * (1/(n+1)) * (n/(n+1))^n
~ (n/(n+1)) * e^{-1} -> 1/e ~ 0.368.
M_n -> 1/e != 0 => NOT uniform: a bump of fixed height 1/e
slides toward 0 but never flattens.
Sequence C: h_n(x) = x/n on [0,1]. Pointwise limit 0.
M_n = sup |x/n| = 1/n -> 0 => UNIFORM.