Move the quantifier: demand a single stage beyond which every function of the sequence is within the tolerance at every point at once. That is uniform convergence, and it is exactly the repair the failures of the last lesson needed. The working test is a single number per stage — the largest distance between the function and the limit — and uniform convergence is that number tending to zero.

A committee has finished when every member has settled, not merely when each member eventually settles. The difference is a moment at which the whole room is quiet, and in an infinite committee such a moment need not exist however certainly each member falls silent.

The bars are $\sup_x\lvert f_n(x)-f(x)\rvert$, the worst error at stage $n$. Uniform convergence is exactly this sequence of numbers tending to zero; pointwise convergence says nothing about it, and for $x^n$ on $[0,1]$ every one of these bars is $1$.

$f_n\to f$ uniformly on $A$ when for every $\varepsilon>0$ there is an $$ such that for every and every . Equivalently . Uniform convergence implies pointwise convergence and not conversely, and the limit must be computed pointwise first before uniformity can be tested.

• The test: find the pointwise limit, form $d_n=\sup_x\lvert f_n-f\rvert$, and ask whether that sequence of numbers tends to zero.

Both directions unpack the definitions. If the convergence is uniform then past some stage every point has its error below the tolerance, so the supremum of the errors is at most the tolerance — which makes that sequence of numbers converge to zero. Conversely, if the worst error is eventually below the tolerance then so is the error at every individual point, since each is at most the worst. The only care needed is that a supremum may be approached and not attained, which is why the inequalities are arranged with room to spare.

Apply the definition at half the tolerance.

The supremum of numbers below half the tolerance is at most half the tolerance.

Conversely, suppose the worst errors tend to zero.

Each individual error is at most the worst one.

Move the quantifier over the point past the quantifier over the stage, and the notion changes completely.

Uniformity cannot be tested until the pointwise limit is known.

• Compute the pointwise limit.
• Form the supremum of the difference.
• Ask whether that sequence of numbers tends to zero.

• Uniform convergence chooses the stage before the point.
• It is equivalent to the worst error $d_n$ tending to zero.
• It implies pointwise convergence and is strictly stronger.
• Uniformity depends on the set, and shrinking the set can rescue it.

Great work! You now know how to:

• State uniform convergence and say which quantifier moved
• Test uniformity by computing the worst error
• Say why uniformity depends on the set
• Use the Cauchy criterion to avoid naming the limit