§2. Uniform convergence
207
Another simple case, but much less frequently met, is that where the
functions fn are equicontinuous at the point a; this means that for all r > 0
there exists an r' > 0 such that
(5.6)
d(x, a) < r' => d [fn(a), fn(x)] < r for all n.
If this condition is satisfied, then passing to the limit, with x and a fixed, as
n increases indefinitely, shows that 5
d(x, a) < r' => d[f(a), f(x)] ~ r,
whence the continuity of f at a without even having to use the arguments
leading to (4).
If we agree to say that a family of functions defined on X is equicontinuous
on X if it is equicontinuous at all points of X, we finally obtain the following
result:
Theorem 8. Let Un) be a sequence of functions defined and continuous on
a subset X ofe and suppose that limfn(x) = f(x) exists for all x E X. For
f to be continuous in X it suffices that one of the two following conditions be
satisfied: (i) the given sequence converges uniformly on X; (ii) the junctions
fn are equicontinuous on X.
Experience has shown that at the beginning innumerable students have
very great trouble in understanding this fundamental concept of uniform
convergence. Yet it is simple: supposing that the functions have real values,
it means that for all r > 0
f(x) - r < fn(x) < f(x) + r for any x E X
for all sufficiently large n. Or again: for all sufficiently large n the graph of
fn lies entirely in the strip in the plane of height 2r contained between the
graphs of the functions f(x) + rand f(x) - r (fig. 5).
The difficulty, apparently, is with logic: one has here a proposition PN(X),
namely the implication
(n > N) => {If(x) - fn(x)1 < r},
which depends simultaneously on N and on x [it does not depend on n, since
we have very deliberately omitted the sign (Vn) which ought to figure there
and to transform n into a phantom variable]. Simple, nonuniform, convergence demands that
for all x, there exists N such that PN(X);
while uniform convergence requires that
5 We use the fact that if Iunl < r for all n then I lim unl ~ r (weak inequality).
207
Another simple case, but much less frequently met, is that where the
functions fn are equicontinuous at the point a; this means that for all r > 0
there exists an r' > 0 such that
(5.6)
d(x, a) < r' => d [fn(a), fn(x)] < r for all n.
If this condition is satisfied, then passing to the limit, with x and a fixed, as
n increases indefinitely, shows that 5
d(x, a) < r' => d[f(a), f(x)] ~ r,
whence the continuity of f at a without even having to use the arguments
leading to (4).
If we agree to say that a family of functions defined on X is equicontinuous
on X if it is equicontinuous at all points of X, we finally obtain the following
result:
Theorem 8. Let Un) be a sequence of functions defined and continuous on
a subset X ofe and suppose that limfn(x) = f(x) exists for all x E X. For
f to be continuous in X it suffices that one of the two following conditions be
satisfied: (i) the given sequence converges uniformly on X; (ii) the junctions
fn are equicontinuous on X.
Experience has shown that at the beginning innumerable students have
very great trouble in understanding this fundamental concept of uniform
convergence. Yet it is simple: supposing that the functions have real values,
it means that for all r > 0
f(x) - r < fn(x) < f(x) + r for any x E X
for all sufficiently large n. Or again: for all sufficiently large n the graph of
fn lies entirely in the strip in the plane of height 2r contained between the
graphs of the functions f(x) + rand f(x) - r (fig. 5).
The difficulty, apparently, is with logic: one has here a proposition PN(X),
namely the implication
(n > N) => {If(x) - fn(x)1 < r},
which depends simultaneously on N and on x [it does not depend on n, since
we have very deliberately omitted the sign (Vn) which ought to figure there
and to transform n into a phantom variable]. Simple, nonuniform, convergence demands that
for all x, there exists N such that PN(X);
while uniform convergence requires that
5 We use the fact that if Iunl < r for all n then I lim unl ~ r (weak inequality).