§2. Uniform convergence
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(R 2) If two sequences (fn) and (gn) converge uniformly to f and to g,
and if the limit functions f and 9 are bounded, then the sequence (fngn) converges uniformly to fg.
To see this we use the identity
fngn - fg = (fn - f)(gn - g) + f(gn - g) + g(fn - f),
whence, by (5) and (6), omitting the subscripts E,
IIfngn - fgll :-:; IIfn - fll·llgn - gil + IIfll·lIgn - gil + Ilgll·llfn - fll;
thus the left hand side tends to 0 so long as IIfll and Ilgll are finite.
The boundedness hypothesis on f and 9 is essential, as the following
counterexample shows. Take E = [0, +00[, fn(x) = x + lin and gn(x) = lin,
whence f(x) = x, d(fn, f) = lin, g(x) = 0 and d(gn, g) = lin. The functions
fn(x)gn(x) = xln + 1/n 2 converge simply to 0 = f(x)g(x) but not uniformly
since they are not even bounded in E.
(R 3) If a sequence of functions fn converges uniformly on E to a limit
function f, and if inf If(x)1 is strictly positive, then the sequence (11 fn) converges uniformly to 11 f. For
111 fn(x) - 11 f(x)1 = Ifn(x) - f(x)l/lfn(x)I·lf(x)1 :-:;
< dE(fn, f)/lfn(x)I·lf(x)l,
so that to majorise the result one has to minorise the denominator of this
fraction. If there exists a number m > 0 such that If(x)1 ~ m for any x E E,
then the relation Ilfn - fll < m/2, valid for n large, shows that
Ifn(x)1 ~ m/2 for all x E E
if n is sufficiently large, whence dE (ll fn, 11 f) :-:; 2dECfn, f)lm 2 for n large,
qed.
Rule (R 3) requires the limit function f to remain "uniformly away from
zero" for n large or, equivalently, that the function 11 f be bounded. Even if f
and the fn never vanish, the uniform convergence of fn to f does not imply
that of 11 fn to 11 f. If, for example, one takes E = [1, +oo[ and fn(x) =
11x + 1/nx, then f(x) = 11x and d(fn, f) = sup 11/nxl = lin, so that fn
converges uniformly to f; but
11 f(x) - 11 fn(x) = xl(n + 1)
does not converge uniformly to 0 and is not even bounded on E.