An introduction to the study of the elements of the differential and integral calculus. From the German of the late Axel Harnack, With the permission of the author.

~ 51. 52.

Uniform continuity.

87

52. Considering the explicit function s = f(x, y) let us assume it to
be one-valued, and enquire when is it continu o u s in a domain for which
it has determinate values. Conceive any point of the domain inclosed
in a small rectangle having the lengths of its sides parallel to the
axis of abscissae =2 h and of those parallel to the ordinate axis
=c 2k, so that x + h, y + k are the coordinates of its four corners.
Thus the coordinates of any point within this region or on its boundary are x + Oh, y + k,(O~O~, < 0 < < 1). If we denote by
f(x + 0 h, y ~+ i) the corresponding value of the function, then
the function shall be called continuous at x,, only when finite values of
k and k can be found, for which the absolute amount of the difference:
f(x + Oh, y + r k) - f(x, y) is less than a prescribed arbitrarily
small number 8 for every value of the independent variables 0 and l.
For then and only then will every series of numbers obtained from
f(x + Oh, y ~ rh) - f(x, y) by making 0 and r converge to zero
in any manner whatever, have zero as its limit. It is therefore
necessary for continuity, that f(x + Oh, y) - f(x,' y) and likewise
f(x, y + V) - f(x, y) become infinitely small, or in other words:
that f(x, y) be continuous as a function of the variable x alone or of
the variable y alone; but yet this is not sufficient. Therefore to say
f(x, y) is a continuous function of both variables x and y, is different
from saying that f is a continuous function of x as well as of y.
On the other hand we can replace the above definition by its
equivalent: It must be possible to find at the point x, y, a finite
value h and a finite value h, so that for all values equal to or less
than h or k respectively, f(x + Oh, y) shall be a continuous function
of y alone, and f(x, y - r~j) a continuous function of x alone, in
such a way that independently of Oh, we shall have
abs [f(x 4t- Oh, y + rk) - f(x + tOh, y)] < 6
for all values of 'V merely by the value chosen for k, and in like
manner that independently of q^k, we shall have
abs [f(x + Oh, y + 47) - f(x, y +~ k)] < 6
for all values of 0 merely by the value chosen for h.
These conditions are enunciated in the words: f(x, y) must be a
uniform ly continuous function of x as well as of y in the neighbourhood of the point x, y.
For according to this way of putting it, if we assume X = 0 in the
second inequality, we have also for all values of 0
abs [f(x 4+ Oh, y) - f(x, y)] < d.
This inequality added to the first shows that
abs [f(x + 3 h, y + k) - f(x y)]