276
III - Convergence: Continuous variables
For functions of one variable, derivability and differentiability are equivalent properties; not so in the general case, for a good and simple reason.
Consider what happens when one traverses a line passing through (a, b), i.e.
in the direction of a given vector h = (u, v); by (2')
(19.5)
fez + th) = fez) + t· Df(z)h + oCt),
so that f is differentiable in any direction originating at (a, b), with
(19.5')
d
Df(z)h = d/(z + th) at t = 0.
The existence at the point (a, b) of derivatives in all the directions originating
from (a, b) is thus necessary to ensure differentiability. This condition, which
may seem very strict, is still insufficient to ensure the differentiability or even
only the continuity of f at the point in question 4o •
20 - Differentiability of functions of class C l
These difficulties disappear when f is of class C 1 .
Suppose that f is such a function and, following an idea of Euler's in
another context, let us calculate
(20.1) f(a + u, b + v) - f(a, b) =
= (f(a + u, b + v) - f(a, b + v)] + [f(a, b + v) - f(a, b)].
We need to compare this difference to Dd(a, b)u + D2!(a, b)v = c:u + dv
for u and v small. For b and v given, the function g(x) = f(x, b + v) has,
by hypothesis, a derivative g'(x) = Dd(x, b + v) for all x near a, with
g'(a) = Dd(a, b + v). Then, by (16.8) or (16.9),
(20.2)
Ig(a + u) - g(a) - Dd(a, b + v)ul :::;
:::;Iul. sup IDd(a+tu,b+v)-Dd(a,b+v)l·
O~t:51
Since Dd is continuous at the point (a, b), the difference IDd(a + tu, b +
v) - Dd(a, b + v)1 is :::; r for any t E [0,1] if lui and Ivl are sufficiently small.
The right hand side of (2) is thus o(u) when (u,v) tends to (0,0).
Similarly
(20.3)
If(a, b + v) - f(a, b) - D2!(a, b)vl :::;
:::; Ivl sup ID2 f(a, b + tv) - D2!(a, b)l,
09:51
40 Counterexample: I(x,y) = x 2 Y/(X 4 + y2) if (x,y) f= (0,0), 1(0,0) = 0. The
function has derivatives in all directions at the origin [calculate the limit of
I(tu, tv)/t when t --+ 0, paying particular attention to the case where v = 0]. It
is not continuous at the origin since for all a E R, I(x, ax 2 ) tends to (and is even
equal to) a/(l + a 2 ) instead of tending to ° as it would if 1 were continuous. See
the graph in R3 of the function in HOOrer and Wanner, p. 303.
III - Convergence: Continuous variables
For functions of one variable, derivability and differentiability are equivalent properties; not so in the general case, for a good and simple reason.
Consider what happens when one traverses a line passing through (a, b), i.e.
in the direction of a given vector h = (u, v); by (2')
(19.5)
fez + th) = fez) + t· Df(z)h + oCt),
so that f is differentiable in any direction originating at (a, b), with
(19.5')
d
Df(z)h = d/(z + th) at t = 0.
The existence at the point (a, b) of derivatives in all the directions originating
from (a, b) is thus necessary to ensure differentiability. This condition, which
may seem very strict, is still insufficient to ensure the differentiability or even
only the continuity of f at the point in question 4o •
20 - Differentiability of functions of class C l
These difficulties disappear when f is of class C 1 .
Suppose that f is such a function and, following an idea of Euler's in
another context, let us calculate
(20.1) f(a + u, b + v) - f(a, b) =
= (f(a + u, b + v) - f(a, b + v)] + [f(a, b + v) - f(a, b)].
We need to compare this difference to Dd(a, b)u + D2!(a, b)v = c:u + dv
for u and v small. For b and v given, the function g(x) = f(x, b + v) has,
by hypothesis, a derivative g'(x) = Dd(x, b + v) for all x near a, with
g'(a) = Dd(a, b + v). Then, by (16.8) or (16.9),
(20.2)
Ig(a + u) - g(a) - Dd(a, b + v)ul :::;
:::;Iul. sup IDd(a+tu,b+v)-Dd(a,b+v)l·
O~t:51
Since Dd is continuous at the point (a, b), the difference IDd(a + tu, b +
v) - Dd(a, b + v)1 is :::; r for any t E [0,1] if lui and Ivl are sufficiently small.
The right hand side of (2) is thus o(u) when (u,v) tends to (0,0).
Similarly
(20.3)
If(a, b + v) - f(a, b) - D2!(a, b)vl :::;
:::; Ivl sup ID2 f(a, b + tv) - D2!(a, b)l,
09:51
40 Counterexample: I(x,y) = x 2 Y/(X 4 + y2) if (x,y) f= (0,0), 1(0,0) = 0. The
function has derivatives in all directions at the origin [calculate the limit of
I(tu, tv)/t when t --+ 0, paying particular attention to the case where v = 0]. It
is not continuous at the origin since for all a E R, I(x, ax 2 ) tends to (and is even
equal to) a/(l + a 2 ) instead of tending to ° as it would if 1 were continuous. See
the graph in R3 of the function in HOOrer and Wanner, p. 303.