Question

state and prove Schwartz's theorem

Answer 1

SCHWARZ’ THEOREM ABOUT MIXED PARTIAL DERIVATIVES,
WITH SOME PRELIMINARY COMMENTS AND SOME REMARKS
ABOUT WORKING WITH DERIVATIVES ON A COMPUTER.
New version. December 2008.
The theorem of H. A, Schwarz gives sufficient conditions for the two “mixed”
derivatives ∂
2
∂x∂y f and ∂
2
∂y∂x f of a function f to be equal. Most functions which
are defined by some “nice” formula satisfy these conditions and so they satisfy
∂
2
∂x∂y f =
∂
2
∂y∂x f . But there exist functions which do not satisfy the conditions and
also ∂
2
∂x∂y f 6=
∂
2
∂y∂x f at least at some points.
Before we state and prove the theorem, let us try to get some intuitive feeling
for these mixed derivatives. Let us also be explicit about the notation for them.
If we differentiate f first with respect to x and then with respect to y we get the
derivative ∂
∂y
∂f
∂x
(if it exists). It is more usually denoted by ∂
2
f
∂y∂x . Another logical
notation for this same derivative is (f
0
x
)
0
y
or f
00
xy. Alternatively, if we differentiate
first with respect to y and then x we get ∂
2
∂x∂y f = f
00
yx (if it exists).
Let f be a function defined in some open subset E of R
2
. Suppose that the
points A = (x0, y0), H = (x0 +h, y0), K = (x0, y0 +k) and B = (x0 +h, y0 +k) are
all in E. Let us try to approximate the values of the derivatives f
0
x
, f
0
y
, f
00
xy and f
00
yx
at the point A in terms of the values of f at A, B, H and K. We suppose (at least
for now) that x0, y0, h and k are constants. We think of h and k as being “small”
and positive.
Here please draw a picture of the the four points A, B, H and K and the rectangle
they form, of width h and height k. Things will happen later, at the vertices and
also inside this rectangle.
Let us think a little about the meeting between computers and derivatives of
functions: When we work with a computer, it cannot deal completely with func-
tions defined on infinite subsets of R or R
n. The computer has finite memory so it
can store and process the values of our function f only on some finite subset of E.2
One natural subset to use here is a “network” of points of the form (x0+Nh, y0+M k)
where N and M are integers.
Since the computer “knows” the values of the function f only at a finite set of
discrete points, there is no way it can calculate its derivatives exactly. But if h is
small, it is reasonable to expect that f
0
x
(A) = f
0
x
(x0, y0) might be approximately
equal to f(x0+h,y0)−f(x0,y0)
h
. In other words, we can write
(0.1) f
0
x
(A) ≈
f(H) − f(A)
h
.
Similarly, if k also is small, we can also write
(0.2) f
0
y
(A) ≈
f(K) − f(A)
k
.
Furthermore, by the same reasoning,
(0.3) f
0
x
(K) ≈
f(B) − f(K)
h
and
(0.4) f
0
y
(H) ≈
f(B) − f(H)
k
.
Now let us make some similar approximate calculations when, instead of the func-
tion f, we consider the function f
0
x or the function f
0
y
. We get, for example,
(0.5) (f
0
x
)
0
y
(A) ≈
f
0
x
(K) − f
0
x
(A)
k
and also,
(0.6) (f
0
y
)
0
x
(A) ≈
f
0
y
(H) − f
0
y
(A)
h
.
Now let us be very very optimistic, and hope that the approximations in the
above calculations are so good that if in (0.5) and (0.6) if we substitute approximate
values from the previous calculations we will still get expressions which is not too
far from the actual values of the derivatives f
00
xy and f
00
yx at A. If this optimism is
justified we will have
(0.7) (f
0
x
)
0
y
(A) ≈
f
0
x
(K) − f
0
x
(A)
k
≈
f(B)−f(K)
h −
f(H)−f(A)
h
k
,
and also
(0.8) (f
0
y
)
0
x
(A) ≈
f
0
y
(H) − f
0
y
(A)
h
≈
f(B)−f(H)
k −
f(K)−f(A)
k
h
.
But now we can see that the right hand side of both of these expressions is the
same. It equals
f(A) + f(B) − f(H) − f(K)
hk .
This suggests that maybe the two derivatives f
00
xy and f
00
yx might be equal at A
and maybe even at every other point where they both exist. We are now ready to
check this precisely.
Remark: It should be stressed that all the preceding “calculations” are only
approximate and intuitive. If someone asks you or me “What does the symbol ≈
really mean?” how can we answer them?3
As we already said above, there are some “exotic” functions which satisfy f
00
xy 6=
f
00
yx at at least some of the points where these derivatives exist. Let us now consider
the big class of functions for which such problems do not arise.
THEOREM (H. A. Schwarz). Suppose that f is a function of two variables
such that f
00
xy and f
00
yx both exist and are continuous at some point (x0, y0). Then
f
00
xy(x0, y0) = f
00
yx(x0, y0).
Remark: Note that the conditions of the theorem imply that f
00
xy and f
00
yx must
be defined in some neighbourhood of (x0, y0), and so this of course also implies that
f
0
x
, f
0
y and f itself must also be defined at every point of some neighbourhood of
(x0, y0).