558 Real Analysis
501.Using the Leibniz–Newton fundamental theorem of calculus, we can write
f (x, y)−f( 0 , 0 )=∫x0∂f
∂x(s, 0 )ds+∫y0∂f
∂y(x, t)dt.Using the changes of variabless=xσandt=yτ, and the fact thatf( 0 , 0 )=0, we
obtain
f (x, y)=x∫ 1
0∂f
∂x(xσ, 0 )dσ+y∫ 1
0∂f
∂y(x, yτ )dτ.Hence if we setg 1 (x, y) =
∫ 1
0∂f
∂x(xσ,^0 )dσ andg^2 (x, y) =∫ 1
0∂f
∂y(x, yτ )dτ, then
f (x, y)=xg 1 (x, y)+yg 2 (x, y). Areg 1 andg 2 continuous? The answer is yes, and we
prove it only forg 1 , since for the other function the proof is identical. Our argument is
based on the following lemma.
Lemma.Ifφ:[a, b]→Ris continuous, then for every> 0 there isδ> 0 such that
whenever|x−y|<δ,we have|f(x)−f(y)|<.
Proof.The property is called uniform continuity; the word “uniform’’ signifies the fact
that the “δ’’ from the definition of continuity is the same for all points in[a, b].
We argue by contradiction. Assume that the property is not true. Then there exist
two sequences(xn)n≥ 1 and(yn)n≥ 1 such thatxn−yn→0, but|f(xn)−f(yn)|≥for
some>0. Because any sequence in[a, b]has a convergent subsequence, passing to
subsequences we may assume that(xn)nand(yn)nconverge to somecin[a, b]. Then by
the triangle inequality,
≤|f(xn)−f(yn)|≤|f(xn)−f(c)|+|f(c)−f(yn)|,which is absurd because the right-hand side can be made arbitrarily close to 0 by taking
nsufficiently large. This proves the lemma.
Returning to the problem, note that asx′ranges over a small neighborhood ofxand
σranges between 0 and 1, the numbersxσandx′σlie inside a small interval of the real
axis. Note also that|xσ−x′σ|≤|x−x′|when 0≤σ≤1. Combining these two facts
with the lemma, we see that for every>0, there existsδ>0 such that for|x−x′|<δ
we have
∣∣
∣∣∂f
∂x
(xσ, 0 )−
∂f
∂x
(x′σ, 0 )∣∣
∣∣<.
In this case,
∫ 1
0∣
∣∣
∣
∂f
∂x(xσ, 0 )−
∂f
∂x(x′σ, 0 )∣
∣∣
∣dσ < ,