4.6. MIXED PARTIAL DERIVATIVES 97

≤

∣∣∣∣∣∫ 1

0

(∂ f∂xi

(x+

n

∑k=i+1

vkek + tviei

)− ∂ f

∂xi(x)

)∣∣∣∣∣ |v|and by continuity, the integral converges to 0 as |v| → 0. This follows from uniform conti-nuity of ∂ f

∂xion a sufficiently small closed ball containing x. ■

4.6 Mixed Partial DerivativesUnder certain conditions the mixed partial derivatives will always be equal. The simplecondition is that if they exist and are continuous, then they are equal. This astonishingfact is due to Euler in 1734 and was proved by Clairaut although not very well. The firstsatisfactory proof was by Hermann Schwarz in 1873. For reasons I cannot understand,calculus books seldom include a proof of this important result. It is not all that hard. It isbased on the mean value theorem for derivatives. Here it is.

Theorem 4.6.1 Suppose f : U ⊆ R2→ R where U is an open set on which fx, fy,fxy and fyx exist. Then if fxy and fyx are continuous at the point (x,y) ∈ U, it followsfxy (x,y) = fyx (x,y) .

Proof: Since U is open, there exists r > 0 such that B((x,y) ,r)⊆U. Now let |t| , |s|<r/2 and consider

∆(s, t)≡ 1st{

h(t)︷ ︸︸ ︷f (x+ t,y+ s)− f (x+ t,y)−

h(0)︷ ︸︸ ︷( f (x,y+ s)− f (x,y))}. (4.8)

Note that (x+ t,y+ s) ∈U because

|(x+ t,y+ s)− (x,y)|= |(t,s)|=(t2 + s2)1/2 ≤

(r2

4+

r2

4

)1/2

=r√2< r.

As implied above, h(t) ≡ f (x+ t,y+ s)− f (x+ t,y). Then, by the mean value theoremfrom calculus and the (one variable) chain rule,

∆(s, t) =1st(h(t)−h(0)) =

1st

h′ (αt) t =1s( fx (x+αt,y+ s)− fx (x+αt,y))

for some α ∈ (0,1) . Applying the mean value theorem again,

∆(s, t) = fxy (x+αt,y+β s)

where α,β ∈ (0,1).If the terms f (x+ t,y) and f (x,y+ s) are interchanged in 4.8, ∆(s, t) is also unchanged

and the above argument shows there exist γ,δ ∈ (0,1) such that

∆(s, t) = fyx (x+ γt,y+δ s) .

Letting (s, t)→ (0,0) and using the continuity of fxy and fyx at (x,y) ,

lim(s,t)→(0,0)

∆(s, t) = fxy (x,y) = fyx (x,y) .■

The following is obtained from the above by simply fixing all the variables except forthe two of interest.