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But then l = 1. Indeed, if that were not so, we could takea ball B(x(l); r) ⊂ G, in which f (x) = f (x(l)) = f (x0 ), and then by the continuityof the mapping t → x(t) find Δ > 0 such that x(t) ∈ B(x(l); r) for l ≤ t ≤ l + Δ.But then (f ◦ x)(t) = f (x(l)) = f (x0 ) for 0 ≤ t ≤ l + Δ, and so l = sup δ.Thus we have shown that (f ◦ x)(t) = f (x0 ) for any t ∈ [0, 1]. In particular(f ◦ x)(1) = f (x1 ) = f (x0 ), and we have verified that the values of the functionf : G → R are the same at any two points x0 , x1 ∈ G.8.4.2 A Sufficient Condition for Differentiability of a Functionof Several VariablesTheorem 2 Let f : U (x) → R be a function defined in a neighborhood U (x) ⊂ Rmof the point x = (x 1 , .

. . , x m ).∂f∂fIf the function f has all partial derivatives ∂x1 , . . . , ∂x m at each point of theneighborhood U (x) and they are continuous at x, then f is differentiable at x.Proof Without loss of generality we shall assume that U (x) is a ball B(x; r). Then,together with the points x = (x 1 , . . . , x m ) and x + h = (x 1 + h1 , . . .

, x m + hm ),the points (x 1 , x 2 + h2 , . . . , x m + hm ), . . . , (x 1 , x 2 , . . . , x m−1 , x m + hm ) and thelines connecting them must also belong to the domain U (x). We shall use this fact,applying the Lagrange theorem for functions of one variable in the following computation:f (x + h) − f (x) = f x 1 + h1 , . . . , x m + hm − f x 1 , .

. . , x m == f x 1 + h1 , . . . , x m + hm − f x 1 , x 2 + h2 , . . . , x m + hm ++ f x 1 , x 2 + h2 , . . . , x m + hm −− f x 1 , x 2 , x 3 + h3 , . . . , x m + hm + · · · ++ f x 1 , x 2 , . . . , x m−1 , x m + hm − f x 1 , . . . , x m =8.4 Real-valued Functions of Several Variables457= ∂1 f x 1 + θ 1 h1 , x 2 + h2 , .

. . , x m + hm h1 ++ ∂2 f x 1 , x 2 + θ 2 h2 , x 3 + h3 , . . . , x m + hm h2 + · · · ++ ∂m f x 1 , x 2 , . . . , x m−1 , x m + θ m hm hm .So far we have used only the fact that the function f has partial derivatives withrespect to each of its variables in the domain U (x).We now use the fact that these partial derivatives are continuous at x. Continuingthe preceding computation, we obtainf (x + h) − f (x) = ∂1 f x 1 , .

. . , x m h1 + α 1 h1 ++ ∂2 f x 1 , . . . , x m h2 + α 2 h2 + · · · ++ ∂m f x 1 , . . . , x m hm + α m hm ,where the quantities α1 , . . . , αm tend to zero as h → 0 by virtue of the continuity ofthe partial derivatives at the point x.But this means thatf (x + h) − f (x) = L(x)h + o(h) as h → 0,where L(x)h = ∂1 f (x 1 , . . . , x m )h1 + · · · + ∂m f (x 1 , . . . , x m )hm .It follows from Theorem 2 that if the partial derivatives of a function f : G → Rare continuous in the domain G ⊂ Rm , then the function is differentiable at thatpoint of the domain.Let us agree from now on to use the symbol C (1) (G; R), or, more simply, C (1) (G)to denote the set of functions having continuous partial derivatives in the domain G.8.4.3 Higher-Order Partial Derivatives∂fIf a function f : G → R defined in a domain G ⊂ Rm has a partial derivative ∂xi (x)with respect to one of the variables x 1 , .

. . , x m , this partial derivative is a function∂i f : G → R, which in turn may have a partial derivative ∂j (∂i f )(x) with respectto a variable x j .The function ∂j (∂i f ) : G → R is called the second partial derivative of f withrespect to the variables x i and x j and is denoted by one of the following symbols:∂j i f (x),∂ 2f(x).∂x j ∂x iThe order of the indices indicates the order in which the differentiation is carriedout with respect to the corresponding variables.We have now defined partial derivatives of second order.4588 Differential Calculus in Several VariablesIf a partial derivative of order k∂i1 ···ik f (x) =∂kf(x)∂x i1 · · · ∂x ikhas been defined, we define by induction the partial derivative of order k + 1 by therelation∂ii1 ···ik f (x) := ∂i (∂i1 ···ik f )(x).At this point a question arises that is specific for functions of several variables:Does the order of differentiation affect the partial derivative computed?Theorem 3 If the function f : G → R has partial derivatives∂ 2f(x),∂x i ∂x j∂ 2f(x)∂x j ∂x iin a domain G, then at every point x ∈ G at which both partial derivatives arecontinuous, their values are the same.Proof Let x ∈ G be a point at which both functions ∂ij f : G → R and ∂j i f : G →R are continuous.

From this point on all of our arguments are carried out in thecontext of a ball B(x; r) ⊂ G, r > 0, which is a convex neighborhood of the point x.We wish to verify that∂ 2f ∂ 2f 1x , . . . , xm = j i x1, . . . , xm .ij∂x ∂x∂x ∂xSince only the variables x i and x j will be changing in the computations to follow, we shall assume for the sake of brevity that f is a function of two variablesf (x 1 , x 2 ), and we need to verify that∂ 2f 1 2∂ 2f 1 2x=x ,x ,,x∂x 1 ∂x 2∂x 2 ∂x 1if the two functions are both continuous at the point (x 1 , x 2 ).Consider the auxiliary functionF h1 , h2 = f x 1 + h1 , x 2 + h2 − f x 1 + h1 , x 2 − f x 1 , x 2 + h2 + f x 1 , x 2 ,where the displacement h = (h1 , h2 ) is assumed to be sufficiently small, namely sosmall that x + h ∈ B(x; r).If we regard F (h1 , h2 ) as the differenceF h1 , h2 = ϕ(1) − ϕ(0),8.4 Real-valued Functions of Several Variables459where ϕ(t) = f (x 1 + th1 , x 2 + h2 ) − f (x 1 + th1 , x 2 ), we find by Lagrange’s theorem thatF h1 , h2 = ϕ (θ1 ) = ∂1 f x 1 + θ1 h1 , x 2 + h2 − ∂f x 1 + θ1 h1 , x 2 h1 .Again applying Lagrange’s theorem to this last difference, we find thatF h1 , h2 = ∂21 f x 1 + θ1 h1 , x 2 + θ2 h2 h2 h1 .(8.55)If we now represent F (h1 , h2 ) as the differenceF h1 , h2 = ϕ̃(1) − ϕ̃(0),where ϕ̃(t) = f (x 1 + h1 , x 2 + th2 ) − f (x 1 , x 2 + th2 ), we find similarly thatF h1 , h2 = ∂12 f x 1 + θ̃1 h1 , x 2 + θ̃2 h2 h1 h2 .(8.56)Comparing (8.55) and (8.56), we conclude that∂21 f x 1 + θ1 h1 , x 2 + θ2 h2 = ∂12 f x 1 + θ̃1 h1 , x 2 + θ̃2 h2 ,(8.57)where θ1 , θ2 , θ̃1 , θ̃2 ∈ ]0, 1[.

Using the continuity of the partial derivatives at thepoint (x 1 , x 2 ), as h → 0, we get the equality we need as a consequence of (8.57).∂21 f x 1 , x 2 = ∂12 f x 1 , x 2 .We remark that without additional assumptions we cannot say in general that∂ij f (x) = ∂j i f (x) if both of the partial derivatives are defined at the point x (seeProblem 2 at the end of this section).Let us agree to denote the set of functions f : G → R all of whose partial derivatives up to order k inclusive are defined and continuous in the domain G ⊂ Rm bythe symbol C (k) (G; R) or C (k) (G).As a corollary of Theorem 3, we obtain the following.Proposition 1 If f ∈ C (k) (G; R), the value ∂i1 ...ik f (x) of the partial derivative isindependent of the order i1 , .

. . , ik of differentiation, that is, remains the same forany permutation of the indices i1 , . . . , ik .Proof In the case k = 2 this proposition is contained in Theorem 3.Let us assume that the proposition holds up to order n inclusive. We shall showthat then it also holds for order n + 1.But ∂i1 i2 ···in+1 f (x) = ∂i1 (∂i2 ···in+1 f )(x). By the induction assumption the indicesi2 , . . . , in+1 can be permuted without changing the function ∂i2 ···in+1 f (x), and hencewithout changing ∂i1 ···in+1 f (x). For that reason it suffices to verify that one canalso permute, for example, the indices i1 and i2 without changing the value of thederivative ∂i1 i2 ···in+1 f (x).4608 Differential Calculus in Several VariablesSince∂i1 i2 ···in+1 f (x) = ∂i1 i2 (∂i3 ···in+1 f )(x),the possibility of this permutation follows immediately from Theorem 3. By theinduction principle Proposition 1 is proved.Example 1 Let f (x) = f (x 1 , x 2 ) be a function of class C (k) (G; R).Let h = (h1 , h2 ) be such that the closed interval [x, x + h] is contained in thedomain G.

We shall show that the functionϕ(t) = f (x + th),which is defined on the closed interval [0, 1], belongs to class C (k) [0, 1] and find itsderivative of order k with respect to t.We haveϕ (t) = ∂1 f x 1 + th1 , x 2 + th2 h1 + ∂2 f x 1 + th1 , x 2 + th2 h2 ,ϕ (t) = ∂11 f (x + th)h1 h1 + ∂21 f (x + th)h2 h1 ++ ∂12 f (x + th)h1 h2 + ∂22 f (x + th)h2 h2 = 2 2= ∂11 f (x + th) h1 + 2∂12 f (x + th)h1 h2 + ∂22 f (x + th) h2 .These relations can be written as the action of the operator (h1 ∂1 + h2 ∂2 ):ϕ (t) = h1 ∂1 + h2 ∂2 f (x + th) = hi ∂i f (x + th),2ϕ (t) = h1 ∂1 + h2 ∂2 f (x + th) = hi1 hi2 ∂i1 i2 f (x + th).By induction we obtainkϕ (k) (t) = h1 ∂1 + h2 ∂2 f (x + th) = hi1 · · · hik ∂i1 ···ik f (x + th)(summation over all sets i1 , . .

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