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The set of all points of R2satisfying this condition is the unit circle (Fig. 8.4).8.5 The Implicit Function Theorem481Fig. 8.4The presence of the relation (8.80) shows that after fixing one of the coordinates,for example, x, we can no longer choose the second coordinate arbitrarily. Thusrelation (8.80) determines the dependence of y on x. We are interested in the question of the conditions under which the implicit relation (8.80) can be solved as anexplicit functional dependence y = y(x).Solving Eq. (8.80) with respect to y, we find that#(8.81)y = ± 1 − x2,that is, to each value of x such that |x| < 1, there are actually two admissible valuesof y.
In forming a functional relation y = y(x) satisfying relation (8.80) one cannotgive preference to either of the values (8.81) without invoking √additional requirements. For example, the function y(x) that assumes the√value + 1 − x 2 at rationalpoints of the closed interval [−1, 1] and the value − 1 − x 2 at irrational pointsobviously satisfies (8.80).It is clear that one can create infinitely many functional relations satisfying (8.80)by varying this example.The question whether the set defined in R2 by (8.80) is the graph of a functiony = y(x) obviously has a negative answer, since from the geometric point of viewit is equivalent to the question whether it is possible to establish a one-to-one directprojection of a circle into a line.But observation (see Fig.
8.4) suggests that nevertheless, in a neighborhood of aparticular point (x0 , y0 ) the arc projects in a one-to-one manner into the x-axis, andthat it can be represented uniquely as y = y(x), where x → y(x) is a continuousfunction defined in a neighborhood of the point x0 and assuming the value y0 at x0 .In this aspect, the only bad points are (−1, 0) and (1, 0), since no arc of the circlehaving them as interior points projects in a one-to-one manner into the x-axis. Evenso, neighborhoods of these points on the circle are well situated relative to the yaxis, and can be represented as the graph of a function x = x(y) that is continuousin a neighborhood of the point 0 and assumes the value −1 or 1 according as the arcin question contains the point (−1, 0) or (1, 0).How is it possible to find out analytically when a geometric locus of points defined by a relation of the type (8.80) can be represented in the form of an explicitfunction y = y(x) or x = x(y) in a neighborhood of a point (x0 , y0 ) on the locus?4828 Differential Calculus in Several VariablesWe shall discuss this question using the following, now familiar, method.
Wehave a function F (x, y) = x 2 + y 2 − 1. The local behavior of this function in aneighborhood of a point (x0 , y0 ) is well described by its differentialFx (x0 , y0 )(x − x0 ) + Fy (x0 , y0 )(y − y0 ),sinceF (x, y) = F (x0 , y0 ) + Fx (x0 , y0 )(x − x0 ) ++ Fy (x0 , y0 )(y − y0 ) + o |x − x0 | + |y − y0 |as (x, y) → (x0 , y0 ).If F (x0 , y0 ) = 0 and we are interested in the behavior of the level curveF (x, y) = 0of the function in a neighborhood of the point (x0 , y0 ), we can judge that behaviorfrom the position of the (tangent) lineFx (x0 , y0 )(x − x0 ) + Fy (x0 , y0 )(y − y0 ) = 0.(8.82)If this line is situated so that its equation can be solved with respect to y, then,since the curve F (x, y) = 0 differs very little from this line in a neighborhood ofthe point (x0 , y0 ), we may hope that it also can be written in the form y = y(x) insome neighborhood of the point (x0 , y0 ).The same can be said about local solvability of F (x, y) = 0 with respect to x.Writing Eq.
(8.82) for the specific relation (8.80), we obtain the following equation for the tangent line:x0 (x − x0 ) + y0 (y − y0 ) = 0.This equation can always be solved for y when y0 = 0, that is, at all points of thecircle (8.80) except (−1, 0) and (1, 0). It is solvable with respect to x at all pointsof the circle except (0, −1) and (0, 1).8.5.2 An Elementary Version of the Implicit Function TheoremIn this section we shall obtain the implicit function theorem by a very intuitive, butnot very constructive method, one that is adapted only to the case of real-valuedfunctions of real variables. The reader can become familiar with another methodof obtaining this theorem, one that is in many ways preferable, and with a moredetailed analysis of its structure in Chap.
10 (Part 2), and also in Problem 4 at theend of the section.The following proposition is an elementary version of the implicit function theorem.8.5 The Implicit Function Theorem483Proposition 1 If the function F : U (x0 , y0 ) → R defined in a neighborhoodU (x0 , y0 ) of the point (x0 , y0 ) ∈ R2 is such that10 F ∈ C (p) (U ; R), where p ≥ 1,20 F (x0 , y0 ) = 0,30 Fy (x0 , y0 ) = 0,then there exist a two-dimensional interval I = Ix × Iy whereIx = x ∈ R |x − x0 | < α ,Iy = y ∈ R |y − y0 | < β ,that is a neighborhood of the point (x0 , y0 ) contained in U (x0 , y0 ), and a functionf ∈ C (p) (Ix ; Iy ) such thatF (x, y) = 0 ⇔ y = f (x),(8.83)for any point (x, y) ∈ Ix × Iy and the derivative of the function y = f (x) at thepoints x ∈ Ix can be computed from the formula*−1 ) *) Fx x, f (x) .(8.84)f (x) = − Fy x, f (x)Before taking up the proof, we shall give several possible reformulations of theconclusion (8.83), which should bring out the meaning of the relation itself.Proposition 1 says that under hypotheses 10 , 20 , and 30 the portion of the setdefined by the relation F (x, y) = 0 that belongs to the neighborhood Ix × Iy of thepoint (x0 , y0 ) is the graph of a function f : Ix → Iy of class C (p) (Ix ; Iy ).In other words, one can say that inside the neighborhood I of the point (x0 , y0 )the equation F (x, y) = 0 has a unique solution for y, and the function y = f (x) isthat solution, that is, F (x, f (x)) ≡ 0 on Ix .It follows in turn from this that if y = f˜(x) is a function defined on Ix that isknown to satisfy the relation F (x, f˜(x)) ≡ 0 on Ix , f˜(x0 ) = y0 , and this function iscontinuous at the point x0 ∈ Ix , then there exists a neighborhood Δ ⊂ Ix of x0 suchthat f˜(Δ) ⊂ Iy , and then f˜(x) ≡ f (x) for x ∈ Δ.Without the assumption that the function f˜ is continuous at the point x0 and thecondition f˜(x0 ) = y0 , this last conclusion could turn out to be incorrect, as can beseen from the example of the circle already studied.Let us now prove Proposition 1.Proof Suppose for definiteness that Fy (x0 , y0 ) > 0.
Since F ∈ C (1) (U ; R), it follows that Fy (x, y) > 0 also in some neighborhood of (x0 , y0 ). In order to avoid introducing new notation, we can assume without loss of generality that Fy (x, y) > 0at every point of the original neighborhood U (x0 , y0 ).Moreover, shrinking the neighborhood U (x0 , y0 ) if necessary, we can assumethat it is a disk of radius r = 2β > 0 with center at (x0 , y0 ).Since Fy (x, y) > 0 in U , the function F (x0 , y) is defined and monotonically increasing as a function of y on the closed interval y0 −β ≤ y ≤ y0 +β.
Consequently,F (x0 , y0 − β) < F (x0 , y0 ) = 0 < F (x0 , y0 + β).4848 Differential Calculus in Several VariablesBy the continuity of the function F in U , there exists a positive number α < β suchthat the relationsF (x, y0 − β) < 0 < F (x, y0 + β)hold for |x − x0 | ≤ α.We shall now show that the rectangle I = Ix × Iy , whereIx = x ∈ R |x − x0 | < α ,Iy = y ∈ R |y − y0 | < β ,is the required two-dimensional interval in which relation (8.83) holds.For each x ∈ Ix we fix the vertical closed interval with endpoints (x, y0 − β),(x, y0 + β). Regarding F (x, y) as a function of y on that closed interval, we obtain a strictly increasing continuous function that assumes values of opposite signat the endpoints of the interval. Consequently, for each x ∈ Ix , there is a uniquepoint y(x) ∈ Iy such that F (x, y(x)) = 0.
Setting y(x) = f (x), we arrive at relation(8.83).We now establish that f ∈ C (p) (Ix ; Iy ).We begin by showing that the function f is continuous at x0 and that f (x0 ) = y0 .This last equality obviously follows from the fact that for x = x0 there is a uniquepoint y(x0 ) ∈ Iy such that F (x0 , y(x0 )) = 0. At the same time, F (x0 , y0 ) = 0, andso f (x0 ) = y0 .Given a number ε, 0 < ε < β, we can repeat the proof of the existence of thefunction f (x) and find a number δ, 0 < δ < α such that in the two-dimensionalinterval I˜ = I˜x × I˜y , whereI˜x = x ∈ R |x − x0 | < δ ,I˜y = y ∈ R |y − y0 | < ε ,the relation F (x, y) = 0 in I˜ ⇔ y = f˜(x), x ∈ I˜x(8.85)holds with a new function f˜ : I˜x → I˜y .But I˜x ⊂ Ix , I˜y − ⊂ Iy , and I˜ ⊂ I , and therefore it follows from (8.83) and (8.85)that f˜(x) ≡ f (x) for x ∈ I˜x ⊂ Ix . We have thus verified that |f (x) − f (x0 )| =|f (x) − y0 | < ε for |x − x0 | < δ.We have now established that the function f is continuous at the point x0 .
Butany point (x, y) ∈ I at which F (x, y) = 0 can also be taken as the initial point ofthe construction, since conditions 20 and 30 hold at that point. Carrying out thatconstruction inside the interval I , we would once again arrive via (8.83) at the corresponding part of the function f considered in a neighborhood of x. Hence thefunction f is continuous at x. Thus we have established that f ∈ C(Ix ; Iy ).We shall now show that f ∈ C (1) (Ix ; Iy ) and establish formula (8.84).Let the number Δx be such that x + Δx ∈ Ix . Let y = f (x) and y + Δy =f (x + Δx).
Applying the mean-value theorem to the function F (x, y) inside the8.5 The Implicit Function Theorem485interval I , we find that0 = F x + Δx, f (x + Δx) − F x, f (x) == F (x + Δx, y + Δy) − F (x, y) == Fx (x + θ Δx, y + θ Δy)Δx + Fy (x + θ Δx, y + θ Δy)Δy(0 < θ < 1),from which, taking account of the relation Fy (x, y) = 0 in I , we obtainΔyF (x + θ Δx, y + θ Δy)= − x.ΔxFy (x + θ Δx, y + θ Δy)(8.86)Since f ∈ C(Ix ; Iy ), it follows that Δy → 0 as Δx → 0, and, taking account ofthe relation F ∈ C (1) (U ; R), as Δx → 0 in (8.86), we obtainf (x) = −Fx (x, y),Fy (x, y)where y = f (x). Thus formula (8.84) is now established.By the theorem on continuity of composite functions, it follows from formula(8.84) that f ∈ C (1) (Ix ; Iy ).If F ∈ C (2) (U ; R), the right-hand side of formula (8.84) can be differentiatedwith respect to x, and we find thatf (x) = − + F · f (x)]F − F [F + F · f (x)][Fxxxyyx xyyy(Fy )2,(8.84 ) , F , and F are all computed at the point (x, f (x)).where Fx , Fy , FxxxyyyThus f ∈ C (2) (Ix ; Iy ) if F ∈ C (2) (U ; R).
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