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5.18 on p. 251).Example 10 Let f (x) = a x , 0 < a, a = 1. Since f (x) = a x ln2 a > 0, the exponential function a x is strictly convex (downward) on R for any allowable value ofthe base a (see Fig. 5.12).lExample 11 For the function f (x) = loga x we have f (x) = − x 2 ln, so that theafunction is strictly convex (downward) if 0 < a < 1, and strictly convex upward if1 < a (see Fig.
5.13).Example 12 Let us study the convexity of f (x) = sin x (see Fig. 5.14).Since f (x) = − sin x, we have f (x) < 0 on the intervals π · 2k < x <π(2k + 1) and f (x) > 0 on π(2k − 1) < x < π · 2k, where k ∈ Z. It follows2445Differential CalculusFig. 5.12Fig. 5.13Fig. 5.14from this, for example, that the arc of the graph of sin x on the closed interval0 ≤ x ≤ π2 lies above the chord it subtends everywhere except at the endpoints;therefore sin x > π2 x for 0 < x < π2 .We now point out another characteristic of a convex function, geometricallyequivalent to the statement that a convex region of the plane lies entirely on oneside of a tangent line to its boundary.Proposition 6 A function f : ]a, b[ → R that is differentiable on the open interval]a, b[ is convex (downward) on ]a, b[ if and only if its graph contains no pointsbelow any tangent drawn to it.
In that case, a necessary and sufficient condition forstrict convexity is that all points of the graph except the point of tangency lie strictlyabove the tangent line.Proof Necessity. Let x0 ∈ ]a, b[. The equation of the tangent line to the graph at(x0 , f (x0 )) has the formy = f (x0 ) + f (x0 )(x − x0 ),so thatf (x) − y(x) = f (x) − f (x0 ) − f (x0 )(x − x0 ) = f (ξ ) − f (x0 ) (x − x0 ),5.4 Differential Calculus Used to Study Functions245where ξ is a point between x and x0 .
Since f is convex, the function f (x) isnondecreasing on ]a, b[ and so the sign of the difference f (ξ ) − f (x0 ) is the sameas the sign of the difference x − x0 . Therefore f (x) − y(x) ≥ 0 at each point x ∈]a, b[. If f is strictly convex, then f is strictly increasing on ]a, b[ and so f (x) −y(x) > 0 for x ∈ ]a, b[ and x = x0 .Sufficiency.
If the inequalityf (x) − y(x) = f (x) − f (x0 ) − f (x0 )(x − x0 ) ≥ 0(5.94)holds for any points x, x0 ∈ ]a, b[, thenf (x) − f (x0 )≤ f (x0 )x − x0for x < x0 ,f (x) − f (x0 )≥ f (x0 )x − x0for x0 < x.Thus, for any triple of points x1 , x, x2 ∈ ]a, b[ such that x1 < x < x2 we obtainf (x) − f (x1 ) f (x2 ) − f (x),≤x − x1x2 − xand strict inequality in (5.94) implies strict inequality in this last relation, which, aswe see, is the same as the definition (5.93) for convexity of a function.Let us now consider some examples.Example 13 The function f (x) = ex is strictly convex. The straight line y = x + 1is tangent to the graph of this function at (0, 1), since f (0) = e0 = 1 and f (0) =ex |x=0 = 1.
By Proposition 6 we conclude that for any x ∈ Rex ≥ 1 + x,and this inequality is strict for x = 0.Example 14 Similarly, using the strict upward convexity of ln x, one can verify thatthe inequalityln x ≤ x − 1holds for x > 0, the inequality being strict for x = 1.In constructing the graphs of functions, it is useful to distinguish the points ofinflection of a graph.Definition 3 Let f : U (x0 ) → R be a function defined and differentiable on aneighborhood U (x0 ) of x0 ∈ R.
If the function is convex downward (resp. upward)on the set Ů − (x0 ) = {x ∈ U (x0 ) | x < x0 } and convex upward (resp. downward) onŮ + (x0 ) = {x ∈ U (x0 ) | x > x0 }, then (x0 , f (x0 )) is called a point of inflection ofthe graph.2465Differential CalculusThus when we pass through a point of inflection, the direction of convexity ofthe graph changes. This means, in particular, that at the point (x0 , f (x0 )) the graphof the function passes from one side of the tangent line to the other.An analytic criterion for the abscissa x0 of a point of inflection is easy to surmise,if we compare Proposition 5 with Proposition 3. To be specific, one can say that if fis twice differentiable at x0 , then since f (x) has either a maximum or a minimumat x0 , we must have f (x0 ) = 0.Now if the second derivative f (x) is defined on U (x0 ) and has one sign everywhere on Ů − (x0 ) and the opposite sign everywhere on Ů + (x0 ), this is sufficient forf (x) to be monotonic in Ů − (x0 ) and monotonic in Ů + (x0 ) but with the oppositemonotonicity.
By Proposition 5, a change in the direction of convexity occurs at(x0 , f (x0 )), and so that point is a point of inflection.Example 15 When considering the function f (x) = sin x in Example 12 we foundthe regions of convexity and concavity for its graph. We shall now show that thepoints of the graph with abscissas x = πk, k ∈ Z, are points of inflection.Indeed, f (x) = − sin x, so that f (x) = 0 at x = πk, k ∈ Z. Moreover, f (x)changes sign as we pass through these points, which is a sufficient condition for apoint of inflection (see Fig. 5.14 on p.
244).Example 16 It should not be thought that the passing of a curve from one side ofits tangent line to the other at a point is a sufficient condition for the point to be apoint of inflection. It may, after all, happen that the curve does not have any constantconvexity on either a left- or a right-hand neighborhood of the point. An example iseasy to construct, by improving Example 5, which was given for just this purpose.Let!2x 3 + x 3 sin x12 for x = 0,f (x) =0for x = 0.Then x 3 ≤ f (x) ≤ 3x 3 for 0 ≤ x and 3x 3 ≤ f (x) ≤ x 3 for x ≤ 0, so that thegraph of this function is tangent to the x-axis at x = 0 and passes from the lowerhalf-plane to the upper at that point.
At the same time, the derivative of f (x)!6x 2 + 3x 2 sin x12 − 2 cos x12 for x = 0,f (x) =0for x = 0is not monotonic in any one-sided neighborhood of x = 0.In conclusion, we return again to the definition (5.92) of a convex function andprove the following proposition.Proposition 7 (Jensen’s inequality)19 If f : ]a, b[ → R is a convex function,x1 , . . . , xn are points of ]a, b[, and α1 , .
. . , αn are nonnegative numbers such that19 J.L.Jensen (1859–1925) – Danish mathematician.5.4 Differential Calculus Used to Study Functions247α1 + · · · + αn = 1, thenf (α1 x1 + · · · + αn xn ) ≤ α1 f (x1 ) + · · · + αn f (xn ).(5.95)Proof For n = 2, condition (5.95) is the same as the definition (5.92) of a convexfunction.We shall now show that if (5.95) is valid for n = m − 1, it is also valid for n = m.For the sake of definiteness, assume that αn = 0 in the set α1 , . . . , αn .
Then β =α2 + · · · + αn > 0 and αβ2 + · · · + αβn = 1. Using the convexity of the function, wefindα2αn≤x2 + · · · + xnf (α1 x1 + · · · + αn xn ) = f α1 x1 + βββα2αn≤ α1 f (x1 ) + βfx2 + · · · + xn ,ββsince α1 + β = 1 and ( αβ2 x1 + · · · + αβn xn ) ∈ ]a, b[.By the induction hypothesis, we now haveα2α2αnαnfx2 + · · · + xn ≤ f (x2 ) + · · · + f (xn ).ββββConsequentlyα2αnx2 + · · · + xn ≤f (α1 x1 + · · · + αn xn ) ≤ α1 f (x1 ) + βfββ≤ α1 f (x1 ) + α2 f (x2 ) + · · · + αn f (xn ).By induction we now conclude that (5.95) holds for any n ∈ N. (For n = 1, relation(5.95) is trivial.)We remark that, as the proof shows, a strict Jensen’s inequality corresponds tostrict convexity, that is, if the numbers α1 , . .
. , αn are nonzero, then equality holdsin (5.95) if and only if x1 = · · · = xn .For a function that is convex upward, of course, the opposite relation to inequality(5.95) is obtained:f (α1 x1 + · · · + αn xn ) ≥ α1 f (x1 ) + · · · + αn f (xn ).(5.96)Example 17 The function f (x) = ln x is strictly convex upward on the set of positive numbers, and so by (5.96)α1 ln x1 + · · · + αn ln xn ≤ ln(α1 x1 + · · · + αn xn )or,2485Differential Calculusx1α1 · · · xnαn ≤ α1 x1 + · · · + αn xnfor xi ≥ 0, αi ≥ 0, i = 1, . . . , n, and ni=1 αi = 1.In particular, if α1 = · · · = αn = n1 , we obtain the classical inequality√x1 + · · · + xnnx1 · · · xn ≤n(5.97)(5.98)between the geometric and arithmetic means of n nonnegative numbers. Equalityholds in (5.98), as noted above, only when x1 = x2 = · · · = xn .
If we set n = 2,α1 = p1 , α2 = q1 , x1 = a, x2 = b in (5.97), we again obtain the known equality(5.86).Example 18 Let f (x) = x p , x ≥ 0, p > 1. Since such a function is convex, we have n pnpαi xi≤αi xi .i=1i=1q q−1/(p−1) nqp, αi = bi ( ni=1 bi )−1 , and xi = ai biSetting q = p−1i=1 bi here, we obtain Hölder’s inequality (5.88): n 1/p n 1/qn p qai bi ≤aibi,i=1i=1i=1where p1 + q1 = 1 and p > 1.For p < 1 the function f (x) = x p is convex upward, and so analogous reasoningcan be carried out in Hölder’s other inequality (5.89).5.4.4 L’Hôpital’s RuleWe now pause to discuss a special, but very useful device for finding the limit of aratio of functions, known as l’Hôpital’s rule.20Proposition 8 (l’Hôpital’s rule) Suppose the functions f : ]a, b[ → R and g :]a, b[ → R are differentiable on the open interval ]a, b[ (−∞ ≤ a < b ≤ +∞) withg (x) = 0 on ]a, b[ andf (x)→ A as x → a + 0 (−∞ ≤ A ≤ +∞).g (x)20 G.F. de l’Hôpital’s (1661–1704) – French mathematician, a capable student of Johann Bernoulli,a marquis for whom the latter wrote the first textbook of analysis in the years 1691–1692.
Theportion of this textbook devoted to differential calculus was published in slightly altered form byl’Hôpital’s under his own name. Thus “l’Hôpital’s rule” is really due to Johann Bernoulli.5.4 Differential Calculus Used to Study Functions249Thenf (x)→Ag(x)as x → a + 0in each of the following two cases:10 (f (x) → 0) ∧ (g(x) → 0) as x → a + 0,or20 g(x) → ∞ as x → a + 0.A similar assertion holds as x → b − 0.L’Hôpital’s rule can be stated succinctly, but not quite accurately, as follows. Thelimit of a ratio of functions equals the limit of the ratio of their derivatives if thelatter exists.Proof If g (x) = 0, we conclude on the basis of Rolle’s theorem that g(x) is strictlymonotonic on ]a, b[.
Hence, shrinking the interval ]a, b[ if necessary by shiftingtoward the endpoint a, we can assume that g(x) = 0 on ]a, b[. By Cauchy’s theorem,for x, y ∈ ]a, b[ there exists a point ξ ∈ ]a, b[ such thatf (x) − f (y) f (ξ )= .g(x) − g(y)g (ξ )Let us rewrite this equality in a form convenient for us at this point:g(y)f (x) f (y) f (ξ )=+ 1−.g(x)g(x)g (ξ )g(x)As x → a + 0, we shall make y tend to a + 0 in such a way thatf (y)→ 0 andg(x)g(y)→ 0.g(x)This is obviously possible under each of the two hypotheses 10 and 20 that we areconsidering. Since ξ lies between x and y, we also have ξ → a + 0.
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