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5.10Now consider two such media, and suppose that light propagates from point A1to A2 , as shown in Fig. 5.10.If c1 and c2 are the velocities of light in these media, the time required to traversethe path is11t (x) =h21 + x 2 +h22 + (a − x)2 .c1c2We now find the extremum of the function t (x):t (x) =xa−x11− = 0,c1 h2 + x 2 c2 h2 + (a − x)212which in accordance with the notation of the figure, yields c1−1 sin α1 = c2−1 sin α2 .It is clear from physical considerations, or directly from the form of the functiont (x), which increases without bound as x → ∞, that the point where t (x) = 0 is anabsolute minimum of the continuous function t (x).
Thus Fermat’s principle impliesα1c1the law of refraction sinsin a2 = c2 .Example 7 We shall show that for x > 0x α − αx + α − 1 ≤ 0,when 0 < α < 1,(5.84)x α − αx + α − 1 ≥ 0,when α < 0 or 1 < α.(5.85)Proof Differentiating the function f (x) = x α − αx + α − 1, we find f (x) =α(x α−1 − 1) and f (x) = 0 when x = 1. In passing through the point 1 the derivativepasses from positive to negative values if 0 < α < 1 and from negative to positivevalues if α < 0 or α > 1.
In the first case the point 1 is a strict maximum, and inthe second case a strict minimum (and, as follows from the monotonicity of f onthe intervals 0 < x < 1 and 1 < x, not merely a local minimum). But f (1) = 0 andhence both inequalities (5.84) and (5.85) are established.
In doing so, we have evenshown that both inequalities are strict if x = 1.We remark that if x is replaced by 1 + x, we find that (5.84) and (5.85) areextensions of Bernoulli’s inequality (Sect. 2.2; see also Problem 2 below), whichwe already know for a natural-number exponent α.5.4 Differential Calculus Used to Study Functions239By elementary algebraic transformations one can obtain a number of classicalinequalities of great importance for analysis from the inequalities just proved. Weshall now derive these inequalities.a. Young’s Inequalities16If a > 0 and b > 0, and the numbers p and q such that p = 0, 1, q = 0, 1 and11p + q = 1, thena 1/p b1/q ≤11a + b,pqif p > 1,(5.86)a 1/p b1/q ≥11a + b,pqif p < 1,(5.87)and equality holds in (5.86) and (5.87) only when a = b.Proof It suffices to set x =notation1q=1−aband α =1pin (5.84) and (5.85), and then introduce the1p.b.
Hölder’s Inequalities17Let xi ≥ 0, yi ≥ 0, i = 1, . . . , n, andnxi yi ≤ ni=1andni=1+1qni=1pxi= 1. Then 1/p ni=1xi yi ≥pxi1p 1/qqyifor p > 1,(5.88)for p < 1, p = 0.(5.89)i=1 1/p n 1/qqyii=1In the case p < 0 it is assumed in (5.89) that xi > 0 (i = 1, . . . , n). Equality ispppqpossible in (5.88) and (5.89) only when the vectors (x1 , . .
. , xn ) and (y1 , . . . , yn )are proportional.pProof Let us verify the inequality (5.88). Let X = ni=1 xi > 0 and Y =pqnxiyiqi=1 yi > 0. Setting a = X and b = Y in (5.86), we obtainpqxi yi1 yi1 xi+.≤1/p1/qp Xq YX Y16 W.H.17 O.Young (1882–1946) – British mathematician.Hölder (1859–1937) – German mathematician.2405Differential CalculusSumming these inequalities over i from 1 to n, we obtainni=1 xi yi≤ 1,X 1/p Y 1/qwhich is equivalent to relation (5.88).We obtain (5.89) similarly from (5.87).
Since equality occurs in (5.86) and (5.87)only when a = b, we conclude that it is possible in (5.88) and (5.89) only when apqqpproportionality xi = λyi or yi = λxi holds.c. Minkowski’s Inequalities18Let xi ≥ 0, yi ≥ 0, i = 1, . . . , n. Then n 1/p n 1/p n 1/p p pp(xi + yi )≤xi+yii=1i=1and n 1/p n 1/p n 1/p p pp(xi + yi )≥xi+yii=1i=1when p > 1,(5.90)i=1when p < 1, p = 0. (5.91)i=1Proof We apply Hölder’s inequality to the terms on the right-hand side of the identitynnn(xi + yi )p =xi (xi + yi )p−1 +yi (xi + yi )p−1 .i=1i=1i=1The left-hand side is then bounded from above (for p > 1) or below (for p < 1)in accordance with inequalities (5.88) and (5.89) by the quantity 1/p n 1/q n 1/p n 1/q n p pppxi(xi + yi )+yi(xi + yi ).i=1i=1i=1i=1After dividing these inequalities by ( ni=1 (xi + yi )p )1/q , we arrive at (5.90) and(5.91).Knowing the conditions for equality in Hölder’s inequalities, we verify thatequality is possible in Minkowski’s inequalities only when the vectors (x1 , .
. . , xn )and (y1 , . . . , yn ) are collinear.For n = 3 and p = 2, Minkowski’s inequality (5.90) is obviously the triangleinequality in three-dimensional Euclidean space.18 H. Minkowski (1864–1909) – German mathematician who proposed a mathematical modeladapted to the special theory of relativity (a space with a sign-indefinite metric).5.4 Differential Calculus Used to Study Functions241Fig.
5.11Example 8 Let us consider another elementary example of the use of higher-orderderivatives to find local extrema. Let f (x) = sin x. Since f (x) = cos x and f (x) =− sin x, all the points where f (x) = cos x = 0 are local extrema of sin x, sincef (x) = − sin x = 0 at these points.
Here f (x) < 0 if sin x > 0 and f (x) > 0if sin x < 0. Thus the points where cos x = 0 and sin x > 0 are local maxima andthose where cos x = 0 and sin x < 0 are local minima for sin x (which, of course,was already well-known).5.4.3 Conditions for a Function to be ConvexDefinition 1 A function f : ]a, b[ → R defined on an open interval ]a, b[ ⊂ R isconvex if the inequalitiesf (α1 x1 + α2 x2 ) ≤ α1 f (x1 ) + α2 f (x2 )(5.92)hold for any points x1 , x2 ∈ ]a, b[ and any numbers α1 ≥ 0, α2 ≥ 0 such that α1 +α2 = 1.
If this inequality is strict whenever x1 = x2 and α1 α2 = 0, the function isstrictly convex on ]a, b[.Geometrically, condition (5.92) for convexity of a function f : ]a, b[ → R meansthat the points of any arc of the graph of the function lie below the chord subtendedby the arc (see Fig. 5.11).In fact, the left-hand side of (5.92) contains the value f (x) of the function atthe point x = α1 x1 + α2 x2 ∈ [x1 , x2 ] and the right-hand side contains the value atthe same point of the linear function whose (straight-line) graph passes through thepoints (x1 , f (x1 )) and (x2 , f (x2 )).Relation (5.92) means that the set E = {(x, y) ∈ R2 | x ∈ ]a, b[, f (x) < y} of thepoints of the plane lying above the graph of the function is convex; hence the term“convex”, as applied to the function itself.Definition 2 If the opposite inequality holds for a function f : ]a, b[ → R, thatfunction is said to be concave on the interval ]a, b[, or, more often, convex upwardin the interval, as opposed to a convex function, which is then said to be convexdownward on ]a, b[.2425Differential CalculusSince all our subsequent constructions are carried out in the same way for afunction that is convex downward or convex upward, we shall limit ourselves tofunctions that are convex downward.We first give a new form to the inequality (5.92), better adapted for our purposes.In the relations x = α1 x1 + α2 x2 , α1 + α2 = 1, we haveα1 =x2 − x,x2 − x1α2 =x − x1,x2 − x1so that (5.92) can be rewritten asf (x) ≤x − x1x2 − xf (x1 ) +f (x2 ).x2 − x1x2 − x1Taking account of the inequalities x1 ≤ x ≤ x2 and x1 < x2 , we multiply by x2 − x1and obtain(x2 − x)f (x1 ) + (x1 − x2 )f (x) + (x − x1 )f (x2 ) ≥ 0.Remarking that x2 − x1 = (x2 − x) + (x − x1 ) we obtain from the last inequality,after elementary transformations,f (x) − f (x1 ) f (x2 ) − f (x)≤x − x1x2 − x(5.93)for x1 < x < x2 and any x1 , x2 ∈ ]a, b[.Inequality (5.93) is another way of writing the definition of convexity of thefunction f (x) on an open interval ]a, b[.
Geometrically, (5.93) means (see Fig. 5.11)that the slope of the chord I joining (x1 , f (x1 )) to (x, f (x)) is not larger than (andin the case of strict convexity is less than) the slope of the chord II joining (x, f (x))to (x2 , f (x2 )).Now let us assume that the function f : ]a, b[ → R is differentiable on ]a, b[.Then, letting x in (5.93) tend first to x1 , then to x2 , we obtainf (x1 ) ≤f (x2 ) − f (x1 )≤ f (x2 ),x2 − x1which establishes that the derivative of f is monotonic.Taking this fact into account, for a strictly convex function we find, using Lagrange’s theorem, thatf (x1 ) < f (ξ1 ) =f (x) − f (x1 ) f (x2 ) − f (x)= f (ξ2 ) ≤ f (x2 )<x − x1x2 − xfor x1 < ξ1 < x < ξ2 < x2 , that is, strict convexity implies that the derivative isstrictly monotonic.Thus, if a differentiable function f is convex on an open interval ]a, b[, then f is nondecreasing on ]a, b[; and in the case when f is strictly convex, its derivativef is increasing on ]a, b[.5.4 Differential Calculus Used to Study Functions243These conditions turn out to be not only necessary, but also sufficient for convexity of a differentiable function.In fact, for a < x1 < x < x2 < b, by Lagrange’s theoremf (x) − f (x1 )= f (ξ1 ),x − x1f (x2 ) − f (x)= f (ξ2 ),x2 − xwhere x1 < ξ1 < x < ξ2 < x2 ; and if f (ξ1 ) ≤ f (ξ2 ), then condition (5.93) forconvexity holds (with strict convexity if f (ξ1 ) < f (ξ2 )).We have thus proved the following proposition.Proposition 5 A necessary and sufficient condition for a function f : ]a, b[ → Rthat is differentiable on the open interval ]a, b[ to be convex (downward) on thatinterval is that its derivative f be nondecreasing on ]a, b[.
A strictly increasing f corresponds to a strictly convex function.Comparing Proposition 5 with Proposition 3, we obtain the following corollary.Corollary A necessary and sufficient condition for a function f : ]a, b[ → R havinga second derivative on the open interval ]a, b[ to be convex (downward) on ]a, b[ isthat f (x) ≥ 0 on that interval. The condition f (x) > 0 on ]a, b[ is sufficient toguarantee that f is strictly convex.We are now in a position to explain, for example, why the graphs of the simplestelementary functions are drawn with one form of convexity or another.Example 9 Let us study the convexity of f (x) = x α on the set x > 0.
Since f (x) =α(α − 1)x α−2 , we have f (x) > 0 for α < 0 or α > 1, that is, for these values ofthe exponent α the power function x α is strictly convex (downward). For 0 < α < 1we have f (x) < 0, so that for these exponents it is strictly convex upward. Forexample, we always draw the parabola f (x) = x 2 as convex downward. The othercases α = 0 and α = 1 are trivial: x 0 ≡ 1 and x 1 = x. In both of these cases thegraph of the function is a ray (see Fig.
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