Some basic inequalities Definition. Let V be a vector space over the complex numbers. An inner product is given by a function h·; ·i V × V ! C (x; y) 7! hx; yi satisfying the following properties (for all x 2 V , y 2 V and c 2 C) (1) hx +x; ~ yi = hx; yi + hx;~ yi (2) hcx; yi = chx; yi (3) hy; xi = hx; yi (4) hx; xi ≥ 0 and hx; xi = 0 if and only of x = 0. Note that if h·; ·i is an inner product then for each y the function x 7! hx; yi is a linear function. Also we have hx; cyi = chx; yi and hx; y +y ~i = hx; yi + hx; y~i. Remark: We can also define inner products for vector spaces over R, but then the third axiom is changed to the symmetry axiom hy; xi = hx; yi for all x; y 2 V . Thus if V is a vector space over the real numbers then then for each y the function x 7! hx; yi is a linear function, and for each x the function y 7! hx; yi is a linear function. The latter statement for y 7! hx; yi fails in vector spaces over C. Definition. A semi-norm on a vector space over C (or over R) is a function k·k : V ! [0; 1) satisfying the following properties for all x; y 2 V . (1) kxk ≥ 0 (2) For scalars c, kcxk = jcjkxk. (3) kx + yk ≤ kxk + kyk (the triangle inequality). If in addition we also have the property that and kxk = 0 only if x = 0 then we call k · k a norm. 1. The Cauchy-Schwarz inequality Theorem. Let h·; ·i be an inner product on V . Then for all x; y 2 V jhx; yij ≤ phx; xiphy; yi: Proof. The inequality is immediate if one of the two vectors is 0. We may thus assume that y 6= 0 and therefore hy; yi > 0. We shall first show the weaker inequality (1.1) Re hx; yi ≤ phx; xiphy; yi Let t 2 R. We shall use that hx + ty; x + tyi ≥ 0: 1 2 Then compute hx + ty; x + tyi = hx; xi + thx; yi + thy; xi + t2hy; yi = hx; xi + 2t Re hx; yi + t2hy; yi: Here we used that for the complex number z = hx; yi we have z+z = 2 Re (z). We have seen that for all t 2 R hx; xi + 2t Re hx; yi + t2hy; yi ≥ 0: Re (hx;yi) We use this inequality for the special choice t = − hy;yi (which happens to be the choice of t that minimizes the quadratic polynomial). Plugging in this value of t yields the inequality (Re hx; yi)2 hx; xi − ≥ 0 hy; yi which gives (Re hx; yi)2 ≤ hx; xihy; yi and (1.1) follows. Finally let z := hx; yi. If z = 0 there is nothing to prove, so assume z 6= 0. Then we can write z in polar form, i.e. z = jzj(cos φ+i sin φ) for some angle φ. Let c = cos φ − i sin φ. Then cz = jzj and cz is real and positive. 1 Also jcj = 1. Hence we get jhx; yij = chx; yi = hcx; yi = Re hcx; yi: Applying the already proved inequality (1.1) for the vectors cx and y we see that the last expression is ≤ phcx; cxiphy; yi = pcchx; xiphy; yi = phx; xiphy; yi : This finishes the proof. Exercise: Show that equality in Cauchy-Schwarz, jhx; yij = phx; xiphy; yi, only happens if x and y are linearly dependent (i.e. one of the two is a scalar multiple of the other). Definition. We set kxk = phx; xi. Theorem The map x 7! phx; xi defines a norm on V . Proof. Setting kxk := phx; xi we clearly have that kxk ≥ 0 and kxk = 0 if and only if x = 0, by property (4) for the inner product. Also phcx; cxi = pcchx; xi = jcjphx; xi. It remains to prove the triangle inequality. 1If you prefer not to use polar notation, another equivalent way to define c, given z = a+bi with z 6= 0 is to set c = pa−ib , i.e. c = z=jzj. Note that cz = zz=jzj = jzj2=jzj = jzj. a2+b2 3 We compute kx + yk2 = hx + y; x + yi = hx; xi + hx; yi + hy; xi + hy; yi = kxk2 + 2 Re (hx; yi) + kyk2 and by the Cauchy-Schwarz inequality the last expression is ≤ kxk2 + 2kxkkyk + kyk2 = (kxk + kyk)2: So we have shown kx + yk2 ≤ (kxk + kyk)2 and the triangle inequality follows. 2. Generalized arithmetic and geometric means p Given two nonnegative numbers a; b we call ab the geometric mean of a and b. The geometric significance is that the rectangle withp sides of length a and b has the same area as the square with sidelength ab. The arithmetic a+b mean is 2 . The arithmetic mean exceeds the geometric mean: p a + b ab ≤ : 2 p p p p This follows immediately from ( a − b)2 ≥ 0, i.e. a + b − 2 a b ≥ 0 (for nonnegative a; b). A useful generalization is Theorem. Let a; b be nonnegative numbers and let 0 < # < 1. Then (2.1) a1−#b# ≤ (1 − #)a + #b : Proof. If one of a, b is zero then the inequality is immediate. Let's assume that a 6= 0. Then setting c = b=a the assertion is equivalent with (2.2) c# ≤ (1 − #) + #c; for c ≥ 0. To prove (2.2) we set f(c) := (1 − #) + #c − c# and observe that f 0(c) = #(1 − c#−1). Since by assumption 0 < # < 1 we see that f 0(c) ≤ 0 for 0 ≤ c ≤ 1 and f 0(c) ≥ 0 for c ≥ 1. Hence f must have a minimum at c = 1. Clearly f(1) = 0 and therefore f(c) ≥ 0 for all c ≥ 0. Thus (2.2) holds. 3. The inequalities by Holder¨ and Minkowski n n For vectors x = (x1; : : : ; xn) in R (or in C ) we define n 1=p X p kxkp = jxij : i=1 It is our intention to show that kxkp defines a norm ehen p > 1. We shall use the following result (H¨older'sinequality) to prove this. 4 For p > 1 we define the conjugate number p0 by 1 1 + = 1 p p0 0 p i.e. p = p−1 . Theorem: (H¨older'sinequality): Let 1 < p < 1, 1=p + 1=p0 = 1. For n x; y 2 C , n n n 0 1=p 0 1=p X X p X p xiyi ≤ jxij jyij i=1 i=1 i=1 or in the above notation n X xiyi ≤ kxkpkykp0 : i=1 Remark. When p = 2, then p0 = 2 and H¨older's inequality becomes the Cauchy-Schwarz inequality for the standard scalar product hx; yi = Pn n Pn n i=1 xiyi on R (or the standard scalar product hx; yi = i=1 xiyi on C ). Proof of H¨older'sinequality. If we replace x with x=kxkp and y with y=kykp0 then we see that it is enough to show that n X (3.1) xiyi ≤ 1 provided that kxkp = 1 and kykp0 = 1 i=1 Also it is clearly sufficient to do this for vectors x and y with nonnegative entries (simply replace xi with jxij etc.) Thus for the rest of the proof we assume that x, y are vectors with non- negative entries satisfying kxkp = 1, kykp0 = 1. p p0 Set ai = xi , bi = yi . And set # = 1 − 1=p. Since we assume p > 1 we see that 0 < # < 1. By the inequality for the generalized arithmetic and 1−# # geometric means we have ai bi ≤ (1 − #)ai + #bi i.e. 1 1 1 1 0 x y = a1=pb1−1=p ≤ a + (1 − )b = xp + (1 − )yp i i i i p i p i p i p i Thus n n n X 1 X 1 X 0 x y ≤ xp + (1 − ) yp i i p i p i i=1 i=1 i=1 1 1 0 1 1 = kxkp + (1 − )kykp = + (1 − ) = 1; p p p p0 p p here we have used that kxkp = 1, kykp0 = 1. Remark: H¨older'sinequality has extensions to ther settings. One is in Problem 6 on the first homework assignment. Here note that Riemann integrals can be approximated by sums, and so the H¨older inequality with n summands may be useful for similar versions for integrals as well. 5 The following result called Minkowski's inequality2 establishes the triangle inequality for k · kp. n Theorem: For x; y 2 C n n n 1=p 1=p 1=p X p X p X p jxi + yij ≤ jxij + jyij i=1 i=1 i=1 or shortly, kx + ykp ≤ kxkp + kykp. Proof. If x + y = 0 the inequality is trivial, thus we assume that x + y 6= 0 and hence kx + ykp > 0 Write n n p X p X p−1 kx + ykp = jxi + yij = jxi + yij jxi + yij i=1 i=1 n n n X p−1 X p−1 X p−1 ≤ jxi + yij (jxij + jyij) = jxijjxi + yij + jyijjxi + yij i=1 i=1 i=1 By H¨older'sinequality n n n 1=p 1=p0 X p−1 X p X (p−1)p0 jxijjxi + yij ≤ jxij jxi + yij i=1 i=1 i=1 p−1 = kxkpkx + ykp since (p − 1)p0 = p. The same calculation yields n X p−1 p−1 jyijjxi + yij ≤ kykpkx + ykp : i=1 We add the two inequalities and we get p p−1 kx + ykp ≤ kx + ykp (kxkp + kykp): p−1 Divide by kx + ykp and the asserted inequality follows.