distr2

February 23, 1999 On some especially interesting distributions Christer O. Kiselman Contents : 1. Introduction 2. The principal value 3. Pseudofunctions 4. On Plemelj’s formulas Resumo: Pri iuj aparte interesaj distribucioj Por miaj kursanoj mi faros detalan pritrakton de distribucioj difinitaj per la pre- cipa valoro kaj la finia parto de diverˆ gaj integraˆ joj. Krome mi studos ecojn pri kontinuo de la distribucioj difinitaj per holomorfaj funkcioj. Abstract: For my students I shall consider in detail distributions defined by the principal value and the finite part of divergent integrals. I shall also consider continuity properties of distributions defined by holomorphic functions. 1. Introduction The purpose of this note is to discuss the notions of principal part of a divergent integral, the finite part of a divergent integral, and, finally, the continuity propertiesof the distributions that appear in Plemelj’s formulas. 2. The principal value Letfbe a continuous function on the real line, or more generally a locally integrable function. The integral/integraldisplay Rf(x)dx is said to exist in the generalized sense if the limit lim a,b→+∞/integraldisplayb −af(x)dx exists. We say that the principal value of the integral exists if the limit exists when we impose the condition b=a,t h u s vp/integraldisplay Rf(x)dx= lim a→+∞/integraldisplaya −af(x)dx. Example 2.1. Iff(x)=( s i n x)/x, then its integral exists in the generalized sense, although the function is not Lebesgue integrable. If f(x)=x/(1 +x2), then its integral does not exist in the generalized sense, but the principal value exists (and is zero). 2C . O . K i s e l m a n We can replace the singularity at infinity by a singularity at any other point. Letf∈L1 loc(R/integerdivide{0}). Suppose for simplicity that fhas compact support so that there is no difficulty at infinity. Then we say that its integral exists in the generalized sense if the limit lim a,b→0+/parenleftbigg/integraldisplay−a −∞f(x)dx+/integraldisplay∞ bf(x)dx/parenrightbigg exists. We say that the principal value exists if the limit exists when we restrict a andbto satisfy a=b,t h u s vp/integraldisplay Rf(x)dx= lim ε→0+/integraldisplay |x|>εf(x)dx. T h ei d e ai st h u st h a tw er e m o v ea symmetric neighborhood of a singular point and then pass to the limit. Large negative and positive values of the function can balanceeach other. Exercise 2.2. Prove that a function f∈L 1 loc(R/integerdivide{0}) such that its integral over [−1,1] exists in the generalized sense defines a distribution of order at most 1. Example 2.3. The principal values vp/integraldisplay1 −1xmdx exist whenever mis an odd integer; the value is of course zero. Example 2.4. The principal values vp/integraldisplay Rϕ(x) xdx exists if ϕ∈D(R); more generally if f∈C1 0(R), for we can write /integraldisplay |x|>εϕ(x) xdx=/integraldisplay∞ εϕ(x)−ϕ(−x) xdx=/integraldisplay∞ εdx/integraldisplay1 −1ϕ/prime(tx)dt. The existence of the limit is now obvious, and we can define a distribution vp(1/x) by the formula (2.1)/parenleftbigg vp1 x/parenrightbigg (ϕ)=/integraldisplay∞ 0ϕ(x)−ϕ(−x) xdx=/integraldisplay∞ 0dx/integraldisplay1 −1ϕ/prime(tx)dt, ϕ ∈D(R). From the last expression we can estimate the values as follows: (2.2)/vextendsingle/vextendsingle/vextendsingle/vextendsingle/parenleftbigg vp1 x/parenrightbigg (ϕ)/vextendsingle/vextendsingle/vextendsingle/vextendsingle/lessorequalslant/integraldisplay A 0dx2/bardblϕ/prime/bardbl∞=2A/bardblϕ/prime/bardbl∞, ifAis so large that the support of ϕis contained in [ −A, A]. This shows that vp(1/x) is a distribution of order at most 1. (Show that it is of order at least 1, i.e., that Especially interesting distributions 3 it is not a measure!) However, the estimate (2.2) is not translation invariant. We therefore subdivide the interval [0 ,+∞[ into two and use different estimates for each part: (2.3)/vextendsingle/vextendsingle/vextendsingle/vextendsingle/parenleftbigg vp1 x/parenrightbigg (ϕ)/vextendsingle/vextendsingle/vextendsingle/vextendsingle/lessorequalslant/integraldisplay c 0/parenleftbigg/integraldisplay1 −1|ϕ/prime(tx)|dt/parenrightbigg dx+/integraldisplay∞ c|ϕ(x)|+|ϕ(−x)| xdx /lessorequalslant2c/bardblϕ/prime/bardbl∞+c−1/bardblϕ/bardbl1, which is a translation-invariant estimate. The translation invariance implies that we can estimate also convolutions: (2.4)/vextenddouble/vextenddouble/vextenddouble/vextenddouble/parenleftbigg vp1 x/parenrightbigg ∗ϕ/vextenddouble/vextenddouble/vextenddouble/vextenddouble ∞/lessorequalslant2c/bardblϕ/prime/bardbl∞+c−1/bardblϕ/bardbl1. The best choice of chere, by the way, is c=/radicalBigg /bardblϕ/bardbl1 2/bardblϕ/prime/bardbl∞, which yields (2.5)/vextendsingle/vextendsingle/vextendsingle/vextendsingle/parenleftbigg vp1 x/parenrightbigg (ϕ)/vextendsingle/vextendsingle/vextendsingle/vextendsingle/lessorequalslant2/radicalbig 2/bardblϕ/prime/bardbl∞/bardblϕ/bardbl1,ϕ∈D(R). Exercise 2.5. How sharp is the estimate (2.5)? Prove that it can be improved to /vextendsingle/vextendsingle/vextendsingle/vextendsingle/parenleftbigg vp1 x/parenrightbigg (ϕ)/vextendsingle/vextendsingle/vextendsingle/vextendsingle/lessorequalslant2/radicalbig /bardblϕ/prime/bardbl∞/bardblϕ/bardbl1,ϕ∈D(R). Prove that, on the other hand, in any estimate /vextendsingle/vextendsingle/vextendsingle/vextendsingle/parenleftbigg vp1 x/parenrightbigg (ϕ)/vextendsingle/vextendsingle/vextendsingle/vextendsingle/lessorequalslantC/radicalbig /bardblϕ/prime/bardbl∞/bardblϕ/bardbl1,ϕ∈D(R), we must have C/greaterorequalslant2l o g2 >1.3862. Exercise 2.6. Prove the estimate /vextendsingle/vextendsingle/vextendsingle/vextendsingle/parenleftbigg vp1 x/parenrightbigg (ϕ)/vextendsingle/vextendsingle/vextendsingle/vextendsingle/lessorequalslantC/bardblϕ/prime/bardbl1/3 ∞/bardblϕ/bardbl2/3 2. Exercise 2.7. Prove that xvp(1/x) = 1. The solutions to the equation xu=1a r e u=vp(1/x)+Cδ,C∈C. Example 2.8. Is it possible to define a distribution vp(1/x3)? No, for the balance between negative and positive values in Example 2.3 for m=−3 is not sufficiently stable to allow multiplication by a smooth function. This follows from the formula /integraldisplay |x|>εϕ(x) x3dx=/integraldisplay∞ εϕ(x)−ϕ(−x) x3dx, 4C . O . K i s e l m a n where the limit does not exist if we take ϕsuch that ϕ(x)=xnear the origin. More generally, we see that we can define vpfiffis an odd function which is locally integrable in R/integerdivide{0}and such that for some α>−2,|f(x)|/lessorequalslantCxα,0<x< 1. For such functions we can also change variables in the integral. In conclusion, odd symmetry can kill singularities, but only the mild ones. 3. Pseudofunctions The pseudofunction defined by 1 /x2is a distribution defined as follows: /parenleftbigg pf1 x2/parenrightbigg (ϕ) = lim ε→0/parenleftBigg/integraldisplay |x|>εϕ(x) x2dx−C ε/parenrightBigg , where Cis the constant, if any, such that the limit exists. Of course the limit cannot exist for more that one choice of C. The notation pfis chosen to make us think not only of a pseudofunction but also of the finite part ( la partie finie ) of Hadamard. We write /integraldisplay |x|>εϕ(x) x2dx=/integraldisplayA εϕ(x)+ϕ(−x) x2dx =/integraldisplayA εϕ(x)−2ϕ(0) + ϕ(−x) x2dx+2ϕ(0)/parenleftbigg1 ε−1 A/parenrightbigg , where Ais so large that [ −A, A] contains the support of the test function. Hence C=2ϕ(0) is the only choice, and we define (3.1)/parenleftbigg pf1 x2/parenrightbigg (ϕ)=/integraldisplayA 0ϕ(x)−2ϕ(0) + ϕ(−x) x2dx−2ϕ(0) A,ϕ∈D(R) Note that the right-hand side is independent of Aas long as Ais large (calculate the derivative of the right-hand side with respect to A). Since the integral of 1 /x2is convergent at infinity, we can let Atend to infinity here, thus the definition can also be written: (3.2)/parenleftbigg pf1 x2/parenrightbigg (ϕ)=/integraldisplay∞ 0ϕ(x)−2ϕ(0) + ϕ(−x) x2dx, ϕ ∈D(R). To estimate the integral we now write (3.3)ϕ(x)−2ϕ(0) + ϕ(−x) x2=/integraldisplay1 0sd s/integraldisplay1 −1ϕ/prime/prime(stx)dt, which shows that the pseudofunction can be estimated as /vextendsingle/vextendsingle/vextendsingle/vextendsingle/parenleftbigg pf1 x2/parenrightbigg (ϕ)/vextendsingle/vextendsingle/vextendsingle/vextendsingle/lessorequalslant/integraldisplay A 0/integraldisplay1 0sds/integraldisplay1 −1/bardblϕ/prime/prime/bardbl∞dt=A/bardblϕ/prime/prime/bardbl∞, Especially interesting distributions 5 where Ais so large that the support of ϕis contained in [ −A, A]. This shows that pf(1/x2) is a distribution of order at most two. (Show that the order is two!) Since the estimate is not translation invariant, we shall modify it as in the case of the principal value. We use the representation (3.2) only for x∈[0,c]: /parenleftbigg pf1 x2/parenrightbigg (ϕ)=/integraldisplayc 0/integraldisplay1 0sd s/integraldisplay1 −1ϕ/prime/prime(stx)dt+/integraldisplay∞ cϕ(x)−2ϕ(0) + ϕ(−x) x2dx, which can be estimated as /vextendsingle/vextendsingle/vextendsingle/vextendsingle/parenleftbigg pf1 x2/parenrightbigg (ϕ)/vextendsingle/vextendsingle/vextendsingle/vextendsingle/lessorequalslantc/bardblϕ/prime/prime/bardbl∞+2c−1/bardblϕ/bardbl∞+c−2/bardblϕ/bardbl1, or perhaps simpler as /vextendsingle/vextendsingle/vextendsingle/vextendsingle/parenleftbigg pf1 x2/parenrightbigg (ϕ)/vextendsingle/vextendsingle/vextendsingle/vextendsingle/lessorequalslantc/bardblϕ/prime/prime/bardbl∞+4c−1/bardblϕ/bardbl∞; both are translation-invariant estimates. What is the best choice of c? Exercise 3.1. Prove that x2pf(1/x2) = 1. The solutions to x2u=1a r e u= pf(1/x2)+C0δ+C1δ/prime. In contrast to the principal value, we can consider pseudofunctions defined by one-sided integrals. Thus the distribution pf(1/x+) is defined by /parenleftbigg pf1 x+/parenrightbigg (ϕ) = lim ε→0/parenleftbigg/integraldisplay∞ εϕ(x) xdx+Clogε/parenrightbigg , where again Cis the only constant such that the limit exists. We can write /integraldisplayA εϕ(x) xdx+Clogε=/integraldisplayA εϕ(x)−C xdx+ClogA, which makes it obvious that the only choice is C=ϕ(0). Thus /parenleftbigg pf1 x+/parenrightbigg (ϕ)=/integraldisplayA 0ϕ(x)−ϕ(0) xdx+ϕ(0) log A, ϕ ∈D(R). (Here we cannot let Atend to infinity.) Similarly for pf(1/x2 +): /parenleftbigg pf1 x2 +/parenrightbigg (ϕ) = lim ε→0/parenleftBigg/integraldisplayA εϕ(x) x2dx−C0 ε+C1logε/parenrightBigg . Again only one choice of constants is possible, and we can write /integraldisplayA εϕ(x) x2dx−C0 ε+C1logε=/integraldisplayA εϕ(x)−C0−C1x x2dx−C0 A+C1logA, 6C . O . K i s e l m a n which shows that the constants must be C0=ϕ(0),C1=ϕ/prime(0), so that /parenleftbigg pf1 x2 +/parenrightbigg (ϕ)=/integraldisplayA 0ϕ(x)−ϕ(0)−ϕ/prime(0)x x2dx−ϕ(0) A+ϕ/prime(0) log A, ϕ ∈D(R). We can perhaps formulate a rule for these pseudofunctions as follows: we sub- tract a part of the Taylor expansion of the function to make the integral convergent at the singular point (the origin in our case), but then we must compensate by afunction of A, the upper limit of the integral, to make the whole expression indepen- dent of A. This sounds sloppy, but if we add the information that the right-hand side shall be zero for ϕ= 0, then it is actually uniquely determined. 4. On Plemelj’s formulas Let us consider the distributions usdefined for real nonzero sby (4.1) us(ϕ)=/integraldisplay Rϕ(x) x+isdx, ϕ ∈D(R). Thus they are defined by smooth functions, but the interesting question is what happens when the parameter stends to zero. We rewrite the definition as follows: (4.2) us(ϕ)=/integraldisplay∞ 0ϕ(x)−ϕ(−x) x2+s2xdx−is/integraldisplay∞ 0ϕ(x)+ϕ(−x) x2+s2dx. Now it is easy to see that usconverges to a distribution u0+asstends to zero while being positive: (4.3) u0+(ϕ)=/integraldisplay∞ 0ϕ(x)−ϕ(−x) xdx−iπϕ(0),ϕ∈D(R), thus (4.4) lim s→0+us=u0+=vp/parenleftbigg1 x/parenrightbigg −iπδ, a relation known as Plemelj’s formula, or Sohockij–Plemelj’s formula. (Do the neces- sary deliberations.) Note that the passage to the limit works for any function which is of class C1in a neighborhood of the origin and such that xεϕ(x) is bounded for some positive ε. We can estimate the integrals defining usas follows, taking a number Aso large that [ −A, A] contains the support of ϕ: |us(ϕ)|/lessorequalslant2/bardblϕ/prime/bardbl∞/integraldisplayA 0x2 x2+s2dx+2/bardblϕ/bardbl∞/integraldisplayA 0s x2+s2dx. Here the first integral is not larger than A, the second not larger than π/2; thus |us(ϕ)|/lessorequalslant2A/bardblϕ/prime/bardbl∞+π/bardblϕ/bardbl∞,ϕ∈D(R); Especially interesting distributions 7 cf. (2.2). As we noted several times already, such estimates containing a number A depending on the support of the test function are not translation invariant. Insteadwe write (4.5) |u s(ϕ)|/lessorequalslant/integraldisplayc 0|ϕ(x)−ϕ(−x)| xx2 x2+s2dx+/integraldisplay∞ c|ϕ(x)−ϕ(−x)| xx2 x2+s2dx+π/bardblϕ/bardbl∞ /lessorequalslant2c/bardblϕ/prime/bardbl∞+c−1/bardblϕ/bardbl1+π/bardblϕ/bardbl∞. This holds for all c>0a n da l l s∈R,t h ec a s e s= 0 including both u0+andu0−. We can also estimate in terms of the Lp-norm for any p>1; in particular we get for theL2-norm, (4.6) |us(ϕ)|/lessorequalslant2c/bardblϕ/prime/bardbl∞+c−1/2/bardblϕ/bardbl2+π/bardblϕ/bardbl∞,ϕ ∈D(R),s∈R. These estimates make it possible to let usact on more general functions than the usual test functions in D(R). We formulate this result as a lemma. Lemma 4.1. LetC1 /diamondmath(R)denote the space of all f∈C1(R)such that f∈L2∩L∞ andf/prime∈L∞.W ee q u i p C1 /diamondmath(R)with the norm /bardblϕ/bardbl/diamondmath=/bardblϕ/bardbl2+π/bardblϕ/bardbl∞+2/bardblϕ/prime/bardbl∞. Then the distributions us,s∈R/integerdivide{0},a n d u0+,u0−can be extended from D(R)to C1 /diamondmath(R)as defined by (4.2) and (4.3) and they satisfy the estimate |us(ϕ)|/lessorequalslant/bardblϕ/bardbl/diamondmath, ϕ∈C1 /diamondmath(R); equivalently (4.7) /bardblus∗ϕ/bardbl∞/lessorequalslant/bardblϕ/bardbl/diamondmath,ϕ ∈C1 /diamondmath(R). It can be seen easily that us∗ϕ,u0+∗ϕ,a n d u0−∗ϕall belong to C1 /diamondmath(R), and we shall soon apply (4.7) to such more general functions. Let us now look at the difference us−u0+: (us−u0+)(ϕ)=−s2/integraldisplay∞ 0ϕ(x)−ϕ(−x) x(x2+s2)dx−is/integraldisplay∞ 0ϕ(x)−2ϕ(0) + ϕ(−x) x2+s2dx. We can estimate these integrals by the methods already used in the proof of (4.5) and get (4.8) |(us−u0+)(ϕ)|/lessorequalslants/parenleftbig 4c−1/bardblϕ/bardbl∞+π/bardblϕ/prime/bardbl∞+c/bardblϕ/prime/prime/bardbl∞/parenrightbig . This gives a quantitative idea of how fast usconverges to u0+. We may extend the validity of (4.8) to functions fsuch that ϕ, ϕ/prime∈C1 /diamondmath(R), or even a little farther: Lemma 4.2. LetC2 ∞(R2)denote the space of all functions ϕ∈C2(R)such that ϕ, ϕ/prime,ϕ/prime/prime∈L∞with norm /bardblϕ/bardbl2,∞=4/bardblϕ/bardbl∞+π/bardblϕ/prime/bardbl∞+/bardblϕ/prime/prime/bardbl∞. 8C . O . K i s e l m a n Then us−u0+can be extended from D(R)toC2 ∞(R)and satisfies (4.9) /bardbl(us−u0+)∗ϕ/bardbl∞/lessorequalslants/bardblϕ/bardbl2,∞,ϕ ∈C2 ∞(R). We can also see that the derivative with respect to sexists: us−u0+ s→πδ/prime−ipf/parenleftbigg1 x2/parenrightbigg ,s→0+. One can prove, using the residue formula, that us∗ut=0i f st <0. (Do so! We have a ring with zero divisors!) If we could pass to the limit in this equation we would obtain u0+∗u0−= 0; hence, in view of Plemelj’s formula (4.4), (4.10) vp/parenleftbigg1 x/parenrightbigg ∗vp/parenleftbigg1 x/parenrightbigg =−π2δ. What does the Titchmarsh support theorem say? We shall now see that passage to the limit is legitimate. We state the result as ap r o p o s i t i o n . Proposition 4.3. Letu0+be defined by (4.3) and let u0−be its complex conjugate. Then u0+∗u0−=0. Proof. We write (4.11) −u0+∗u0+=us∗ut−u0+∗u0−=us∗(ut−u0−)+(us−u0+)∗u0−. To estimate the first term we use (4.9) with sreplaced by t,a n d ϕ∗usas a test function: /bardbl(ϕ∗us)∗(ut−u0−)/bardbl∞/lessorequalslant|t|/bardblϕ∗us/bardbl2,∞=|t|(4/bardblus∗ϕ/bardbl∞+π/bardblus∗ϕ/prime/bardbl∞+/bardblus∗ϕ/prime/prime/bardbl∞). Here the right-hand side can be estimated using (4.7); it does not exceed |t|(4/bardblϕ/bardbl/diamondmath+π/bardblϕ/prime/bardbl/diamondmath+/bardblϕ/prime/prime/bardbl/diamondmath), which is independent of s. To take care of the second term in (4.11) we shall use (4.9) with u0+∗ϕas a test function: /bardbl(us−u0+)∗(us∗ϕ)/bardbl∞/lessorequalslants/bardblu0+∗ϕ/bardbl2,∞/lessorequalslants(4/bardblϕ/bardbl/diamondmath+π/bardblϕ/prime/bardbl/diamondmath+/bardblϕ/prime/prime/bardbl/diamondmath). These estimates for the two terms in (4.10) show that for s>0>t,w eh a v e /bardbl(u0+∗u0−)∗ϕ/bardbl∞=/bardbl(us∗ut−u0+∗u0−)∗ϕ/bardbl∞/lessorequalslantCϕ(s+|t|), where Cϕ=4/bardblϕ/bardbl/diamondmath+π/bardblϕ/prime/bardbl/diamondmath+/bardblϕ/prime/prime/bardbl/diamondmathis a constant depending on ϕ∈D(R). Thus u0+∗u0−is zero. Author’s address: Uppsala University, Department of Mathematics, P. O. Box 480, SE-751 06 Uppsala, Sweden. Telephone: +46 18 4713216 (office); +46 18 3007 08 (home) Electronic mail: kiselman @ math.uu.se Fax:+46 18 4713201