2 How to norm a sum of normed vector spaces

When is L ^r (R) contained in L ^p (R) + L ^q (R) ?

Jean-Baptiste Hiriart-Urruty and Lass` ere Patrice

Abstract

We prove a necessary and sufficient condition on the exponentsp, q, r≥ 1 such thatL^r(R)⊂L^p(R)+L^q(R). In doing so, we explore the structure ofL^p(R) +L^q(R) as a normed vector space.

1 Introduction.

In a recent mathematical note aimed at undergraduate students and their teach- ers ([3]), J.-B. Hiriart-Urruty and M. Pradel proposed a way to extend the Fourier transformation to all the spacesL^r(R) with 1≤r≤2 in the following manner.

– First, they classically define the Fourier transformation onL¹(R). Then they define it on L²(R) using in that case the much less known Wiener’s ap- proach which relies on a specific Hilbertian basis made of the so-called Bernstein functions.

– After having checked the coherence of both definitions onL¹(R)∩L²(R), they extend the definition of the Fourier transformation to the spaceL¹(R) + L²(R).

– Finally, and this is the key-point, they prove the inclusion L^r(R) ⊂ L¹(R) +L²(R) for all 1 ≤ r ≤ 2, so that the Fourier transformation can be extended to all the Lebesgue spacesL^r(R).

An obvious question which arises is, what happens ifr6∈[1,2] ? For example, is the inclusion L³(R) ⊂ L¹(R) +L²(R) true or not ? The objective of the present mathematical note is to answer this question. We provide a necessary and sufficient condition for the inclusionL^r(R)⊂L^p(R) +L^q(R) to hold true.

For that purpose, concentrate on the sum L^p(R) +L^q(R) and see how it is structured as a normed vector space.

2 How to norm a sum of normed vector spaces

?

LetV and W be two vector subspaces of a ”holdall” vector space, E. IfV is equipped with a normk · k1 and W with a norm k · k2, is there a natural way to define a norm on the vector subspaceV +W ? Of course, we do not assume

thatV∩W ={OE}. If this was the case, a natural way to define a normN on V +W would be

N(u) =kvk1+kwk2,

wheneveru∈V +W is (uniquely) decomposed asu=v+w, withv∈V and w∈W.

What we have in mind is indeed V =L^p(R), W =L^q(R) andE =L(R) as the “holdall” vector space (L(R) stands for the set of all Lebesgue classes of mesurable functions onR).

The theorem below answers the question posed above. It does not seem to be well-known, except by people who have to deal with the interpolation of functional spaces (like in [1]).

Theorem 1 LetN be defined on V +W as follows

N(u) := inf{kvk1+kwk2 ; u=v+w withv∈V andw∈W}. (1) Then,N is a semi-norm onV +W. It is a norm under the following “compat- ibility” assumption :

(T)

zk ∈V ∩W,

z_k −→ain (V,||.||1) z_k −→b in (W,||.||2)







=⇒a=b.

Proof : To check thatN(O_E) = 0, N(λu) =|λ|N(u) for allλ∈R,u∈V+W, andN(u₁+u₂)≤N(u₁) +N(u₂) for allu₁ andu₂ in V +W, does not raise any difficulty. It suffices to use the definition of the lower bound (or infimum) of a set of real numbers.

To prove that N(u) = 0 implies that u = OE is a bit more tricky. Our experience with that question with undergraduate students shows that they usually fail to answer it correctly. Their common mistake is to deduce that a sequence (vk+wk)k converges toOEusing the fact that (vk)k converges toOE

inV and (wk)k converges toOE inW. We therefore provide a proof here.

We first begin by observing that

ν : V ∩W 3u7→ν(u) := max(kuk1,kuk2) (2) is a norm onV ∩W ; this is an easy result to prove.

Consider therefore,u∈V +W such thatN(u) = 0. We take for example u=v+w, withv∈V and w∈W. (3) Due to the definition (1) ofN(u), for all positive integersk, there existsv_k∈V andwk ∈W such that

u=vk+wk, and (4)

kvkk1+kwkk2≤N(u) +1 k = 1

k. (5)

Thus, (vk)k converges toOE in V and (wk)k converges toOE in w. But what about (vk +wk)k ? Recall that no norm, hence, no topology, has yet been defined onV +W. We infer from (3) and (4) that

v+w=vk+wk, and thus

v−v_k=w_k−w. (6)

As a consequence, this common vectorz_k:=v−v_k=w_k−wlies inV∩W and, sinceν(z_k) =kv−v_kk₁ orkw_k−wk₂,

zk→v in (V,k · k1) andzk→ −win (W,k · k2).

The assumption (T) then ensures that

v=−w, that is u=v+w=OE.

Note that the technical “compatibility” assumption (T) is satisfied

– trivially ifV ∩W ={OE} (in that case it amounts to 0 = 0).

– in the cases whereV =L^p(R), W =L^q(R) (indeed, convergence of (f_k)_k towardsf inL^p(R) implies convergence almost everywhere of a subsequence of (f_k)_k towardsf).

– when the “holdall” vector spaceEis a Hausdorff topological vector space in whichV andW are continuously imbedded.

We suppose that (T) is in force for the rest of the section.

The vector spaceV+W, equipped with the normNas defined in (1), inherits some properties of (V,k · k1) and (W,k · k2). Here is one.

Theorem 2 If(V,k·k1)and(W,k·k2)are Banach spaces, then so is(V+W, N).

Proof : The proof of this theorem offers the opportunity to use a caracter- ization of completeness of normed vector spaces which is not well-known. Let (X,k · k) be a normed space. We have indeed

((X,k · k) is complete ) ⇐⇒





Every series inX, of general termak,for which P∞

k=0kakk<+∞ does converges inX (toward a sum denoted as P∞

k=0ak)



. (7) The implication (⇒) is classical, it is the most often used. The converse im- plication (⇐) is not often used, but we have an opportunity to do that here.

For a proof of the equivalence (7), see for example ([6], Theorems 2-XIV-2.1

and 2.2, pages 164-165), ([4], page 20) or ([7], pages 262 and 270); it is also sketched in ([1], page 24). The proof is not very difficult however, but readers are encouraged to work through it themselves. We now proceed in this manner to a proof of Theorem 2.

Consider a series inV+W, of general termuk for whichP∞

k=0N(uk)<∞.

We have to prove thatPn

k=0u_k converges as n →+∞, to some element u∈ V +W. In view of the definition (1) ofN, for all positive integerk, there exist v_k∈V andw_k ∈W satisfying

u_k=v_k+w_k, and (8)

kvkk1+kwkk2≤N(uk) + 1

k². (9)

Thus, P∞

k=0kvkk1 <+∞ and P∞

k=0kwkk2 <+∞. Since both (V,k · k1) and (W,k · k2) have been assumed to be complete, there exist v ∈ V and w ∈ W such that

n

X

k=0

vk −→

n→+∞vin V, and

n

X

k=0

wk −→

n→+∞w inW. (10)

Letu:=v+w∈V+W. Let us check that, as expected,Pn

k=0uk converges to uin (V +W, N). Indeed,

N u−

n

X

k=0

uk

!

=N u+v−

n

X

k=0

(vk+wk)

!

≤

v−

n

X

k=0

vk

₁

+

w−

n

X

k=0

wk

₂

.

It remains to apply (10) to get the desired result,N(u−Pn

k=0uk) −→

n→+∞0.

Comments : The construction of the norm ν on V ∩W and N on V +W deserves some geometrical interpretation. Even ifk · k₁ (resp. k · k₂) is only defined onV ⊂E (resp. on W ⊂E), we can extend it to the whole of E by settingkuk1= +∞ifu6∈V (resp. kuk2= +∞ifu6∈W). We still denote by k · k1 andk · k2 the extended functions.

Clearly, k · k1 and k · k2 are convex positively homogeneous functions on E. Modern convex analysis accepts and can handle convex functions possibly taking the value +∞([5]). An important geometrical object associated with a convex functionf : E→R∪ {+∞} is its so-called strict epigraph

epi_sf :={(x, r)∈E×R : f(x)< r}

(literally, what is strictly above the graph off). In our situation,K₁:= epi_sk·k1

andK₂:= epi_sk·k2are open convex cones ofE. So, what are the strict epigraphs of the norm functionsν andN ? One easily checks the following

epi_sν = (epi_sk · k₁)∩(epi_sk · k₂), and

epi_sN = (epi_sk · k1) + (epi_sk · k2). (11)

The sets where ν (resp. N) is finite, called the domain of ν (resp. of N) in convex analysis, is justV ∩W (resp. V +W).

The binary operation which builds a convex functionf from two other ones f1 andf2, via the geometric construction

epi_sf = epi_sf₁+ epi_sf₂

is called the infimal convolution of f1 and f2 ([2],[5]). This operation enjoys properties similar to the usual (integral) convolution in classical analysis.

In brief, the norm N has been designed as an infimal convolution of the normsk · k1 andk · k2.

Returning to our particular settingE=L(R), V =L^p(R) andW =L^q(R) with 1≤p, q <+∞, the vector spaceL^p(R) +L^q(R) can be equipped with a norm that we denotek · kp,q as follows,

kfkp,q= inf{kgkp+khkq ; f =g+h withg∈L^p(R) andh∈L^q(R)}. (12) As proved in Theorem 2, (L^p(R) +L^q(R),k · kp,q) is a Banach space.

3 Comparing L

(R) with L

(R) + L

(R)

We know that the Lebesgue spacesL^r(R) and L^s(R) (for 1≤ r 6=s < +∞) cannot be compared. NeitherL^r(R) is contained inL^s(R) nor the converse. A direct comparison is however possible if we deal with the sum of these spaces.

Here is the main result of this section.

Theorem 3 If1≤p < q <+∞, then we have the following : 1. L^r(R)is contained inL^p(R) +L^q(R)wheneverr∈[p, q],

2. if 1 ≤ r < +∞ does not lie in [p, q], then L^r(R) is not contained in L^p(R) +L^q(R).

Proof : (1) Let f ∈ L^r(R) and consider X := {x ∈ R : |f(x)| > 1} (a measurable set defined within a set of null measure) as well asX^c=R\X (the complementary set ofX in R). We decomposef as follows :

f =f1+f2, withf1=f ·1X andf2=f·1X^c. (13) We claim that (13) provides an explicit decomposition off in L^p(R) +L^q(R), that is to sayf₁∈L^p(R) andf₂∈L^q(R). We first prove that,

Z

R

|f1(x)|^pdλ(x) = Z

X

|f(x)|^pdλ(x) = Z

X

|f(x)|^p−r· |f(x)|^rdλ(x). (14)

Forx∈X,|f(x)|>1 and, since the exponentp−ris nonpositive,|f(x)|^p−r≤1.

Consequently, the last integral in the string of equalities (14) is bounded above byR

X|f(x)|^rdλ(x). Finally, (14⁰)

Z

R

|f1(x)|^pdλ(x)≤ Z

X

|f(x)|^rdλ(x)≤ Z

R

|f(x)|^rdλ(x)<+∞.

We thus have proved thatf1∈L^p(R).

Second, we prove thatf2∈L^q(R). Indeed, Z

R

|f2(x)|^qdλ(x) = Z

X^c

|f(x)|^qdλ(x) = Z

X^c

|f(x)|^q−r· |f(x)|^rdλ(x). (15) Forx∈X^c,|f(x)| ≤1 and, since the exponentq−ris nonegative,|f(x)|^q−r≤1.

Again, the last integral in the string of equalities (15) is bounded above by R

X^c|f(x)|^rdλ(x). As a result, Z

R

|f2(x)|^qdλ(x) = Z

X^c

|f(x)|^rdλ(x) = Z

R

|f(x)|^rdλ(x)<+∞. (16) We therefore, have proved thatf2∈L^q(R).

(2) The second part of Theorem 3 is a bit harder to prove (like most of the negative results in mathematics). We actually have to distinguish two cases for rin the segment [p, q] : r < pandr > q.

Case 1 : r < p. Chooseαsatisfying 1/p < α <1/rand let f be defined on Rbyf(x) =x^−α1_(0,1](x).

Since|f(x)|^r=x^−αr for x∈(0,1] and 0 elsewhere, the choice ofαimplies that f ∈ L^r(R) (since αr < 1). The same argument shows that f 6∈ L^p(R) (sinceαp >1).

Suppose now thatL^r(R)⊂L^p(R) +L^q(R). Then

f ∈L^r(R)⊂L^p(R) +L^q(R)⊂L^p([0,1]) +L^q([0,1]). (17) But since [0,1] is of Lebesgue finite measure andp < q,L^q([0,1]) is contained inL^p([0,1]), so that (17) yields thatf ∈L^p([0,1]). This is not the case.

Thus we have proved thatL^r(R) is not contained inL^p(R) +L^q(R).

Case 2 : r > q. Our proof in this case relies on a technical lemma that we present separately.

Lemma 1 Let 1 ≤ p < q < +∞, let Ω be a mesurable subset of R, and let f ∈L^p(Ω) +L^q(Ω). Thenf ∈L^q(Ω) whenever it is essentially bounded onΩ.

Proof of the Lemma : Let f be decomposed as f = fp+fq, with fp ∈ L^p(Ω) andfq ∈L^q(Ω). So, to prove thatf ∈ L^q(Ω) amounts to proving that fp∈L^q(Ω).

LetX :={x∈R : |fp(x)| >1}. To show thatR

Ω |fp(x)|^qdλ(x) is finite, we cut it into two pieces : R

Ω∩X^c |fp(x)|^qdλ(x) andR

Ω∩X |fp(x)|^qdλ(x).

Consider the first piece. Sincefp∈L^p(R), the setX is of finite (Lebesgue) measure. Now with the definition ofX and the fact thatq−p >0, we obtain

Z

Ω∩X^c

|f_p(x)|^qdλ(x) = Z

Ω∩X^c

|f_p(x)|^q−p· |f_p(x)|^pdλ(x)

≤ Z

Ω∩X^c

|fp(x)|^pdλ(x)≤ Z

R

|fp(x)|^pdλ(x)<+∞.

This concludes the argument for the first piece.

Consider now the second piece. Sincefhas been assumed essentially bounded on Ω,

|f_p(x)| ≤ |f(x)−f_q(x)| ≤ kfk_∞+|f_q(x)| for almost allxin Ω.

Consequently, Z

Ω∩X

|fp(x)|^qdλ(x)≤ Z

Ω∩X

(kfk∞+|fq(x)|)^qdλ(x). (18) The convexity of the functiont7→t^qon [0,+∞) implies that (kfk∞+|fq(x)|)^q ≤ 2^q−1(kfk^q_∞+|fq(x)|^q). So, we pursue the string of inequalites (18) with

Z

Ω∩X

|fp(x)|^qdλ(x)≤2^q−1 Z

Ω∩X

(kfk^q_∞+|fq(x)|^q)dλ(x)

≤2^q−1[λ(X)kfk^q_∞+ (kfqkq)^q]. To summarize, we have proved that

Z

Ω

|fp(x)|^qdλ(x) = Z

Ω∩X^c

|fp(x)|^qdλ(x) + Z

Ω∩X

|fp(x)|^qdλ(x)<+∞.

Thus, fp ∈ L^p(Ω), which was our objective. That concludes the proof of the

technical lemma.

Let us go back to the second part of the proof of Theorem 3, the case where r > q. Choose α satisfying 1/r < α < 1/q and let f be defined on R by f(x) =x^−α1_[1,+∞)(x).

Since|f(x)|^r=x^−αrforx∈[1,+∞) and 0 elsewhere, the choice ofαimplies that f ∈ L^r(Ω) (sinceαr >1). But also f is essentially bounded onR. If f were inL^p(R) +L^q(R), the technical lemma would imply thatf ∈L^q(R). But, this is not the case since|f(x)|^q =x^−αq forx∈[1,+∞) and 0 elsewhere, the choice ofαimplies thatαq <1.

Thus, again in the case where r > q, L^r(R) is not contained in L^p(R) +

L^q(R).

We end this note with the following observation, which links sections 2 and 3. In the first part of Theorem 3, we have proved thatL^r(R)⊂L^p(R) +L^q(R)

wheneverr∈[p, q]. In the course of its proof, a simple explicit decomposition of f ∈L^r(R) as f =f1+f2, withf1∈L^p(R) andf2∈L^q(R) has been provided (see (13) and the upper bounds (14’) and (16)). Indeed, as a consequence of (14) and (16),

kfkp,q≤ kf1kp+kf2kq ≤ kfk^r/p_r +kfk^r/q_r . (19) Hence, the injection of L^r(R) into L^p(R) +L^q(R) is continuous ; therefore, there existsC >0 such that

kfkp,q ≤Ckfkr. (20)

Indeed, using the inequality (19), we can get an upper bound for the norm of this injection (a somewhat complicated expression in terms ofp, q, r). This result complements a more classical one which says that, whenr∈[p, q],L^p(R)∩L^q(R) is contained inL^r(R), then the injection is continuous.

References

[1] J. Bergh; J.-B. L˝ofstr˝om,Interpolation Spaces, Grundlehren der mathema- tischen Wissenschaften,223, Springer-Verlag, New-York, 1976.

[2] J.-B. Hiriart-Urruty; C. Lemarechal, Fundamentals of Convex Analysis, Grundlehren TextEditions, Springer-Verlag, New-York, 2001.

[3] J.-B. Hiriart-Urruty; M. Pradel, La transformation de Fourier de fonctions : au-del`a deL¹ etL², Revue de Math´ematiques Sp´eciales,113no.2 (2003) 25-33.

[4] R.E. Megginson,An Introduction to Banach Space Theory, Graduate Texts in Mathematics,183. Springer-Verlag, New-York, 1998.

[5] R.T. Rockafellar,Convex Analysis, Princeton University Press, 1970.

[6] L. Schwartz,Topologie G´en´erale et Analyse Fonctionnelle, Hermann, Paris, 1970.

[7] Y. Sonntag,Topologie et Analyse Fonctionnelle, Ellipses, Paris, 1998.

JEAN-BAPTISTE HIRIART-URRUTYHas been a professor of Math- ematics at the Paul Sabatier University since 1981. He recieved his Doctorat `es sciences math´ematiques from the Blaise Pascal University (Clermont-Ferrand) in 1977. His current research interests include Variational analysis (convex, nonsmooth, applied) and all the aspects of Optimization.

Institute of Mathematics, Paul Sabatier University, TOULOUSE, FRANCE jbhu@cict.fr.

www.math.univ-toulouse.fr/ jbhu/

PATRICE LASS `ERE received his Doctorat `es sciences math´ematiques from the University Paul Sabatier, Toulouse, France, in 1991. He currently teaches at the University Paul Sabatier. His research interests includes complex analysis in one or several variables and functional analysis.

Institute of Mathematics, Paul Sabatier University, TOULOUSE, FRANCE lassere@math.ups-tlse.fr

http://www.math.univ-toulouse.fr/ lassere/