near operators

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1 Perturbations and small perturbations: definitions andsome results

We start our discussion on the Campanato theory of near operatorswith some preliminary tools.

Let $X,Y$ be two sets and let a metric $d$ be defined on $Y$. If$F:X\to Y$ is an injective map, we can define a metric on $X$ byputting

$$d_F(x^{\prime },x^{\prime \prime })=d(F(x^{\prime }),F(x^{\prime \prime })).$$

Indeed, $d_F$ is zero if and only if $x^{\prime }=x^{\prime \prime }$ (since $F$ isinjective); $d_F$ is obviously symmetric and the triangleinequality follows from the triangle inequality of $d$.

Moreover, if $F(X)$ is a complete subspace of $Y$, then $X$ iscomplete with respect to the metric $d_F$.

Indeed, let $(u_n)$ be a Cauchy sequence in $X$. By definition of$d$, then $(F(u_n))$ is a Cauchy sequence in $Y$, and inparticular in $F(X)$, which is complete. Thus, there exists $y_0=F(x_0)\in F(X)$ which is limit of the sequence $(F(u_n))$.$x_0$ is the limit of $(x_n)$ in $(X,d_F)$, which completes theproof.

A particular case of the previous statement is when $F$ is onto(and thus a bijection) and $(Y,d)$ is complete.

Similarly, if $F(X)$ is compact in $Y$, then $X$ is compact withthe metric $d_F$.

Definition 1.1

Let $X$ be a set and $Y$ be a metric space. Let $F,G$ be two mapsfrom $X$ to $Y$. We say that $G$ is a perturbation of $F$ if thereexist a constant $k>0$ such that for each $x^{\prime },x^{\prime \prime }\in X$ one has:

$$d(G(x^{\prime }),G(x^{\prime \prime }))\le kd(F(x^{\prime }),F(x^{\prime \prime }))$$

Remark 1.2

In particular, if $F$ is injective then $G$ is a perturbation of$F$ if $G$ is uniformly continuous with respect to the metric induced on$X$ by $F$.

Definition 1.3

In the same hypothesis as in the previous definition, we say that$G$ is a small perturbation of $F$ if it is a perturbation ofconstant .

We can now prove this generalization of the Banach-Caccioppolifixed point theorem:

Theorem 1.4

Let $X$ be a set and $\mathrm{(}Y\mathrm{,}d\mathrm{)}$ be a complete metric space. Let $F\mathrm{,}G$ be two mappings from $X$ to $Y$ such that:

1. $F$ is bijective;

2. $G$ is a small perturbation of $F$.

Then, there exists a unique $u\mathrm{\in }X$ such that $G\mathit{}\mathrm{(}u\mathrm{)}\mathrm{=}F\mathit{}\mathrm{(}u\mathrm{)}$

Proof.

The hypothesis (1) ensures that the metric space $(X,d_F)$ is complete. If we now consider the function $T:X\to X$defined by

$$T(x)=F^{-1}(G(x))$$

we note that, by (2), we have

$$d(G(x^{\prime }),G(x^{\prime \prime }))\le kd(F(x^{\prime }),F(x^{\prime \prime }))$$

where $k\in (0,1)$ is the constant of the small perturbation;note that, by the definition of $d_F$ and applying $F\circ F^{-1}$ to the first side, the last equation can be rewritten as

$$d_F(T(x^{\prime }),T(x^{\prime \prime }))\le k{d}_F(x^{\prime },x^{\prime \prime });$$

in other words, since , $T$ is a contraction in thecomplete metric space $(X,d_F)$; therefore (by the classicalBanach-Caccioppoli fixed point theorem) $T$ has a unique fixedpoint: there exist $u\in X$ such that $T(u)=u$; by definitionof $T$ this is equivalent to $G(u)=F(u)$, and the proof is hencecomplete.∎

Remark 1.5

The hypothesis of the theorem can be generalized as such: let $X$be a set and $Y$ a metric space (not necessarily complete); let$F,G$ be two mappings from $X$ to $Y$ such that $F$ is injective,$F(X)$ is complete and $G(X)\subseteq F(X)$; then there exists $u\in X$ such that $G(u)=F(u)$.

(Apply the theorem using $F(X)$ instead of $Y$ as target space.)

Remark 1.6

The Banach-Caccioppoli fixed point theorem is obtained when$X=Y$ and $F$ is the identity.

We can use theorem 1.4 to prove a result that appliesto perturbations which are not necessarily small (i.e. for whichthe constant $k$ can be greater than one). To prove it, we mustassume some supplemental structure on the metric of $Y$: inparticular, we have to assume that the metric $d$ is invariant bydilations, that is that $d(\alpha y^{\prime },\alpha y^{\prime \prime })=\alpha d(y^{\prime },y^{\prime \prime })$ for each $y^{\prime },y^{\prime \prime }\in Y$. The most common case of sucha metric is when the metric is deduced from a norm (i.e. when $Y$is a normed space, and in particular a Banach space). The resultfollows immediately:

Corollary 1.7

Let $X$ be a set and $\mathrm{(}Y\mathrm{,}d\mathrm{)}$ be a complete metric space witha metric $d$ invariant by dilations. Let $F\mathrm{,}G$ be two mappingsfrom $X$ to $Y$ such that $F$ is bijective and $G$ isa perturbation of $F$, with constant $K\mathrm{>}\mathrm{0}$.

Then, for each $M\mathrm{>}K$ there exists a unique $u_M\mathrm{\in }X$ such that$G\mathit{}\mathrm{(}u\mathrm{)}\mathrm{=}M\mathit{}F\mathit{}\mathrm{(}u\mathrm{)}$

Proof.

The proof is an immediate consequence of theorem 1.4given that the map $\tilde{G}(u)=G(u)/M$ is a small perturbationof $F$ (a property which is ensured by the dilation invariance ofthe metric $d$).∎

We also have the following

Corollary 1.8

Let $X$ be a set and $\mathrm{(}Y\mathrm{,}d\mathrm{)}$ be a complete, compact metric spacewith a metric $d$ invariant by dilations. Let $F\mathrm{,}G$ be twomappings from $X$ to $Y$ such that $F$ is bijective and $G$ isa perturbation of $F$, with constant $K\mathrm{>}\mathrm{0}$.

Then there exists at least one $u_K\mathrm{\in }X$ such that$G\mathit{}\mathrm{(}{u}_{\mathrm{\infty }}\mathrm{)}\mathrm{=}K\mathit{}F\mathit{}\mathrm{(}{u}_{\mathrm{\infty }}\mathrm{)}$

Proof.

Let $(a_n)$ be a decreasing sequence of real numbers greater thanone, converging to one ($a_n\downarrow 1$) and let $M_n=a_{n}K$for each $n\in \mathbb{N}$. We can apply corollary 1.7 toeach $M_n$, obtaining a sequence $u_n$ of elements of $X$ forwhich one has

$$G(u_n)=M_{n}F(u_n).$$(1)

Since $(X,d_F)$ is compact, there exist a subsequence of $u_n$which converges to some $u_{\mathrm{\infty }}$; by continuity of $G$ and $F$we can pass to the limit in (1), obtaining

$$G(u_{\mathrm{\infty }})=KF(u_{\mathrm{\infty }})$$

which completes the proof.∎

Remark 1.9

For theorem 1.8 we cannot ensure uniqueness of $u_{\mathrm{\infty }}$,since in general the sequence $u_n$ may change with the choice of$a_n$, and the limit might be different. So the corollary can onlybe applied as an existence theorem.

2 Near operators

We can now introduce the concept of near operators and discusssome of their properties.

A historical remark: Campanato initially introduced the concept inHilbert spaces; subsequently, it was remarked that most of thetheory could more generally be applied to Banach spaces; indeed,it was also proven that the basic definition can be generalized tomake part of the theory available in the more general environmentof metric vector spaces.

We will here discuss the theory in the case of Banach spaces,with only a couple of exceptions: to see some of the extraproperties that are available in Hilbert spaces and to discussa generalization of the Lax-Milgram theorem to metric vector spaces.

2.1 Basic definitions and properties

Definition 2.1

Let $X$ be a set and $Y$ a Banach space. Let $A,B$ be twooperators from $X$ to $Y$. We say that $A$ is near $B$ if and onlyif there exist two constants $\alpha >0$ and $k\in (0,1)$ suchthat, for each $x^{\prime },x^{\prime \prime }\in X$ one has

$$\parallel B(x^{\prime })-B(x^{\prime \prime })-\alpha (A(x^{\prime })-A(x^{\prime \prime }))\parallel \le k\parallel B(x^{\prime })-B(x^{\prime \prime })\parallel$$

In other words, $A$ is near $B$ if $B-\alpha A$ is a smallperturbation of $B$ for an appropriate value of $\alpha$.

Observe that in general the property is not symmetric: if $A$ isnear $B$, it is not necessarily true that $B$ is near $A$; as wewill briefly see, this can only be proven if , or inthe case that $Y$ is a Hilbert space, by using an equivalentcondition that will be discussed later on. Yet it is possible todefine a topology with some interesting properties on the space ofoperators, by using the concept of nearness to form a base.

The core point of the nearness between operators is that it allowsus to “transfer” many important properties from $B$ to $A$; inother words, if $B$ satisfies certain properties, and $A$ is near$B$, then $A$ satisfies the same properties. To prove this, and toenumerate some of these “nearness-invariant” properties, we willemerge a few important facts.

In what follows, unless differently specified, we will alwaysassume that $X$ is a set, $Y$ is a Banach space and $A,B$ are twooperators from $X$ to $Y$.

Lemma 2.2

If $A$ is near $B$ then there exist two positive constants $M_{\mathrm{1}}\mathrm{,}{M}_{\mathrm{2}}$ such that

$$\parallel B(x^{\prime })-B(x^{\prime \prime })\parallel \le M_1\parallel A(x^{\prime })-A(x^{\prime \prime })\parallel$$

$$\parallel A(x^{\prime })-A(x^{\prime \prime })\parallel \le M_2\parallel B(x^{\prime })-B(x^{\prime \prime })\parallel$$

Proof.

We have:

$$\begin{array}{c}\parallel B(x^{\prime })-B(x^{\prime \prime })\parallel \le \hfill \\ \hfill \le \parallel B(x^{\prime })-B(x^{\prime \prime })-\alpha (A(x^{\prime })-A(x^{\prime \prime }))\parallel +\alpha \parallel A(x^{\prime })-A(x^{\prime \prime })\parallel \le \\ \hfill \le k\parallel B(x^{\prime })-B(x^{\prime \prime })\parallel +\alpha \parallel A(x^{\prime })-A(x^{\prime \prime })\parallel \end{array}$$

and hence

$$\parallel B(x^{\prime })-B(x^{\prime \prime })\parallel \le \frac{\alpha }{1-k}\parallel A(x^{\prime })-A(x^{\prime \prime })\parallel$$

which is the first inequality with $M_1=\alpha /(1-k)$ (which ispositive since ).

But also

$$\begin{array}{c}\parallel A(x^{\prime })-A(x^{\prime \prime })\parallel \le \hfill \\ \hfill \le \frac{1}{\alpha }\parallel B(x^{\prime })-B(x^{\prime \prime })-\alpha (A(x^{\prime })-A(x^{\prime \prime }))\parallel +\frac{1}{\alpha }\parallel B(x^{\prime })-B(x^{\prime \prime })\parallel \le \\ \hfill \le \frac{k}{\alpha }\parallel B(x^{\prime })-B(x^{\prime \prime })\parallel +\frac{1}{\alpha }\parallel B(x^{\prime })-B(x^{\prime \prime })\parallel \end{array}$$

and hence

$$\parallel A(x^{\prime })-A(x^{\prime \prime })\parallel \le \frac{1+k}{\alpha }\parallel B(x^{\prime })-B(x^{\prime \prime })\parallel$$

which is the second inequality with $M_2=(1+k)/\alpha$.∎

The most important corollary of the previous lemma is thefollowing

Corollary 2.3

If $A$ is near $B$ then two points of $X$ have the same imageunder $A$ if and only if the have the same image under $B$.

We can express the previous concept in the following formal way:for each $y$ in $B(X)$ there exist $z$ in $Y$ such that$A(B^{-1}(y))=\{z\}$ and conversely. In yet other words: eachfiber of $A$ is a fiber (for a different point) of $B$, andconversely.

It is therefore possible to define a map $T_A:B(X)\to Y$ byputting $T_A(y)=z$; the range of $T_A$ is $A(X)$. Conversely, itis possible to define $T_B:A(X)\to Y$, by putting $T_B(z)=y$;the range of $T_B$ is $B(X)$. Both maps are injective and, ifrestricted to their respective ranges, one is the inverse of theother.

Also observe that $T_B$ and $T_A$ are continuous. This followsfrom the fact that for each $x\in X$ one has

$$T_A(B(x))=A(x),T_B(A(x))=B(x)$$

and that the lemma ensures that given a sequence $(x_n)$ in $X$,the sequence $(B(x_n))$ converges to $B(x_0)$ if and only if$(A(x_n))$ converges to $A(x_0)$.

We can now list some invariant properties of operators withrespect to nearness. The properties are given in the form “ifand only if” because each operator is near itself (thereforeensuring the “only if” part).

1. a map is injective if and only if it is near aninjective operator;

2. a map is surjective if and only if it is near a surjectiveoperator;

3. a map is open if and only if it is near an open map;

4. a map has dense range if and only if it is near a map withdense range.

To prove (2) it is necessary to usetheorem 1.4.

Another important property that follows from the lemma is that ifthere exist $y\in Y$ such that $A^{-1}(y)\cap B^{-1}(y)\ne \mathrm{\varnothing }$, then it is $A^{-1}(y)=B^{-1}(y)$: intersectingfibers are equal. (Campanato only stated this property for thecase $y=0$ and called it “‘the kernel property”; I prefer tocall it the “fiber persistence” property.)

2.1.1 A topology based on nearness

In this section we will show that the concept of nearness betweenoperator can indeed be connected to a topological understanding ofthe set of maps from $X$ to $Y$.

Let $ℳ$ be the set of maps between $X$ and $Y$. For each$F\in ℳ$ and for each $k\in (0,1)$ we let $U_k(F)$ the set ofall maps $G\in ℳ$ such that $F-G$ is a small perturbation of $F$with constant $k$. In other words, $G\in U_k(F)$ if and only if $G$ is near$F$ with constants $1,k$.

The set satisfies the axioms ofthe set of fundamental neighbourhoods. Indeed:

1. $F$ belongs to each $U_k(F)$;

2. $U_k(F)\subset U_h(F)$ if and only if , and thus theintersection property of neighbourhoods is trivial;

3. for each $U_k(F)$ there exist $U_h(F)$ such that for each $G\in U_h(F)$ there exist $U_j(G)\subseteq U_k(F)$.

This last property (permanence of neighbourhoods) is somewhat lesstrivial, so we shall now prove it.

Proof.

Let $U_k(F)$ be given.

Let $U_h(F)$ be another arbitrary neighbourhood of $F$ and let$G$ be an arbitrary element in it. We then have:

$$\parallel F(x^{\prime })-F(x^{\prime \prime })-(G(x^{\prime })-G(x^{\prime \prime }))\parallel \le h\parallel F(x^{\prime })-F(x^{\prime \prime })\parallel .$$(2)

but also (lemma 2.2)

$$\parallel (G(x^{\prime })-G(x^{\prime \prime }))\parallel \le (1+h)\parallel F(x^{\prime })-F(x^{\prime \prime })\parallel .$$(3)

Let also $U_j(G)$ be an arbitrary neighbourhood of $G$ and $H$ anarbitrary element in it. We then have:

$$\parallel G(x^{\prime })-G(x^{\prime \prime })-(H(x^{\prime })-H(x^{\prime \prime }))\parallel \le j\parallel G(x^{\prime })-G(x^{\prime \prime })\parallel .$$(4)

The nearness between $F$ and $H$ is calculated as such:

$$\begin{array}{c}\parallel F(x^{\prime })-F(x^{\prime \prime })-(H(x^{\prime })-H(x^{\prime \prime }))\parallel \le \hfill \\ \hfill \parallel F(x^{\prime })-F(x^{\prime \prime })-(G(x^{\prime })-G(x^{\prime \prime }))\parallel +\parallel G(x^{\prime })-G(x^{\prime \prime })-(H(x^{\prime })-H(x^{\prime \prime }))\parallel \le \\ \hfill h\parallel F(x^{\prime })-F(x^{\prime \prime })\parallel +j\parallel G(x^{\prime })-G(x^{\prime \prime })\parallel \le (h+j(1+h))\parallel F(x^{\prime })-F(x^{\prime \prime })\parallel .\end{array}$$(5)

We then want $h+j(1+h)\le k$, that is $j\le (k-h)/(1+h)$;the condition is always satisfied on the right side, andthe left side gives us .∎

It is important to observe that the topology generated this way isnot a Hausdorff topology: indeed, it is not possible to separate$F$ and $F+y$ (where $F\in ℳ$ and $y$ is a constant element of $Y$).On the other hand, the subset of all maps with with a fixed valuedat a fixed point ($F(x_0)=y_0$) is a Hausdorff subspace.

Another important characteristic of the topology is that the set$\mathscr{H}$ of invertible operators from $X$ to $Y$ is open in $ℳ$(because a map is invertible if and only if it is near an invertible map).This is not true in the topology of uniform convergence, as iseasily seen by choosing $X=Y=ℝ$ and the sequence withgeneric element $F_n(x)=x^3-x/n$: the sequence converges (inthe uniform convergence topology) to $F(x)=x^3$, which isinvertible, but none of the $F_n$ is invertible. Hence $F$ is anelement of $\mathscr{H}$ which is not inside $\mathscr{H}$, and $\mathscr{H}$ is not open.

2.2 Some applications

As we mentioned in the introduction, the Campanato theory of nearoperators allows us to generalize some important theorems; we willnow present some generalizations of the Lax-Milgram theorem, anda generalization of the Riesz representation theorem.

[TODO]