Some more notes on some Real and Functional Analysis text

Updating this as I am working.  Sort of interested in seeing if these notes are at all useful to myself or anyone else studying.

From pg 87. (Real and Functional Analysis, Serge Lange, Springer Verlag, 1993, 3rd edition)

Notes on Theorem 1.3: Stated ...Let c be a right end point of this interval, namely of the convex and compact set λ(S0) . To show that λ(S0) consists of one point. The the set λ−1(c)⋂S0 is non empty, convex and compact. We contend that it lies in F. Let x be an element in λ−1(c)⋂S0 and suppose that we can write

x = ty1 +(1−t)y2 with y1,y2 ∈ S and 0S

0 ∈ F , we get y1,y2 ∈ S0 . Applying λ we find that λ(x) = c = tλ(y1)+(1−t)λ(y2). Since c is an end point of the interval λ(y1) = λ(y2) = c.

And it is here where I begin a bit of notes since at least to me it hadn’t appeared immediately clear that λ(y1) = λ(y2) = c.

First we’d presume the interval including c is ordered and that c is a maximal element in such interval being a chosen right end point of the interval. Secondly. We could presume for now that λ(y1) ≤ c, λ(y2) ≤ c since by definition c is an end point or maximal element for the set of λ(S0).

Now if λ(y2) < c means t(λ(y1) − λ(y2)) > 0 and since t > 0, then λ(y1) > λ(y2). But c = tλ(y1) +(t −1)λ(y2) < tλ(y1) + (1−t)λ(y1) = λ(y1)or

λ(y1) > c, contradicting our earlier assumption (since c is a maximal element on λ(S0)). Thus λ(y2) = c. and from this relation we see

λ(y1) = λ(y2) = c

Pg. 89 ­90 Lemma 2.2 We have E = C ⋃(−C). To see this, let x ∈ E and suppose x ∈/ C, x ∈/ −C. Since C is maximal, the cone consisting of all elements c+tx with c ∈ C, t ≥ 0. Hence we can write

−a = c1 +t1x = c2 −t2x

with c1,c2 ∈ C and t1,t2 ≥ 0. Consequently

c1 +(t1 +t2)x ∈ C,

However, c1 +t1x = −a is on the line segment between c1and c1 +(t1 +t2)x , and thus lies in C, a contradiction which proves that E = C ⋃(−C). Noting here...that the line defined by points c1 and c1 +(t1 +t2)x, are in C...paying attention to the above C is not to be confused immediately as the cone of elements c+tx which were a point I initially overlooked. Also by the previous lines where −a ∈/ C the cone c+tx and c+t(­x) generates the elements not in C which should include ­a which in turn leads into the contradictions found above where we have x must belong to C and/or ­C. Its worth noting the role that C plays in defining its compliment cone which were necessary in demonstrating contradiction here. Since c2 ∈ C and since t1,t2 ≥ 0 where t1 +t2 form yet another positive real greater than or equal to 0 means that c1 +(t1 +t2)x ∈ C (mostly c2 is the crux for which this is true). Thus with elements of the compliment cone on a given line of elements found in C leads to our given contradiction from this point, or in other words, no such compliment exists as a non empty set.

Further comments on the strategy of this proof. H forming the closed hyperplane also represents the defined kernel of the desired sought after functional on E. In this case the generator of H is found by the convex application of elements of C and the element a, (1−τ)x −τa = h ∈ H.

In application to Mazur’s Theorem, we are looking for a functional λ of our given hyperplane H separating it, such that λ(x) ≥ c where c is a given number determined by the hyperplane and its functional which in turn provides for the definition of the closed half space, and that in defining this functional we had done so by use of intersections of closed half spaces containing A. This is actually done in the remainder of the proof with its other remaining subpart of the proof...along these lines, it would appear that the intersection of closed half spaces containing A are used in defining the functional necessary to supply a given equivalent closed half space.

Hermitian Forms chapter. Theorem 1.6 (pg. 100), Not sure if there is an error in the text. Namely where a reading … “We let α = t < z,y > with t real and not equal to 0. We can then cancel t and get a contradiction for small t, if < y,z >/= 0 . I believe this should correctly read with large negative t as in a previous proof given, but that is apparently if the following logic is true:

First it would be worth noting that under expansions: |z +αy|2 =< z +αy,z +αy >= < z,z >+< αy,z >+< z,αy >+< αy,αy >

=< z,z >+α < y,z >+αˉ < z,y >+ααˉ < y,y > and where |z|2 =< z,z > provides 0 ≤ α < y,z >+αˉ < z,y >+ααˉ < y,y >

using the α = t < z,y >, we have 0 ≤ t < z,y >< y,z >+t < y,z >< z,y >+t2 < y,z >< z,y >< y,y > where dividing by t yields

0 ≤ < z,y >< y,z >+< y,z >< z,y >+t < y,z >< z,y >< y,y > obviously at this point using small t appears to yield nothing obviously of immediate contradiction(?) since 0 ≤ < z,y >< y,z > is positive definite by convention applied for this inequality (recall positive definite means that we examine the real component of the hermitian inner product and this is also assumed to be positive with inner products if not equal to zero, and keep in mind with examination using inequalities, the book is neglecting in notation Re() of all the resultant right hand side of of such inner product and summations, or rather assumes this into the inequalities expression). However, if taking t negative still being real and large we can see contradictory problems arising in the given inequality since the negative t term predominates and over the set of reals we can make t large enough so that the right hand side of the inequality is negative contradicting our initial assumptions regarding the inequality. We could assume here additionally that y is not an element of the null vector space or that < y,y >/= 0 .

pg.103 we have v = v −∑anvn +∑anvn here we apply theorem due to pythagoras as usual assuming v −∑anvn is orthogonal at least to ∑anvn , and

|v|2 = |v −∑anvn +∑anvn|2 = |v −∑anvn|2 +|∑anvn|2 means 0 ≤ |v −∑anvn +∑anvn|2 = |v|2 −|∑anvn|2 means ∑|an|2 = |∑anvn|2 ≤ |v|2 obviously by the

orthogonality of vi = vj where i =/ j which provides for the bessel inequality. It would be noted that an orthonormal vector neither expressed fully in the context of a hilbert basis leaves as in this an orthogonal component of v remaining, and consequently why the bessel inequality is true.

pg.104 Theorem 2.1 Some notes on this proof: Theorem 1.6 provides that there exists z ∈ E, z =/ 0 such that z is perpendicular to F. And further, we contend that some scalar multiple of z achieves our purpose, say αz. A necessary condition on αis that < z,αz > = λ(z) or in other words,

αˉ = λ(z)/ < z,z >. This is also sufficient. Indeed, for any x ∈ E , we can write

x = x −λ(x)/λ(z) z + λ(x)/λ(z) z

and

x −λ(x)/λ(z) z

lies in F. Taking the product with αz , we obtain < x,αz > = λ(x) thus proving our theorem.

Additional notes: First, while Theorem 1.6 says there is a point perpendicular to z ∈F, and from this coupled with another point x ∈ E , we can find with suitable choice of scalar ultimately providing for a vector of desired magnitude in the sense of the unit norm of z such that a resultant vector (from the difference of x and the scalar times the unit vector z) provides for a point in F (or kernel of λ). It would appear outside of additional assumptions or having shown by method what such a scalar should be, beyond abstraction we hadn’t the conveniences of exactly confirming this, although the consequence of this fact, should reside on the existence of any scalar α that should be suitable and that such existence should be inferred from Theorem 1.6 (?), or in other words finding a resultant vector which lands in F is merely picking a suitable magnitude in the direction of unit vector of z such this subtracted from x gives us the appropriate result (our theorem indicates this and this is not outside of reason working in vector algebra, remembering that from the outside z is in direction perpendicular to F, we should have an intercept by choice of scalar on the unit norm of z, thus regardless of where x results on a given coordinate systems relative to F, we should be in the direction of intercepting the line F in the direction of z multiplied by its scalar. More in depth, to convince yourself of this, I recommend graphically sketching this. First we use from a local coordinate system defining some point y0 in F such that y0intercepts our given origin, and having constructed from the point a perpendicular line running from y0 to the set E and intercepting a point z, defines the direction of our vector z, then having this and having chosen any point x in E such that the vector of x is drawn from the origin y0 means that we have a resultant vector intersecting and in the direction of F (parallel and intercept to F), in using suitable x −λ(x)/λ(z) z and thus perpendicular to z. Next from our given x,z ∈ E and x,z ∈/ F we can make use of the the following to aid us. Namely, that

< x −λ(x)/λ(z) z,z >	= 0	since z and x −λ(x)/λ(z) z are perpendicular.
< x −λ(x)/λ(z) z,z >	=	< x,z > −λ(x)/λ(z) < z,z > = < x,z > −λ(x)/(αˉ < z,z >)*< z,z > means < x,z > = λ(x)/αˉ

Then < x,αz > = αˉ < x,z > = λ(x) α/αˉ ˉ = λ(x) .

Some key notes of Chapter 6:

The General Integral

We want two things from an integral which are not provided by the standard Riemann integral of bounded functions:

(1)We want to integrate unbounded functions.

(2)We want to be able to take limits under the integral sign, of a fairly general nature, more general than uniform limits.

To achieve this, we proceed in a manner entirely similar to the manner used when extending the integral to the completion of a space of step functions, except that instead of the sup norm we use the L1 ­ norm.

pg 117 Theorem M7: Let f : X → Y be a mapping of X into a metric space. Let {fn} be a sequence of measurable mappings X into Y which converges pointwise to f . Then f is measurable.

Proof notes: Generally the proof is subdivided into two different parts, and important assumption highlights here {fn} is a sequence of measurable mappings X into Y which converges pointwise to f . Where some element of {fn}is a measurable mapping, and the sequence of these elements converge pointwise to f. It is important also noting, implicitly (and is used in the second part of this proof) that a σ ­algebra of X and Y are used to the definition of a measurable mapping. Here if (X,M) and (Y ,N) are measurable spaces, and f : X → Y is a map, we define f to be measurable if for every B ∈ N the set f−1(B) is in M . Hence for a sequence of measurable maps as indicated above, we should have a σ­algebras defined on the sets X and Y for each measurable map. In this way, if we can show that an element of the inverse mapping of f belongs also to the inverse mapping of an element in {fn}, and vice versa elements in the inverse mapping of {fn} are in the set of elements in inverse mapping of f, then we have that f must be measurable. The first part of the proof, uses convergence of {fn} to f pointwise indicating that from an open set U of Y, f −1(U)is contained in

∞ ∞

m∩=1k∪=mfk−1(U). Conversely, with closed sets showing the opposite is true. For part 2 we can use the following illustration in aiding

Here we can see under successive iterations of n that the dotted circle grows larger and larger to

conform to V as n → ∞. n∪=1∞ V n = n∪=1∞ An = V while

for any n , V n ⊂ An. Now we have open and close set relations formulated so that we can inter relate the open and closed set inclusions shown in the proof (in part 1). While I mentioned formally the definition of a measurable map here, really there weren’t a formal demonstration which showed exactly the definition given for measurable maps and f, but instead by showing equality in the relation of the inclusions above, that we have a link between measurable maps and f means that f is a measurable. (in this case being a countable/denumerable intersection and union of measurable maps)...additionally σ­ algebras on X and Y should make this clear in the given formulation.

Key points in this proof really use the topology of our given metric space, coupled with convergence of f from its sequence of measurable maps to demonstrate the inter relation between f and its sequences of measurable maps, because of this inter relation, f is measurable (coupled with definitions provided by the sigma algebra).

pg. 118 M8 A map f : X → E of X into a finite dimensional space is measurable if and only if it is a pointwise limit of simple maps.

Notes on proof: On simple maps. A simple map f : X → Z into any set Z is said to be a simple map if it takes on only a finite number of values, and if, for each v ∈ Z the inverse image f−1(v) is measurable. The converse assumption is given by M7. So it is to be shown that in the case (because of reducibility. I believe that since the case of the complex plane being nothing more than a R x R problem and by extension any finite dimension of Rn or Cn similarly reducible to one of the same type namely demonstrating in each dimension the problem of the E = R order type) of E = R. For the construction of the simple map, for an integer n ≥ 1 cut up the interval [­n,n] into intervals of equal length 1/n and denote these intervals by J1,...,JN . We take each Jk to be closed on the left and open on the right. We let JN+1consist of all t such that |t| ≥ n. Let

Let Ak = f−1(Jk) for k = 1, …,N+1 so that each Ak is measurable, the sets Ak (k = 1,...,N +1) are disjoint, and their union is X. On each Ak we define a constant map ψn by ψn(Ak) = infAKf if k = 1, …, N. It that these values for the given set Ak should be mapped to the least upper bound of the cut interval (which is the left value set closure element) under ψn . We can write AN+1 = B ⋃B′ where B consists of those numbers t such that f(t) <−n. We define ψn(B) = n and ψn(B′) = −n . Then the sequence {ψn} converges pointwise to f, and each ψn is a simple function. Keeping in mind that as n tends to infinity the given interval size diminishes to zero or namely to each point of X as ψn maps from X into R. The corresponding interval size of Akdiminishes likewise to a point sized interval, or that for any ε > 0 there exists δ > 0 and N>0, such that for m ≥ N so that distance between the sequence ψm and f is less then ε and an corresponding distance interval in X exists less than δwhich is to say nothing more than how convergence is defined for f by a given simple map sequence which converges to f.

A measurable space together with a measure is called a measured space. When we want to specify all data in the notation, we write the full triple (X,M,μ) for a measured space.