§ Reisez Lemma
- Let be a closed proper subspace of a normed linear space . Then for all , there exists a (dependent on ), such that . That is, .
- This is easy to establish for say . pick a unit orthogonal vector, it will be at least unit apart (or more) by pythogoras.
- This lemma provides a convenient substitute for orthogonality.
§ Proof via Hahn Banach
- Hahn banach is also a substitute for orthogonality.
- Pick a point . Thus, . Note that is:
- (a) a sublinear function on .
- (b) vanishes on .
- (c) equals the projection onto on .
- By Hahn Banach, the portion of it that is linear extends to a linear functional on all of , and is dominated above by .
- Now, normalize the bounded linear functional so obtained to get a functional such that . Note that noramalization does not change the fact that .
- Next, we build an "approximate normer" . This is an element of unit norm such that . Such an element exists by definition of norm: . See that since , we must surely have that , thus . We claim that . (This is clear by clear and distinct perception since "behaves differently" along ). This must happen, for if not, then for all . This is patently untrue since , thus the unit vector along must vanish at the very least.
- Now, consider .
- Next, estimate .
- This gives .
- The first inequality follows from the defn of norm , and thus , or .
- The second inequality follows from the fact that .
§ What about ?
- does not even hold in . If I pick , there is no unit vector that is units away from the -axis. A vector is at most unit away from the axis.
§ What about ?
- Apparently, this case holds for any reflexive space (double dual equals original space).
- To show counterexample, we need a non-reflexive space. consider , space of sequences under max norm.
- Alternatively, pick under max norm.
- We begin by picking a subspace . So is continuous and .
- Let be the subspace of such that .
- We want to show that there exists no function such that (a) , (that is, and (b) for all .
§ Pedestrian proof when .
- If , then we must have . This means that , or that .
- Intuitively, since , we know that , and since is continuous, it must "spend some time" around , thereby losing some of the integral. Furthermore, since we know that , the maximum integral any such function can attain is .
- Since is continuous and (as ), pick . Then there exists a such that for all , we have that . Thus, we can upper bound the integral of over by . Ie, we surround by two rectangles, one of height , one of height , since we have the bounds. Since , we can see that from the above estimate.
- Thus, this means that , thereby violating the claim that we can find such a such that . Hence proved!
§ Slightly more sophisticated proof when .
- The same setup. We consider the integral operator , defined as .
- We note that .
- We note that . (In fact, , since the function is maximized by being evaluated on which lies on the unit sphere).
- We note that . That is, the distance of a point to the kernel of an operator is the norm of rescaled by the norm of .
- We need an estimate on . By the above argument, we know that by continuity of , , and that for all (as ).
- Combine the two estimates to see tha , which is indeed less than . Done.