Download Banach Spaces Lecture 15: Hahn-Banach Theorem and Invertibility and more Study notes Mathematics in PDF only on Docsity!
B4a Banach spaces Michaelmas 2010 lectured by Bernd Kirchheim based on notes by CJK Batty
Lecture 15
5.Corollary The Hahn-Banach Theorem is true for every separable real normed space (X, ‖ · ‖):
Y subspace of X, g ∈ Y ′^ ⇒ ∃G ∈ X′^ : G|Y = g and ‖G‖X′ = ‖g‖Y ′. 6.Proposition Let (X, ‖ · ‖) be a complex normed space. We consider the real linear space XR obtained from X when allowing multiplication only by α ∈ R, the (XR, ‖ · ‖) is a real normed space. Moreover, if the Hahn- Banach Theorem holds true for (XR, ‖ · ‖) then it holds also for (X, ‖ · ‖). In particular, we have proven it for all separable complex spaces. Proof It is a fact from linear algebra, and easily checked, that every complex vector space X is also a real vector space, denoted XR when consid- ered over the scalars R only1). Moreover, the norm of X is one-homogeneous with respect to the reals, so also a norm on XR. Now, if Y is a subspace of X, then αx+y ∈ YR for all x, y ∈ YR and α ∈ R, so YR is a subspace of XR. Similar, since Re(αz) = α Re(z) for all α ∈ R and z ∈ C, we see that for each g ∈ Y ′^ is Re(g) : x → Re(g(x)) linear over R and | Re(g(x))| ≤ |g(x)| implies Re(g) ∈ (YR)′^ with ‖ Re(g)‖Y (^) R′ ≤ ‖g‖. Using the
HB-T for XR we find G˜ ∈ (XR)′^ such that G˜|YR = Re(g) and ‖ G˜‖X R′ ≤ ‖g‖.
However, G˜ : X → R is not C-linear, so not in X′^! But is G˜ the real part of a complex G ∈ X′, i.e. G˜(x) = Re(G(x)) for all x ∈ X? Then necessarily
G(x) = Re(G(x))+iIm(G(x)) = Re(G(x))−i Re(iG(x)) = Re(G(x))−i Re(G(ix)).
We take the last expression as a definition of G : X → C, then surely Re(G) = G.˜ Now, G is clearly R-linear (as was G˜), moreover
G(ix) = G˜(ix) − i G˜(−x) = G˜(ix) + i G˜(x) = i( G˜(x) − i G˜(ix)) = iG(x)
for all x ∈ X, hence G is even C-linear. Similarly, we calculate for y ∈ Y
g(y) = Re(g(y)) − i Re(g(iy)) = G(y), so G|Y = g,
since iy ∈ Y and Re(g)|Y = G. Finally, for all x ∈ X there is t ∈ R s.t.
|G(x)| = eitG(x) = G(eitx) = Re(G(eitx)) = G˜(eitx) ≤ ‖ G˜‖‖eitx‖ ≤ ‖g‖‖x‖,
and hence ‖G‖ ≤ ‖g‖ as needed.
1)and if B was a basis for X then B ∪ iB, i = √−1, is a basis for XR
Seperation of convex sets
7.Proposition Let (X, ‖ · ‖) be a separable real normed space and let 0 ∈ U ⊂ X be open and convex and xo ∈/ U. Then there is
G ∈ X′^ such that G(y) < G(x 0 ) for all y ∈ U.
Proof There is rU > 0 such that Br(0) ⊂ U , hence
∀y ∈ X : F (y) = inf{t > 0 , y ∈ tU } ≤
‖y‖ rU
the Minkowski functional of U is finite on X. As 0 ∈ U , we have for 0 < α < β that αU ⊂ αU + (β − α)U = βU. So for all y, z ∈ X and ε > 0, t = F (y) + ε and s = F (z) + ε is y ∈ tU and z ∈ sU. Hence if λ ∈ (0, 1) then
λy + (1 − λ)z ∈ λtU + (1 − λ)sU = [λt + (1 − λ)s]U, , hence F (λy + (1 − λ)z) ≤ λt + (1 − λ)s = λF (y) + (1 − λ)F (z) + ε.
Sending ε ց 0 we conclude that F is convex and it is easy to check that for all t (positive) is {y : F (y) < t} = tU open2). Set Y = span({x 0 }) and g(tx 0 ) = tF (x 0 ) for t ∈ R. We notice F (x 0 ) ≥ 1 since x 0 ∈/ U , and F (tx 0 ) = tF (x 0 ) if t > 0 implies g ≤ F on Y. By Corollary 3 and Lemma 4 there is G ∈ X′^ s.t. G ≤ F on X and G(tx 0 ) = tF (x 0 ). Hence, for all y ∈ U we have G(y) ≤ F (y) < 1 ≤ F (x 0 ) = G(x 0 ).
- Corollary Let (X, ‖·‖) be a separable real normed space, and V, W ⊂ X be convex. If V is open and V ∩ W = ∅ then there exists g ∈ X′^ and α ∈ R such that
for all v ∈ V, w ∈ W : g(v) < α ≤ g(w). (α = inf g(W ))
Proof Without loss of generality exist v 0 ∈ V , w 0 ∈ W. Then 0 ∈ U = (V − v 0 ) − (W − w 0 ) =
w∈W V^ + (w^ −^ v^0 −^ w^0 ) open and convex and x 0 = w 0 − v 0 ∈/ U , else 0 ∈ V − W. By Prop 7 exists g′^ ∈ X′^ such that for all v, V, w ∈ W
g((v − v 0 ) − (w − w 0 )) < g(w 0 − v 0 ), hence g(v) < g(w).
Since V is open and g 6 = 0 is g(V ) open and hence α = inf g(W ) ∈/ g(V ).
2)we even have lip(F ) ≤ 1 /rU , DIY
8 Invertibility and spectrum
Let T : X → Y be a bounded linear operator between normed vector spaces X and Y. We say that T is invertible if there exists a bounded linear operator S : Y → X such that ST = IX and T S = IY ; then we write S = T −^1 , the inverse of T. Note that in infinite-dimensional spaces these two equalities are independent of one another (see Example 6.20 with left and right shift T ′T = Il 2 6 = T T ′). If T : X → Y is a bijection, then there is an algebraic inverse. In fact, if X and Y are Banach spaces and T : X → Y is a continuous linear bijection, then the algebraic inverse is automatically bounded! The proof of this, however, is beyond the scope of this course only in 4th year. Nevertheless, we shall assume in this section that X and Y are Banach spaces, and Y = X. There is really no loss in doing so, since an invertible operator extends to an invertible operator on the completions (see Theorem 5.8). Our aim is to generalize the notion/theory of eigenvalues of a T ∈ B(X) from linear finite dimensional to Banach spaces, so to investigate when T − λI is invertible. 1.Definition Given T ∈ B(X) where X is a Banach space, the resolvent set ρ(T ) of T is defined by:
ρ(T ) = {λ ∈ F : λI − T is invertible}.
The spectrum σ(T ) is defined by
σ(T ) = F \ ρ(T ).
The point spectrum σp(T ) is the set of all eigenvalues (i.e. λ ∈ σp(T ) iff ker(λI − T ) 6 = 0) , and the approximate point spectrum σap(T ) is the set of approximate eigenvalues of T , i.e. the set of those λ for which there is a sequence of unit vectors (xn) in X such that ‖(λI − T )xn‖ → 0. Remarks
- Consider T (αj ) = (αj /j), then ker(T ) = { 0 } but as T (ej ) = ej /j → 0 we see that the inclusion σp(T ) ⊂ σap(T ) can be strict.
- The example of the right shift T (ek) = ek+1 show that even if 0 ∈/ σap(T ), T might still fail to be surjective and 0 ∈ σ(T ) is still possible. The next result clarifies that then λ is still in “a” point spectrum. 2.Proposition Let T ∈ B(X), X being Banach. Then T ′^ ∈ B(X) and σ(T ) = σ(T ′). Moreover, σ(T ) = σap(T ) ∪ σp(T ′). Proof The fact that σ(T ) = σ(T ′) will be established in Q4 of Sheet 4, again the HB-Theorem plays an important rule (and can be assumed to hold
for our X). So clearly σ(T ) ⊃ σap(T ) ∪ σp(T ′), and we have to establish the other inclusion. Denoting S = λI − T we have S′^ = λI − T ′^ and suppose λ /∈ σap(T ) ∪ σp(T ′). Then there is an ε > 0 such that ‖Sy‖ ≥ ε‖y‖ for all y ∈ X, since otherwise for each n there is yn ∈ X such that ‖Syn‖ < ‖yn‖/n and hence yn 6 = 0. But then the unit vectors xn = yn/‖yn‖ satisfy ‖Sxn‖ < 1 /n → 0 and show λ ∈ σap(T ). So ε‖x‖ ≤ ‖Sx‖ ≤ ‖S‖‖x‖. Hence, if yn = Sxn → y then ‖xn − xm‖ ≤ ‖yn − ym‖/ε and (xn) must be a Cauchy sequence in the Banach space X, so xn → x for some x ∈ X. Since S is continuous, yn = Sxn → Sx = y ∈ S(X) and hence S(X) is closed3)^ in X. If S(X) = X then the algebraic inverse of S is defined on X and bounded (of norm at most ε−^1 ), so is the inverse of S and λ /∈ σ(T ) as required. Else we find by the Hahn-Banach result Prop 6.9 an g ∈ X′^ \ { 0 } with S(X) ⊂ ker(g), so S′g = gS = 0 – hence g ∈ ker(S′) and λ ∈ σp(T ′), contradiction.
Next, we aim at showing the following. For every T ∈ B(X) is the spectrum σ(T ) a compact set with(in) “radius”
Rad(σ(T )) = max{|λ|, λ ∈ σ(T )} ≤ inf n ‖T n‖^1 /n^ = lim n ‖T n‖^1 /n, (1)
which satisfies the “Spectral mapping Theorem”
p(σ(T )) ⊂ σ(p(T )) for each polynomial p.
Moreover, if X is a complex Banach space then for each T ∈ B(X) σ(T ) 6 = ∅ and Rad(σ(T )) = lim n ‖T n‖^1 /n,
the spectral radius formula holds true(though we will not prove it) and we have also p(σ(T )) = σ(p(T )). Therefore, most of our investigations will be done in complex spaces. The key tool we are using is the so-called Neumann series, which tells us that some operators (beside the identity) are invertible. 3.Theorem Suppose that T ∈ B(X) and ‖T ‖ < 1. Then I − T is invertible and (I − T )−^1 =
r=0 T^
r. Proof Since ‖T r‖ ≤ ‖T ‖r, the series
T r^ is absolutely convergent, and hence convergent in the Banach space B(X) (see Theorem 2.4). Let Sn =
∑n r=0 T^
r, and S = ∑∞ r=0 T^
r, so ‖Sn − S‖ → 0. Now,
Sn(I − T ) = I − T n+1^ = (I − T )Sn. 3)This argument should already be familiar!
Since ker T = { 0 } if and only if 0 ∈ {/ αn : n ≥ 1 }, we immediately see σp(T ) = {αn : n ≥ 1 } and so by Prop 6 σ(T ) ⊃ {αn : n ≥ 1 }. On the other hand if dist(0, {αn : n ≥ 1 }) = ε > 0 then Sx = (α− n 1 xn) defines a bounded operator on c 0 , of norm ε−^1. Since ST = T S = I, we see that T must be invertible. Hence λI −T ∈ BI(c 0 ) if dist(λ, {αn : n ≥ 1 } > 0, so σ(T ) = {αn : n ≥ 1 }. Remark Because every subset of F is separable and hence contains a dense subsequence, we see that each ∅ 6 = K ⊂ F compact is the spectrum of some T of the above type. How about σ(T ) = ∅? We know T (x, y) = (−y, x) in R^2 , so the right angle rotation has no (real) eigenvalue and as 2 is finite σ(T ) = ∅. This showed why the eigenvalues (spectrum) should be studied over C, and we follow this advice. In the sequel F = C, but see the Remark 12
∑As for matrices in Linear algebra, we define for a given polynomial^ p(z) = n j=0 αj^ z
j (^) the operator p(T ) = ∑n j=0 αj^ T^
j (^) , T 0 = I∀T ∈ B(X). How does
applying a polynomial to T change is spectrum? As in Linear Algebra we could show that p(λ) is an (approximate) eigenvalue of p(T ) is λ if the same for T and use Proposition 2 to find σ(p(T )). Our proof below seems more elegant (and convincing). 8.Theorem (Spectral Mapping Theorem) For T ∈ B(X) and any polynomial p(z),
σ(p(T )) = p(σ(T )) = {p(λ) : λ ∈ σ(T )}.
ProofLet μ ∈ C. Then p(z) − μ can be factorised in C as
p(z) − μ = α(z − β 1 )... (z − βn).
with α 6 = 0 (unless p(z) ≡ 0). Then
p(T ) − μI = (T − β 1 I)... (T − βnI),
where the factors commute, and μ = p(λ) if and only if λ ∈ {β 1 ,... , βn}. So μ ∈ p(σ(T )) iff there is λ ∈ σ(T ) : p(λ) = μ iff {β 1 ,... , βn} ∩ σ(T ) 6 = ∅. Let μ 6 ∈ p(σ(T )), then βr ∈/ σ(T ) for each r and thus each T − βrI is invertible, and so is their product. Hence μ /∈ σ(p(T )). If μ /∈ σ(p(T )), then for eachr r ≤ n
p(T ) − μI = (T − βrI)
j 6 =r
(T − βj I) =
j 6 =r
(T − βj I) (T − βrI)
is invertible, and so (T − βrI) is invertible – see Lemma 4.(ii). Hence βr ∈/ σ(T ) for all r, so μ /∈ p(σ(T )). Taking p(z) = zn^ in Theorem 8 and applying Proposition 6, it follows that if λ ∈ σ(T ), then λn^ ∈ σ(T n) and thus |λ|n^ ≤ ‖T ‖n. In fact, 9.Theorem (Spectral Radius Formula). For T ∈ B(X),
Rad(σ(T )) = inf{‖T n‖^1 /n^ : n ≥ 1 } = lim n→∞ ‖T n‖^1 /n.
10.Example Let X = C[a, b] and (T f )(t) =
∫ (^) t a f^ (s)^ ds.^ Then^ ‖T^
n‖ =
(b − a)n/n! (see Problem Sheet 3), so ‖T n‖^1 /n^ → 0. Hence sup{|λ| : λ ∈ σ(T )} = { 0 }, so σ(T ) = { 0 } = σap(T ).
Finally, we want to see why in general σ(T ) 6 = ∅? In Linear algebra every A ∈ Mn(C) has an eigenvalue since the characteristic polynomial χA must have a root in C. This fundemental theorem of algebra is an easy application of Liouville’s Theorem to 1/χA if this would be holomorphic. The same Theorem is useful for us. 11.Theorem If X is a complex Banach space and T ∈ B(X) then σ(B(X)) 6 = ∅. Proof Otherwise the resolvent R(λ, T ) = (T − λI)−^1 is well-defined on C. By Hahn-Banach there is an f ∈ B(X)′^ such that f (R(0, T )) 6 = 0, so φ : λ ∈ C → f (R(λ, T )) ∈ C. Moreover,
R(λ, T ) = (−λ(I − λ−^1 T ))−^1 = −λ−^1 (
︷ →︸︸I ︷
I − λ−^1 T )−^1 → 0 as |λ| → ∞,
so φ(λ) → 0 as |λ| → ∞. We claim that φ is holomorphic on all of C (entire), since it goes to zero at infinity it must be constant and so vanishing also in 0, contradiction. To show the claim, we fix any λ 0 ∈ C and consider small λ
T − (λ + λ 0 )I = (T − λ 0 I) − λI = (I − λR 0 )(T − λ 0 I), if R 0 = (T − λ 0 I)−^1.
Thus R(λ + λ), T ) = R 0 (Id − λR 0 )−^1 = r 0
∑^ ∞
n=
(λR 0 )n^ if |λ|‖R 0 ‖ < 1 ,
so φ(λ + λ 0 ) =
∑^ ∞
n=
λnf (R 0 n +1),
