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B4a Banach spaces Michaelmas 2010 lectured by Bernd Kirchheim based on notes by CJK Batty
Lecture 13
The last still missing concept known from the treatment of duality in Linear Algebra is the adjoint map (operator). 18.Theorem Let X, Y be normed normed vector spaces and let T ∈ B(X, Y ). Fix g ∈ Y ′, then gT ∈ X′^ and hence we can define T ′^ : Y ′^ → X′ by T ′g = gT , i.e. (T ′g)(x) = g(T (x)).
Then T ′^ ∈ B(Y ′, X′) and ‖T ′‖B(Y ′,X′) = ‖T ‖B (X, Y ).
Moreover, if Z is also a normed vector space and S ∈ B(Y, Z) then for S′^ ∈ B(Z′, Y ′) and (ST )′^ ∈ B(Z′, X′) introduced as above, we have
(ST )′^ = T ′S′.
Proof Clearly, by Prop 5.7. gT ∈ B(X, F) and
‖gT ‖ ≤ ‖g‖‖T ‖. (1)
So T ′^ : Y ′^ → X′^ and
T ′(α 1 g 1 + α 2 g 2 )(x) = (α 1 g 1 + α 2 g 2 )(T x) = α 1 g 1 (T x) + α 2 g 2 (T x) = (α 1 T ′g 1 + α 2 T ′g 2 )x
Hence T ′(α 1 g 1 + α 2 g 2 ) = α 1 T ′g 1 + α 2 T ′g 2 so therefore T ′^ is linear and by (1) above, T ′^ is bounded with
‖T ′‖B(Y ′,X′) ≤ ‖T ‖B(X,Y )
. Next we show that ‖T ‖ ≤ ‖T ′‖. Let x ∈ X, so T x ∈ Y. Then by Corollary 6, there exists g 0 ∈ Y ′^ such that ‖g 0 ‖Y ′ = 1 and g 0 (T x) = ‖T x‖Y. Hence,
‖T x‖Y = |g 0 (T x)| = |T ′g 0 (x)| ≤ ‖T ′g 0 ‖X′ ‖x‖X ≤ ‖T ′‖B(Y ′,X′)‖g 0 ‖Y ′^ ‖x‖X = ‖T ′‖B(Y ′,X′)‖x‖X
This hold for all x ∈ X \ { 0 }, and hence
‖T ‖B(X,Y ) ≤ ‖T ′‖B(Y ′,X′)
as required. Finally, if also S ∈ B(Y, Z) is given, then again due to Prop 5.7 ST ∈ B(X, Z) and hence (ST )′^ ∈ B(Z′, X′) by the foregoing. Moreover, for each f ∈ Z′^ we have
(ST )′f = f ST = (f S)T = T ′(f S) = T ′(S′f ) = (T ′S′)f ⇒ (ST )′^ = T ′S′.
19.Definition For T ∈ B(X, Y ), T ′^ ∈ B(Y ′, X′) as defined above is said to be the (Banach Space) adjoint of T.
Remark So with hindsight, the very abstract first part of the proof of Theorem 17 just makes the formal argument X ∼= X′′^ ⇒ X′^ ∼= (X′′)′ rigorous by showing that for the isomorphismˆ: X → X′′^ the adjoint is still an isomorphism from X′′′^ onto X′^ and equals (ˆ: X′^ → X′′′)−^1. 20.Example Let X be lp, p ∈ [1, ∞), or c 0 , Y = lq, q ∈ [1, ∞). or Y = c 0. Then X′^ = lp ′ , Y = lq ′ where p′^ = p/(p − 1) if p ∈ (1, ∞), p′^ = ∞ if p = 1 and p = 1 if X = c 0 , similarly for q′. Let a T ∈ B(X, Y ) be given by
T ((αj )) =
j=
tk,j αj )∞ k=1.
Then T ′^ ∈ B(lq
′ , lp
′ ) is given by
T ((βj )) =
j=
tj,kβj )∞ k=1.
Formally spoken, the matrix representation of T ′^ is the transposed of the matrix representation of T. This claim can be easily shown. Indeed, given any y′^ = J((βj )) ∈ Y ′, (βj ) ∈ lq ′ then x′^ = T ′(y) ∈ X′, so there exists (αk) ∈ lp ′ representing x′^ = J((αk)). Hence for all k ≥ 1 we have
αk = x′(ek) = T ′y′(ek) = y′(T ek) = J((βj ))((tl,k)l) =
j
βj tj,k,
as was to be shown.
Remarks a) f is convex iff the epigraph of f , i.e. the set
epi(f ) = {(x, t) : f (x) ≤ t} ⊂ X × R
is convex. (DIY)
b) clearly each norm on X is convex since λ, 1 − λ > 0 implies
‖λx + (1 − λ)y‖ ≤ ‖λx‖ + ‖(1 − λ)y‖ = λ‖x‖ + (1 − λ)‖y‖.
c) If M ⊂ X is a convex set and α, β > 0 then αM + βM = (α + β)M. Indeed, clearly LHS ⊃ RHS and for the converse let u ∈ αM and v ∈ βM. Then u + v α + β
α α + β
u α
β α + β
v β
= λx + (1 − λ)y ∈ M,
for λ = α/(α + β) ⇒ 1 − λ = β/(α + β) and x = u/α, y = v/β ∈ M.
d) (DIY) Analogous to linear subspaces we see that A + B is convex if A and B are convex. Moreover, any family of convex sets has a convex intersection. So we can define the convex, instead of the linear, hull Aco^ to be a smallest convex set containing A and as in linear algebra show that
Aco^ = {
∑^ N
i=
λixi : N ∈ N, xi ∈ A, λi ≥ 0 and
i
λi = 1}.
2.Proposition Let X be a real linear space and Y its subspace and x 0 ∈ X with X = span(Y ∪ {x 0 }). Given any convex function F : X → R and any linear g : Y → R such that g(y) ≤ F (y) for all y ∈ Y , there is a
G : X → R linear and such that G∣∣Y = g and G(x) ≤ F (x) for all x ∈ X.
Hint: To get a feeling for the proof to come, imagine Y = { 0 } ⊂ X = R draw a picture of F and observe/show that
F (^) −′(0) = lim xր 0
F (x) − F (0) x
= sup x< 0
F (x) − F (0) x
≤ inf x> 0
F (x) − F (0) x
= lim xց 0
F (x) − F (0) x
= F +′(0)
and so any G(x) = ax ≤ ax + F (0) ≤ F (x), where a ∈ [F (^) −′(0), F (^) +′(0)] will do what is required. In the proof to come we just (slightly) generalize this argument. Proof Without loss of generality X 6 = Y and so every x ∈ X can be uniquely written as x = y(x) + α(x)x 0 where y : X → Y and α : X → R are linear. So clearly any linear extension G of g is uniquely determined by its value c 0 = G(x 0 ) since then
G(x) = g(y(x)) + α(x)c 0 for all x ∈ X.
The only additional request we have on G is that G ≤ F which means
g(y) + αc 0 ≤ F (y + αx 0 ) for all y ∈ Y, α ∈ R.
We notice that this inequality is by assumption true if α = 0, the remaining 2 cases α > 0 and α < 0 result in the conditions
c 0 ≤
F (y + αx 0 ) − g(y) α
for all y ∈ Y, α > 0 , and
c 0 ≥
F (y + αx 0 ) − g(y) α
F (y − βx 0 ) − g(y) −β
for all y ∈ Y, β = −α > 0.
which is clearly equivalent to
sup z∈Y,β> 0
g(z) − F (z − βx 0 ) β
= S ≤ c 0 ≤ I = inf y∈Y,α> 0
F (y + αx 0 ) − g(y) α
So if S ≤ I we find a suitable c 0 and hence G, and are done. Else, we will obtain a contradiction. Indeed, I < S means that there is α > 0 , y ∈ Y such that (F (y + αx 0 ) − g(y))/α < S and hence
∃y, z ∈ Y ∃α, β > 0 :
F (y + αx 0 ) − g(y) α
g(z) − F (z − βx 0 ) β
Multiplying with αβ, bringing the ”minus” terms over and dividing by α + β gives
β α + β
F (y + αx 0 ) +
α α + β
F (z − βx 0 ) <
αg(z) + βg(y) α + β
= g
(αz + βy α + β
≤ F
(βy + αz α + β
by our assumption g ≤ F.
Clearly, for λ = β/(α + β), u = y + αx 0 and v = z − βx 0 this contradicts the convexity condition λF (u) + (1 − λ)F (v) ≥ F (λu + (1 − λ)v).
same reason ˜g|Yn satisfies all request put on gn. Since gn was unique, we have ˜g|Yn = gn and by uniqueness of the canonical HB-extension ˜g = gn+1 follows. By the definition of the canonical HB-extension we also have gn ≤ F on Yn for all n. Moreover, again due to the uniqueness of the gn’s we have gn = gm|Yn if n < m. So the map G = ∪ngn, i.e. G(y) = gn(y) if y ∈ Yn is well-defined and satisfies G ≤ F on X˜, GY = g 0 = g. Since for all x, y ∈ X˜ there is n such that x, y ∈ Yn we have
for all λ ∈ F : G(λx + y) = gn(λx + y) = λgn(x) + gn(y) = λG(x) + G(y),
so G is linear on X˜. Remarks We would be in a situation to construct a HB-extension of g even on all of X if we could show could define a similar inductive argument as above along any spanning set (instead of spanning sequence). It can be shown (see FA in the 4th year), however, that any Banach space which is not finite dimensional can not have a countable spanning set! An (so-called “transfinite“) induction procedure along such huge (in particular uncount- able) sets is rather a topic of set-theory and logic and beyond the scope of this course (but treated in the 4th year). We use here an (in principle un- necessary) analytic argument to overcome, at least for separable spaces, the non-existence of countable algebraic bases.
- Lemma Let (X, ‖ · ‖) be a real normed space and X˜ a dense subspace of X. Suppose g : X˜ → R linear and F : X → R satisfy g ≤ F on X˜. If
∀t ∈ R : {x : F (x) < t} is open in X (e.g. if F continuous),
then g ∈ X˜′^ and its unique continuous extension G ∈ X′^ (see Theorem 5.8) satisfies G ≤ F on X. Proof Take t = F (0) + 1 then 0 ∈ {x : F (x) < t} open and hence there is an ε > 0 such that Bε(0) ⊂ {F < t}. Hence, for all x ∈ X˜ with ‖x‖ ≤ 1 we have −εx, εx ∈ {F < t} and hence −εg(x), εg(x) < t implying |g(x)| ≤ |F (0) + 1|/ε and thus ‖g‖ (^) X˜′ ≤ |F (0) + 1|/ε. So g ∈ X˜′^ is continuous and has by Theorem 5.8 a (unique) continuous linear extension G onto X, the closure of X˜. It remains to show that G ≤ F on X. Else there is x ∈ X and t ∈ R such that F (x) < t < G(x). Hence, we find ε, ε′^ > 0 such that
F < t on Bε(x), and G > t on Bε′ (x).
Since x is in the closure of X˜, we find ˜x ∈ X˜ ∩ Bε(x) ∩ Bε′^ (x) so F (˜x) < t < G(˜x) = g(˜x) ≤ F (˜x), contradiction. 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 ′^.
Proof We consider the function F (x) = ‖g‖Y ′^ ‖x‖ on X, since the norm is convex and conntinous on X and ‖g‖Y ′ ≥ 0 we easily see that F is also continuous and convex on X, moreover g ≤ F on Y by definition of ‖g‖. Because X is separable, we find a dense sequence (xi)∞ i=1 in X, cleary the linear space X˜ = span(Y ∪ {xi, i ≥ 1 } ⊃ {xi, i ≥ 1 } is dense in X. By Corollary 3 there is a linear extension G˜ : X˜ → R with G˜ ≤ F on X˜. Hence, by Lemma 4 we find a linear and continous extension G of G˜ to all of X with G ≤ F. Then G|Y = g and for all x ∈ X
G(x), −G(x) = G(−x) ≤ F (±x) = ‖g‖Y ′ ‖ ± x‖ ⇒ |G(x)| ≤ ‖g‖Y ′ ‖x‖.
By definition ‖G‖X′^ ≤ ‖g‖Y ′^ and also ‖G|Y ‖ ≤ ‖G‖ follows.
