MATH414 FUNCTIONAL ANALYSIS HOMEWORK 4 Hw 4: 8,9,10,11,12/126 8,9,10,11,12/126 Problem (8/126) Show that the dual space of the space c 0 is l is l 1 . Solution ∞
∞
It is not difficult to show { ek }k where e where e k = (δ kj kj )j Then every x every x ∈ c 0 has unique representation =1
=1
is a Schauder basis for c for c 0 .
∞
x =
ξ k ek .
k=1
where x where x = (ξ k ). Consider any f any f ∈ c 0 . Then since f since f is is linear and bounded ∞
f ( f (x) =
ξ k f ( f (ek ).
k=1
In the first step, our aim is to show that the sequence ( f ( f (ek )) ∈ l 1 . Lets define the numbers θn as follows θn =
|f (en )| f (en )
f ( f (en ) =0 f ( f (en ) = 0. 0.
0
Now consider the sequence (x ( xn ) where xn = (θ1 , θ2 , · · · , θn , 0, 0, · · · ) = θ 1 e1 + θ + θ2 e2 + θ + θn en . Then clearly x clearly x n ∈ c 0 and || xn || x ∈ c 0 , | f ( f (x)| ≤ ||xn || ||f ||. So
∞
is either 0 or 1 for all n. n . We know that for all
∞
f (xn )| = |f (
n
n
θk f ( f (ek ) =
k =1
Shortly
f (ek )| ≤ ||xn || ||f || ≤ ||f ||. |f ( ∞
k=1
n
f (ek )| ≤ ||f || for all n. |f (
k=1
Since RHS is a constant and and inequality holds for all n, letting n −→ ∞ gives us ∞
f (ek )| ≤ ||f ||. |f (
k=1
1
(1)
So (f (ek )) ∈ l 1 . Secondly we will show that for every b = (β k ) ∈ l1 , we can obtain a corresponding bounded linear operator g on c 0 . If we define g on c 0 by ∞
g(x) =
x = (ξ k ) ∈ c 0 ,
ξ k β k
k=1
then g is linear. And boundedness of g follows from
∞
|g(x)| =
k=1
ξ k β k ≤ sup |ξ j | j
∞
∞
|β k | ≤ ||x||
k =1
∞
|β k |.
k=1
Taking the supremum over all x with norm 1, we get ∞
||g|| ≤
|β k |.
k=1
If we combine this result and (1), we have ∞
||f || =
|f (ek )|
k=1
which is the norm on l1 . So the mapping f −→ (f (ek )) from c0 to l 1 is bijective, linear and preserves norms so it is an isometric isomorphism. Problem (9/126) Show that a linear functional f on a vector space X is uniquely determined by its values on a Hamel basis for X . Solution Let H is a Hamel basis for X . And let f be linear functional on X such that we know f (v) for all v ∈ H . Then this means that we know f (x) for all x ∈ X . Let x be in X . Since H is an Hamel basis, ∃ x 1 ,...,xn in H and scalars α1 ,...,αn such that x = α1 x1 + ... + αn xn . So f (x) = f (α1 x1 + ... + αn xn ) = α1 f (x1 ) + ... + αn f (xn ). Furthermore this determination is unique, that is, if f and g are two linear functionals on X such that f (v) = g(v) for all v ∈ H , then f = g. Lets use above arbitrary x. f (x) = α1 f (x1 ) + ... + αn f (xn ) = α 1 g(x1 ) + ... + αn g(xn ) = g(x).
2
Problem (10/126) Let X and Y = { 0} be normed spaces, where dimX = ∞ . Show that there is at least one unbounded linear operator T : X −→ Y . (Use a Hamel basis.) Solution Let H be an Hamel basis for X (see theorem 4.1-7). We will show ˜ : X → Y . that any function T : H → Y can be extended to a linear operator T Let x be in X . Since H is an Hamel basis, ∃ x 1 ,...,xn in H and scalars α1 ,...,αn such that x = α1 x1 + ... + αn xn . So define ˜ T (x) = α1 T (x1 ) + ... + αn T (xn ). Definition is well-defined since any element x in X has unique representation wrt ˜ is linear because if x, y ∈ X and α, β are scalars (say x = α1 x1 +...+αn xn H . T and y = β 1 y1 + ... + β k yk wrt H ) then ˜ ˜ T (αx + βy) = T (α(α 1 x1 + ... + αn xn ) + β (β 1 y1 + ... + β k yk )) ˜ = T (αα 1 x1 + ... + ααn xn + ββ 1 y1 + ... + ββ k yk ) = αα1 T (x1 ) + ... + ααn T (xn ) + ββ 1 T (y1 ) + ... + ββ k T (yk ) = α[α1 T (x1 ) + ... + αn T (xn )] + β [β 1 T (y1 ) + ... + β k T (yk )] ˜ = αT (x) + β ˜ T (y). Lets turn the question. Since H is an infinite set let v1 , v2 ,... be in H . Fix non-zero y ∈ Y . Define the function T : H → Y by T (vn ) = n ||vn ||y and 0 oth˜ : X → Y . But T can ˜ erwise. Then T can be extended to a linear operator T not ˜ be bounded, given c ∈ R+ choose n > c/||y ||, so || T vn || = n ||vn ||||y || > c||vn ||. Problem (11/126) If X is a normed space and dimX = ∞, show that the dual space X is not identicle with the algebraic dual space X ∗ . Solution Since K = {0}, where K is scalar field of X , by previous result we know that there is at least 1 element f in X ∗ but not in X . So X X ∗ . Problem (12/126) (Completeness) The examples in the text can be used to prove completeness of certain spaces. How? For what spaces? Solution By Theorem 2.10-4, we know that the dual space of any normed space is a Banach space. That is it is complete. From the course book and first problem we know that Rn = R n ,
conjugate of p.
l∞ = l1 ,
l1 = c0 and l p = lq where 1 < q < ∞ and p is
So Rn and l p , where p ∈ [1, ∞] are complete.
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