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Math 6210
Homework 5^1
C´esar Lozano
Problem 1 Let Y be a linear subspace in a normed linear space X. Probe that
dis(x, Y ) = sup{φ(x) : φ ∈ X∗, φ ⊥, ‖φ‖ = 1}.
Here the notation φ ⊥ Y means φ(y) = 0 for all y ∈ Y.
Solution. To see this we need to see the following computation
dist(x, Y ) = inf y∈Y
{‖x − y‖}
= inf max{|φ(x − y)| : φ ∈ X∗, ‖φ‖ = 1} = inf max{|φ(x) − φ(y)| : φ ∈ X∗, ‖φ‖ = 1} = max{|φ(x)| : φ ∈ X∗, φ(y) = 0 y ∈ Y ‖φ‖ = 1}
Problem 2 Let Y be a subset of a normed linear space X. Prove that Y ⊥is a closed linear subspace in X∗
Solution. In order to show this, consider the following map evaluation eva
eva : X∗^ −→ R φ 7 −→ φ(a)
where a ∈ X Now observe that ev− a 1 ({ 0 }) = {φ ∈ X∗^ : φ(a) = 0}
Therefore, we can write the set Y ⊥^ as follows
Y ⊥^ =
a∈Y
= ev− a 1 ({ 0 }) = {φ ∈ X∗^ : φ(a) = 0 a ∈ Y }.
As a result, the question about if the set Y perp^ is closed has become in a question if the map eva is continuous in order to consider the set Y ⊥^ as a intersection of closed sets. It is clearly that eva is linear, then we are going to show it is continuous at zero. Consider a sequence φn → 0.
‖φn‖ = sup a 6 =
{φ ‖na(‖a) } < .
This inequality means that |φn(a)| < ‖a‖ and then eva is continuous at zero. This completes the proof.
(^1) October 30,2008.
Problem 3 Let Y be a subspace of a normed space X. Prove that there is a norm-preserving injective map J : Y ∗^ → X∗^ such that for each φ ∈ Y ∗, Jφ is an extension of φ.
Solution. First consider the functional φ ∈ Y ∗ which is bounded , namely |φ(x)| = ‖φ‖‖x‖ Then by Hahn-Banach we can extend such a functional with the same bound. Such a extension is injective since we consider the map φ(x) = 0 the extension Jφ has de same bound throughout X, then the kernel of this map is only the functional zero hence. Now observer that the Hahn-Banach theorem says that the extension will have the same bound, then ‖Jφ(x)‖ ≤ ‖φ‖‖x‖.
However, for the values ‖x‖ = 1, we get that ‖Jφ(x)‖ ≤ ‖φ‖. Now, by the definition we have that ‖φ(x)‖ ≤ ‖Jφ(x)‖,
same for values ‖x‖ = 1. Therefore, we get the equation
‖φ(x)‖ = ‖Jφ(x)‖.
Problem 4 Prove that in a Banach space, a subspace of second category must be dense.
Solution. Since the space Y is of second category, this implies the interior of Y is not empty. Consider an element suppose x 0 ∈ X arbitrary. We are going to show that Y = X. we can assume 0 ∈ Y ◦ . Observe that there exist r ∈ R such that rx 0 ∈ B(0, ). Now take into account that r = 1 should be still in Y by the subspace property. then Y = X since x ∈ X was an arbitrary element.
Problem 5 Let f ∈ C∞(R). Thus f has derivatives of all orders on R. Suppose that 0 ∈ {f (n)(t) : n = 1, 2 ,.. .} for each t. Then f is a polynomial.
Solution. Observe that f can be written as a Taylor series around t = 0 f (s) = f (0) + f (0)s + f ′′(0)s
2 2! +^ · · ·
Moreover we know there is a neighborhood U 0 of zero 0 ∈ U such that the last Taylor expansion of f (s) converges. Further by hypothesis 0 ∈ {f (n)(t) : n = 1, 2 ,.. .} such a series only have a finitely many non zero terms, i.e., it is a polynomial. As a result, the function f can be thought as a polynomial P 0 on such a neighborhood U 0. Furthermore the function f is locally a polynomial. If the function is not such a polynomial P 0 , then consider a 1 ∈ ∂U 0 and consider the Taylor expansion around a 1. Consequently there exists a neighborhood a 1 ∈ U 1 , where f (s) can be thought as a a polynomial P 1. However observe that U 0 ∩ U 1 6 = ∅ is an open set where the polynomial P 0 (s) − P 1 (s) = 0 s ∈ U 0 ∩ U 1. This fact implies that P 0 = P 1 and as a result f (s) is the polynomial P 0 on the neighbor- hood U 0 ∪ U 1. We can keep doing this in order to get that the function is a polynomial on R.
Then, since G = kerΦ we have that G is closed. Now let us see that gn is Cauchy sequence. To see this note
‖gn − gm‖ =
− 1
|gn − gm|
∫ (^) − 1 /m
− 1 /n
|ns − (^) |ss| | +
∫ (^1) /m
− 1 /m
|ns − (^) |ss| | +
∫ (^1) /n
1 /m
|ns − ms|
Observe that this expression can be arbitrary small. Therefore, gn is a Cauchy sequence. Now the last point, is reflecting that the identity map
(C, ‖ ‖ 1 ) id 1 −→ (C, ‖ ‖∞),
is not continuous with these norms. Now since G = ker Φ it is closed in C[− 1 , 1], ‖ ‖ 1 which does not mean any contradiction due to the sequence {gn} does not converge in such a space. Nevertheless, observe that the function g which is the limit of the sequence gn in L 1 [− 1 , −1] means an equivalence class. This one contains both functions
g˜ 1 (s) =
1 if s ∈ [0, 1] − 1 if s ∈ [− 1 , 0) and
g˜ 2 (s) =
1 if s ∈ (0, 1] − 1 if s ∈ [− 1 , 0]
Then, in this context we can say that the convergence of gn is well defined in L^1. However, this is not longer true when we consider such functions in C[− 1 , 1]. They become, in this space, two distinct functions (points), that are accumulation points. And hence the convergence does not make sense anymore in (C‖ ‖ 1 ).
Problem 8 Let (X, M, μ) be a measure space. For f ∈ L∞(X) and g ∈ L^2 (X) define Mf (g) = f g. Prove that Mf : L^2 (X) → L^2 (X) is a bounded linear map and that
‖Mf ‖ = ‖f ‖ Solution. Linearity. Consider g, h ∈ L^2 (X), and a ∈ R then
Mf (g + h) = f (g + ah) = f g + af h = Mf (f ) + aMf (h).
Boundedness. In order tho prove this, observe that
‖Mf ‖ = sup{‖Mf (g)‖ : ‖g‖ 2 = 1‖}
= sup(
|f g|^2 )^1 /^2
≤ |M |(
|g|^2 )^1 /^2
= ‖f ‖∞
The University of Utah
Department of Mathematics [email protected]
