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Solutions: Problem Set 8
Math 201A, Fall 2006
Problem 1. Recall that a function f : [0, 1] → R is Lipschitz continuous if
its Lipschitz constant
Lip(f ) = sup x 6 =y∈[0,1]
|f (x) − f (y)|
|x − y|
is finite.
(a) For M > 0, let
LM = {f ∈ C([0, 1]) | Lip(f ) ≤ M }.
Show that LM is a closed subset of C([0, 1]) equipped with the sup-norm.
(b) Let L = {f ∈ C([0, 1]) | f is Lipschitz continuous}. Prove that L is a
linear subspace of C([0, 1]).
(c) Is L a closed linear subspace of C([0, 1]) equipped with the sup-norm?
Solution.
- (a) Suppose that fn ∈ LM and fn → f uniformly. For every x, y ∈ [0, 1]
and n ∈ N, we have
|fn(x) − fn(y)|
|x − y|
≤ M.
Taking the limit of this equation as n → ∞, and using the fact that
fn → f pointwise, we get
|f (x) − f (y)|
|x − y|
≤ M,
which implies that f ∈ LM. Thus, LM is closed.
- (b) If λ is a scalar, then
|(λf )(x) − (λf )(y)| = |λ||f (x) − f (y)|.
It follows that if f is Lipschitz continuous, then λf is Lipschitz con-
tinuous, and Lip(λf ) = |λ|Lip(f ). If f , g are Lipschitz continuous,
with
|f (x) − f (y)| ≤ M |x − y|, |g(x) − g(y)| ≤ N |x − y|,
then
|(f + g)(x) − (f + g)(y)| ≤ |f (x) − f (y)| + |g(x) − g(y)|
≤ (M + N )|x − y|.
Hence (f + g) is Lipschitz continuous and Lip(f + g) ≤ Lip(f ) + Lip(g).
It follows that L is a linear space.
- (c) Any polynomial is Lipschitz continuous by the mean value theorem
(since it is differentiable with bounded derivative on [0, 1]). By the
Weierstrass approximation theorem, the closure of L is C([0, 1]), so L
is not closed.
- Alternatively, one can give an explicit sequence (fn) of Lipschitz contin-
uous functions that converges uniformly to a non-Lipschitz continuous
function; for example, fn(x) = (x + 1/n)
1 / 2 .
- The sequence (fn) is not a Schauder basis since an expansion with
respect to this sequence is not unique. For example,
N ∑
n=
2 n^
fn =
2 N^ +^
eN +1 → 0 as N → ∞,
so for any c ∈ R, we have
∞ ∑
n=
c
2 n^
fn.
Problem 3. Suppose that X, Y , Z are normed linear spaces and A : X → Y ,
B : Y → Z are bounded linear operators. Prove that BA : X → Z is a
bounded linear operator, and
‖BA‖ ≤ ‖A‖‖B‖.
Give an example to show that this inequality may be strict.
Solution.
- Using the definitions of ‖B‖ and ‖A‖, we have
‖BAx‖ = ‖B(Ax)‖
≤ ‖B‖ ‖Ax‖
≤ ‖B‖ ‖A‖ ‖x‖.
It follows that BA is bounded and ‖BA‖ ≤ ‖B‖‖A‖.
2 → R
2 be the linear maps with matrices
[A] =
, [B] =
Then ‖A‖ = ‖B‖ = 1 with respect to any norm on R
2 , but BA = 0,
so ‖BA‖ = 0.
