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Analysis
Math 201A, Fall 2006
Solutions: Problem Set 6
- Let C
1 ([0, 1]) denote the space of continuously differentiable functions
f : [0, 1] → R, and define
‖f ‖ = sup x∈[0,1]
|f (x)| + sup x∈[0,1]
|f
′ (x)|.
(a) Show that ‖ · ‖ is a norm on C
1 ([0, 1]).
(b) Prove that C
1 ([0, 1]) is a Banach space with respect to ‖ · ‖.
Solution.
- (a) It is easy to check that ‖ · ‖ is a norm. For example, denoting the
sup-norm by ‖ · ‖∞, we have
‖f + g‖ = ‖f + g‖∞ + ‖f
′
′ ‖∞
≤ ‖f ‖∞ + ‖g‖∞ + ‖f
′ ‖∞ + ‖g
′ ‖∞
≤ ‖f ‖ + ‖g‖.
- (b) Suppose that (fn) is a Cauchy sequence in C
1 ([0, 1]) with respect
to ‖ · ‖. Then (fn), (f
′ n) are Cauchy sequences of continuous functions
with respect to ‖ · ‖∞. Since (C([0, 1]), ‖ · ‖∞) is complete, there exist
f, g ∈ C([0, 1]) such that fn → f , f
′ n →^ g^ uniformly (i.e. with respect
to ‖ · ‖∞).
- Suppose that (fn) is a sequence in C([0, 1]) and fn → f uniformly. Let
Fn(x) =
∫ (^) x
0
fn(t) dt, F (x) =
∫ (^) x
0
f (t) dt.
Then Fn → F uniformly, since
‖Fn − F ‖∞ ≤ sup x∈[0,1]
x
0
|fn(t) − f (t)| dt ≤ ‖fn − f ‖∞.
′ n →^ g^ uniformly, it follows that
fn(x) − fn(0) =
∫ (^) x
0
f
′ n(t)^ dt^ →
∫ (^) x
0
g(t) dt.
Since fn → f uniformly, we conclude that
f (x) = f (0) +
∫ (^) x
0
g(t) dt.
- The fundamental theorem of calculus implies that f is continuously
differentiable and f
′ = g. Thus, fn → f and f
′ n →^ f^
′ uniformly, which
implies that fn → f ∈ C
1 ([0, 1]) with respect to ‖ · ‖. This shows that
(C
1 ([0, 1]), ‖ · ‖) is complete.
- A function f : [0, 1] → R is said to be H¨older continuous with exponent
α if
[f ]α = sup x 6 =y∈[0,1]
|f (x) − f (y)|
|x − y|
α
is finite. Given 0 < α ≤ 1 and M > 0, define
F = {f ∈ C([0, 1]) | ‖f ‖∞ ≤ M, [f ]α ≤ M }.
Prove that F is a compact subset of C([0, 1]) equipped with the sup-norm
‖ · ‖∞.
Solution.
- Suppose that fn ∈ F and fn → f uniformly. Then
‖f ‖∞ = lim n→∞
‖fn‖∞ ≤ M.
For every x 6 = y ∈ [0, 1] and n ∈ N, we have
|fn(x) − fn(y)|
|x − y|
α
≤ 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|α^
≤ M,
which implies that [f ]α ≤ M. Hence f ∈ F, and F is closed.
- If f is H¨older continuous and x ∈ [0, 1], then
|f (x) − f (0)| ≤ [f ]α|x|
α ≤ [f ]α.
Hence,
|f (x)| ≤ |f (x) − f (0)| + |f (0)| ≤ [f ]α + |f (0)|.
Thus, if f ∈ F, we find that |f (x)| ≤ M + M , so ‖f ‖∞ ≤ 2 M , and
therefore F is bounded.
δ =
M
) 1 /α
Then if f ∈ F and |x − y| < δ, we have
|f (x) − f (y)| ≤ [f ]α|x − y|
α ≤ M |x − y|
α < ,
which shows that F is (uniformly) equicontinuous.
- The Arzel`a-Ascoli theorem implies that F is a compact.
- Suppose that
{fn : K → R | n ∈ N}
is an equicontinuous family of functions on a compact metric space K. If
(fn) converges pointwise to a function f , prove that f is continuous. Is the
convergence necessarily uniform?
Solution.
- First, we show that F = {fn | n ∈ N} is a bounded subset of C(K).
- The set F is uniformly equicontinuous since it is equicontinuous and K
is compact. Hence, we can choose δ > 0 such that |fn(x) − fn(y)| < 1
for all n ∈ N and x, y ∈ K such that d(x, y) < δ. Let {x 1 ,... , xN } be a
finite δ-net of K, which exists since K is compact and therefore totally
bounded. The sequences (fn(xi))
∞ n=1 converge for 1^ ≤^ i^ ≤^ N^ , so they
are bounded, by M , say.
- If x ∈ K, then d(x, xi) < δ for some 1 ≤ i ≤ N , and therefore
|fn(x)| ≤ |fn(xi)| + |fn(x) − fn(xi)| ≤ M + 1.
It follows that ‖fn‖∞ ≤ M + 1 for every n ∈ N, so F is bounded.
- By the Arzel`a-Ascoli theorem, F is a precompact subset of C(K), so
(fn) has a uniformly convergent subsequence. The limit of this subse-
quence must be the same as the pointwise limit f of the whole sequence,
so f is continuous, since the uniform limit of continuous functions is
continuous.
- The sequence (fn) is contained in a compact set F and every uniformly
convergent subsequence has the same limit, namely the pointwise limit
f. It follows from the result of Problem 3, Set 4 that the whole sequence
converges uniformly to f.
