0.1 Continuous Functions On Intervals: N N N N N N N N
0.1 Continuous Functions On Intervals: N N N N N N N N
0.1 Continuous Functions On Intervals: N N N N N N N N
Note that supS f could be ∞ and inf S f could be −∞, depending on the
function f and S. For example, f (x) = x2 and S = R. Then supS f = +∞
and inf S f = 0. Now if S = (0, 2), then supS f = 4 and inf S f = 0. There
are no points in (0, 2) where f takes 0 and 4.
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Theorem 0.1.6. (Maximau-Minimum Theorem) Let ) = [a, b] be a closed ad
bounded interval, and f : I → R be continuous on I. Then f has an absolute
maximum and minimum, i.e., there exist points c, d ∈ I such that
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Corollary 0.1.9. Let f be continuous function on [a, b] and define m =
inf I f and M = supI f . Then the range of f is the interval [m, M ], i.e.,
f ([a, b]) = [m, M ].
Proof. We know from above, there exists c, d ∈ [a, b] such that f (c) =
m, f (d) = M . And any number y ∈ (m, M ), there is c ∈ (a, b) such that
f (c) = y. From the definition m, M , f does not have values outside [m, M ].
So the range equals [m, M ].
Theorem 0.1.10. Let I be an interval and let f : IR be continuous on I. If
α < β are numbers in I such that f (α) < 0 < f (β) (or f (α) > 0 > f (β)),
then there exists a number c ∈ (α, β) such that f (c) = 0.
Proof. As f is continuous on I, so f is continuous on [α, β]. Apply the
Intermediate Value Theorem with y = 0.
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Example 0.1.11. Let f (x) = x2 +1
.
1. I1 = (−1, 1). f (I1 ) = ( 21 , 1].
2. I2 = [0, ∞), f (I2 ) = (0, 1].
Lemma 0.1.12. Let S ⊆ R be a nonempty set with the property if x, y ∈ S
with x < y, then [x, y] ⊆ S. Then S is an interval.
Theorem 0.1.13. Let I be an interval and let f : I → R be continuous on
I. Then f (I) is an interval.
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Example 0.2.2. Let f : (0, ∞) → R with f (x) = 1/x. Let x0 = u > 0.
Consider
|x − u|
|f (x) − f (u)| = .
xu
As x → u, so consider only |x − u| < u/2, i.e., u/2 < x < 3u/2. Then
1/x < 2/u. Hence 1/ux < (1/u)(2/u) = 2/u2 . Now given > 0, choose
δ = min{u/2, u2 /2}. So when |x − u| < δ =⇒ |f (x) − f (u)| < .
Here δ depends on both and u. In fact, there is no δ for all u > 0, as
then δ = 0.
See the graph of f (x) = 1/x.
2. ∃0 > 0 such that for every δ > 0 there are points xδ , yδ ∈ A such that
|xδ − yδ | < δ and |f (xδ ) − f (yδ )| ≥ 0 .
3. ∃0 > 0 and two sequences {xn } and {yn } in A such that lim(xn −yn ) =
0 and |f (xn ) − f (yn )| ≥ 0 for all n ∈ N.
Example 0.2.6. Let f : (0, ∞) → R with f (x) = 1/x. Now pick 0 = 1/2,
and choose xn = 1/n and yn = 1/(n + 1). Then lim(xn − yn ) = 0 and
|f (xn ) − f (yn )| = 1 > 1/2 for all n.
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Proof. If f is not uniformly continuous on I. From the above, ∃0 > 0 and
xn , yn ∈ I such that xn − yn → 0 and |f (xn ) − f (yn )| ≥ 0 for all n. As I
is bounded, so by Bolzana-Weierstrass, there is a subsequence {xnk } of {xn }
that converges to z ∈ I, as I is closed interval. In addition, from
ynk → z as well.
Now as f is continuous at z, so we have f (xnk ) → f (z) and f (ynk ) → f (z).
But this is a contradiction, as |f (xn ) − f (yn )| ≥ 0 for all n.
Lipschitz Functions
Definition 0.2.8. Let A ⊂ R and let f : A → R. If there exists a constant
K > 0 such that
Proof. Let > 0, choose δ = /K. Then for all x, y ∈ A with |x − y| < δ, we
have |f (x) − f (y)| < .
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Now for > 0, choose δ = min{1, 1+2|c| } such that for all x satisfying |x−c| <
δ, |f (x) − f (c)| < , i.e., f is continuous on R.
As δ depends on both and c, c is larger and larger, the values of δ is
smaller and smaller (as the graph becomes more steeper). There is no such
δ that works for all points. In fact, inf c δ = 0.
But when we consider only on [−a, a] for a > 0. Then
hence δ = /2a works for all points on [−a, a], i.e., f is uniformly continuous
on [−a, a].