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WEN-CHING LIEN

Department of Mathematics National Cheng Kung University

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## 3.4 Uniform continuity

Definition

Let E be a nonempty subset of R and f :ER.Then f is said to be uniformly continuous on E (notation: f :ER is uniformly continuous) if and only if for every ǫ >0 there is aδ >0 such that

|x −a| < δ and x,aE imply |f(x) −f(a)| < ǫ.

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## 3.4 Uniform continuity

Definition

Let E be a nonempty subset of R and f :ER.Then f is said to be uniformly continuous on E (notation: f :ER is uniformly continuous) if and only if for every ǫ >0 there is aδ >0 such that

|x −a| < δ and x,aE imply |f(x) −f(a)| < ǫ.

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### Example:

Show that f(x) =x2is not uniformly continuous on R.

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### Example:

Show that f(x) =x2is not uniformly continuous on R.

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### Lemma:

Suppose that ER and f :ER is uniformly continuous. If xnE is Cauchy, then f(xn)is Cauchy.

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### Lemma:

Suppose that ER and f :ER is uniformly continuous. If xnE is Cauchy, then f(xn)is Cauchy.

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Theorem

Suppose that I is a closed, bounded interval. If f :IR is continuous on I, then f is uniformly continuous on I.

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Theorem

Suppose that I is a closed, bounded interval. If f :IR is continuous on I, then f is uniformly continuous on I.

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### Proof:

Suppose to the contrary that f is continuous but not uniformly continuous on I Then there is anǫ0 >0 and points xn,ynI such that|xnyn| <1/n and

(10) |f(xn) −f(yn)| ≥ ǫ0, nN.

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### Proof:

Suppose to the contrary that f is continuous but not uniformly continuous on I Then there is anǫ0 >0 and points xn,ynI such that|xnyn| <1/n and

(10) |f(xn) −f(yn)| ≥ ǫ0, nN.

(12)

### Proof:

Suppose to the contrary that f is continuous but not uniformly continuous on I Then there is anǫ0 >0 and points xn,ynI such that|xnyn| <1/n and

(10) |f(xn) −f(yn)| ≥ ǫ0, nN.

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### Proof:

Suppose to the contrary that f is continuous but not uniformly continuous on I Then there is anǫ0 >0 and points xn,ynI such that|xnyn| <1/n and

(10) |f(xn) −f(yn)| ≥ ǫ0, nN.

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By the Bolzano-Weierstrass Theorem and the Comparison Theorem, the sequence {xn}has a subsequence, say xnk, that converges, as k → ∞, to some xI. Similarly, the sequence{ynk}k∈Nhas a convergent subsequence, say ynkj, that converges, as j → ∞, to some y ∈I. Since xnkjx as j → ∞and f is continuous, it follows from (10) that|f(x) −f(y)| ≥ ǫ0; i.e., f(x) 6= f(y). But|xnyn| <1/n for all n∈N, so Theorem 2.9 (the Squeeze Theorem) implies that x =y . Therefore, f(x) = f(y), a contradiction. 2

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By the Bolzano-Weierstrass Theorem and the Comparison Theorem, the sequence{xn}has a subsequence, say xnk, that converges, as k → ∞, to some xI. Similarly, the sequence{ynk}k∈Nhas a convergent subsequence, say ynkj, that converges, as j → ∞, to some y ∈I. Since xnkjx as j → ∞and f is continuous, it follows from (10) that|f(x) −f(y)| ≥ ǫ0; i.e., f(x) 6= f(y). But|xnyn| <1/n for all n∈N, so Theorem 2.9 (the Squeeze Theorem) implies that x =y . Therefore, f(x) = f(y), a contradiction. 2

(16)

By the Bolzano-Weierstrass Theorem and the Comparison Theorem, the sequence {xn}has a subsequence, say xnk, that converges, as k → ∞, to some xI. Similarly,the sequence {ynk}k∈Nhas a convergent subsequence, say ynkj, that converges, as j → ∞, to some y ∈I. Since xnkjx as j → ∞and f is continuous, it follows from (10) that|f(x) −f(y)| ≥ ǫ0; i.e., f(x) 6= f(y). But|xnyn| <1/n for all n∈N, so Theorem 2.9 (the Squeeze Theorem) implies that x =y . Therefore, f(x) =f(y), a contradiction. 2

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By the Bolzano-Weierstrass Theorem and the Comparison Theorem, the sequence {xn}has a subsequence, say xnk, that converges, as k → ∞, to some xI. Similarly, the sequence{ynk}k∈Nhas a convergent subsequence, say ynkj, that converges, as j → ∞, to some y ∈I. Since xnkjx as j → ∞and f is continuous, it follows from (10) that|f(x) −f(y)| ≥ ǫ0; i.e., f(x) 6= f(y). But|xnyn| <1/n for all n∈N, so Theorem 2.9 (the Squeeze Theorem) implies that x =y . Therefore, f(x) = f(y), a contradiction. 2

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By the Bolzano-Weierstrass Theorem and the Comparison Theorem, the sequence {xn}has a subsequence, say xnk, that converges, as k → ∞, to some xI. Similarly,the sequence {ynk}k∈Nhas a convergent subsequence, say ynkj, that converges, as j → ∞, to some y ∈I. Since xnkjx as j → ∞and f is continuous, it follows from (10) that|f(x) −f(y)| ≥ ǫ0; i.e., f(x) 6=f(y). But|xnyn| <1/n for all n∈N, so Theorem 2.9 (the Squeeze Theorem) implies that x =y . Therefore, f(x) =f(y), a contradiction. 2

(19)

By the Bolzano-Weierstrass Theorem and the Comparison Theorem, the sequence {xn}has a subsequence, say xnk, that converges, as k → ∞, to some xI. Similarly, the sequence{ynk}k∈Nhas a convergent subsequence, say ynkj, that converges, as j → ∞, to some y ∈I. Since xnkjx as j → ∞and f is continuous, it follows from (10) that|f(x) −f(y)| ≥ ǫ0; i.e., f(x) 6=f(y). But|xnyn| <1/n for all n∈N, so Theorem 2.9 (the Squeeze Theorem) implies that x =y . Therefore, f(x) =f(y), a contradiction. 2

(20)

By the Bolzano-Weierstrass Theorem and the Comparison Theorem, the sequence {xn}has a subsequence, say xnk, that converges, as k → ∞, to some xI. Similarly, the sequence{ynk}k∈Nhas a convergent subsequence, say ynkj, that converges, as j → ∞, to some y ∈I. Since xnkjx as j → ∞and f is continuous, it follows from (10) that|f(x) −f(y)| ≥ ǫ0;i.e., f(x) 6=f(y). But|xnyn| <1/n for all n∈N, so Theorem 2.9 (the Squeeze Theorem) implies that x =y . Therefore, f(x) =f(y), a contradiction. 2

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By the Bolzano-Weierstrass Theorem and the Comparison Theorem, the sequence {xn}has a subsequence, say xnk, that converges, as k → ∞, to some xI. Similarly, the sequence{ynk}k∈Nhas a convergent subsequence, say ynkj, that converges, as j → ∞, to some y ∈I. Since xnkjx as j → ∞and f is continuous, it follows from (10) that|f(x) −f(y)| ≥ ǫ0; i.e., f(x) 6=f(y). But|xnyn| <1/n for all n∈N, so Theorem 2.9 (the Squeeze Theorem) implies that x =y . Therefore, f(x) =f(y), a contradiction. 2

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By the Bolzano-Weierstrass Theorem and the Comparison Theorem, the sequence {xn}has a subsequence, say xnk, that converges, as k → ∞, to some xI. Similarly, the sequence{ynk}k∈Nhas a convergent subsequence, say ynkj, that converges, as j → ∞, to some y ∈I. Since xnkjx as j → ∞and f is continuous, it follows from (10) that|f(x) −f(y)| ≥ ǫ0;i.e., f(x) 6=f(y). But|xnyn| <1/n for all n∈N,so Theorem 2.9 (the Squeeze Theorem) implies that x =y . Therefore, f(x) = f(y), a contradiction. 2

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By the Bolzano-Weierstrass Theorem and the Comparison Theorem, the sequence {xn}has a subsequence, say xnk, that converges, as k → ∞, to some xI. Similarly, the sequence{ynk}k∈Nhas a convergent subsequence, say ynkj, that converges, as j → ∞, to some y ∈I. Since xnkjx as j → ∞and f is continuous, it follows from (10) that|f(x) −f(y)| ≥ ǫ0; i.e., f(x) 6= f(y). But|xnyn| <1/n for all n∈N, so Theorem 2.9 (the Squeeze Theorem) implies that x =y .Therefore, f(x) =f(y), a contradiction. 2

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By the Bolzano-Weierstrass Theorem and the Comparison Theorem, the sequence {xn}has a subsequence, say xnk, that converges, as k → ∞, to some xI. Similarly, the sequence{ynk}k∈Nhas a convergent subsequence, say ynkj, that converges, as j → ∞, to some y ∈I. Since xnkjx as j → ∞and f is continuous, it follows from (10) that|f(x) −f(y)| ≥ ǫ0; i.e., f(x) 6= f(y). But|xnyn| <1/n for all n∈N,so Theorem 2.9 (the Squeeze Theorem) implies that x =y . Therefore, f(x) = f(y),a contradiction. 2

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By the Bolzano-Weierstrass Theorem and the Comparison Theorem, the sequence {xn}has a subsequence, say xnk, that converges, as k → ∞, to some xI. Similarly, the sequence{ynk}k∈Nhas a convergent subsequence, say ynkj, that converges, as j → ∞, to some y ∈I. Since xnkjx as j → ∞and f is continuous, it follows from (10) that|f(x) −f(y)| ≥ ǫ0; i.e., f(x) 6= f(y). But|xnyn| <1/n for all n∈N, so Theorem 2.9 (the Squeeze Theorem) implies that x =y .Therefore, f(x) =f(y), a contradiction. 2

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By the Bolzano-Weierstrass Theorem and the Comparison Theorem, the sequence {xn}has a subsequence, say xnk, that converges, as k → ∞, to some xI. Similarly, the sequence{ynk}k∈Nhas a convergent subsequence, say ynkj, that converges, as j → ∞, to some y ∈I. Since xnkjx as j → ∞and f is continuous, it follows from (10) that|f(x) −f(y)| ≥ ǫ0; i.e., f(x) 6= f(y). But|xnyn| <1/n for all n∈N, so Theorem 2.9 (the Squeeze Theorem) implies that x =y . Therefore, f(x) = f(y),a contradiction. 2

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By the Bolzano-Weierstrass Theorem and the Comparison Theorem, the sequence {xn}has a subsequence, say xnk, that converges, as k → ∞, to some xI. Similarly, the sequence{ynk}k∈Nhas a convergent subsequence, say ynkj, that converges, as j → ∞, to some y ∈I. Since xnkjx as j → ∞and f is continuous, it follows from (10) that|f(x) −f(y)| ≥ ǫ0; i.e., f(x) 6= f(y). But|xnyn| <1/n for all n∈N, so Theorem 2.9 (the Squeeze Theorem) implies that x =y . Therefore, f(x) = f(y), a contradiction. 2

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Theorem

Let(a,b) be a bounded, open, nonempty interval and f : (a,b) →R. Then f is uniformly continuous on(a,b)if and only if f can be extended continuously to [a,b], i.e., if and only if there is a continuous function g : [a,b] → R that satifies

(11) f(x) =g(x), x ∈ (a,b).

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Theorem

Let(a,b) be a bounded, open, nonempty interval and f : (a,b) →R. Then f is uniformly continuous on(a,b)if and only if f can be extended continuously to [a,b], i.e., if and only if there is a continuous function g : [a,b] → R that satifies

(11) f(x) =g(x), x ∈ (a,b).

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### Proof:

Suppose that f is uniformly continuous on(a,b). Let xn ∈ (a,b)converge to b as n→ ∞. Then{xn}is Cauchy;

hence, by Lemma 3.38, so is{f(xn)}. In particular, g(b) := lim

n→∞f(xn)

exists. This value does not change if we use a different sequence to a approximate b.

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### Proof:

Suppose that f is uniformly continuous on(a,b). Let xn ∈ (a,b)converge to b as n→ ∞. Then{xn}is Cauchy;

hence, by Lemma 3.38, so is{f(xn)}. In particular, g(b) := lim

n→∞f(xn)

exists. This value does not change if we use a different sequence to a approximate b.

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### Proof:

Suppose that f is uniformly continuous on(a,b). Let xn ∈ (a,b)converge to b as n→ ∞. Then{xn}is Cauchy;

hence, by Lemma 3.38, so is{f(xn)}. In particular, g(b) := lim

n→∞f(xn)

exists. This value does not change if we use a different sequence to a approximate b.

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### Proof:

Suppose that f is uniformly continuous on(a,b). Let xn ∈ (a,b)converge to b as n→ ∞. Then{xn}is Cauchy;

hence, by Lemma 3.38, so is{f(xn)}. In particular, g(b) := lim

n→∞f(xn)

exists. This value does not change if we use a different sequence to a approximate b.

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### Proof:

Suppose that f is uniformly continuous on(a,b). Let xn ∈ (a,b)converge to b as n→ ∞. Then{xn}is Cauchy;

hence, by Lemma 3.38, so is{f(xn)}. In particular, g(b) := lim

n→∞f(xn)

exists. This value does not change if we use a different sequence to a approximate b.

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### Proof:

Suppose that f is uniformly continuous on(a,b). Let xn ∈ (a,b)converge to b as n→ ∞. Then{xn}is Cauchy;

hence, by Lemma 3.38, so is{f(xn)}. In particular, g(b) := lim

n→∞f(xn)

exists. This value does not change if we use a different sequence to a approximate b.

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### Proof:

Suppose that f is uniformly continuous on(a,b). Let xn ∈ (a,b)converge to b as n→ ∞. Then{xn}is Cauchy;

hence, by Lemma 3.38, so is{f(xn)}. In particular, g(b) := lim

n→∞f(xn)

exists. This value does not change if we use a different sequence to a approximate b.

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Indeed, let yn ∈ (a,b)be another sequence that converges to b as n→ ∞. Givenǫ >0, chooseδ >0 such that (9) holds for E = (a,b). Since xnyn →0, choose NN so that nN implies|xnyn| < δ. By (9), then,|f(xn) −f(yn)| < ǫfor all nN. Taking the limit of this inequality as n → ∞, we obtain

| lim

n→∞f(xn) − lim

n→∞f(yn)| ≤ ǫ

for all ǫ >0. It follows from Theorem 1.9 that

n→∞lim f(x)= lim

n→∞f(yn).

(38)

Indeed,let yn ∈ (a,b)be another sequence that converges to b as n→ ∞. Given ǫ >0,chooseδ >0 such that (9) holds for E = (a,b). Since xnyn →0, choose NN so that nN implies|xnyn| < δ. By (9), then,|f(xn) −f(yn)| < ǫfor all nN. Taking the limit of this inequality as n → ∞, we obtain

| lim

n→∞f(xn) − lim

n→∞f(yn)| ≤ ǫ

for all ǫ >0. It follows from Theorem 1.9 that

n→∞lim f(x)= lim

n→∞f(yn).

(39)

Indeed, let yn ∈ (a,b)be another sequence that converges to b as n→ ∞. Givenǫ >0, chooseδ >0 such that (9) holds for E = (a,b). Since xnyn →0, choose NN so that nN implies|xnyn| < δ. By (9), then,|f(xn) −f(yn)| < ǫfor all nN. Taking the limit of this inequality as n → ∞, we obtain

| lim

n→∞f(xn) − lim

n→∞f(yn)| ≤ ǫ

for all ǫ >0. It follows from Theorem 1.9 that

n→∞lim f(x)= lim

n→∞f(yn).

(40)

Indeed, let yn ∈ (a,b)be another sequence that converges to b as n→ ∞. Given ǫ >0,chooseδ >0 such that (9) holds for E = (a,b). Since xnyn →0, choose NN so that nN implies|xnyn| < δ. By (9), then,|f(xn) −f(yn)| < ǫfor all nN. Taking the limit of this inequality as n → ∞, we obtain

| lim

n→∞f(xn) − lim

n→∞f(yn)| ≤ ǫ

for all ǫ >0. It follows from Theorem 1.9 that

n→∞lim f(x)= lim

n→∞f(yn).

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Indeed, let yn ∈ (a,b)be another sequence that converges to b as n→ ∞. Given ǫ >0, chooseδ >0 such that (9) holds for E = (a,b). Since xnyn →0, choose NN so that nN implies|xnyn| < δ. By (9), then,|f(xn) −f(yn)| < ǫfor all nN. Taking the limit of this inequality as n → ∞, we obtain

| lim

n→∞f(xn) − lim

n→∞f(yn)| ≤ ǫ

for all ǫ >0. It follows from Theorem 1.9 that

n→∞lim f(x)= lim

n→∞f(yn).

(42)

Indeed, let yn ∈ (a,b)be another sequence that converges to b as n→ ∞. Given ǫ >0, chooseδ >0 such that (9) holds for E = (a,b). Since xnyn →0, choose NN so that nN implies|xnyn| < δ. By (9), then,|f(xn) −f(yn)| < ǫfor all nN. Taking the limit of this inequality as n → ∞, we obtain

| lim

n→∞f(xn) − lim

n→∞f(yn)| ≤ ǫ

for all ǫ >0. It follows from Theorem 1.9 that

n→∞lim f(x)= lim

n→∞f(yn).

(43)

Indeed, let yn ∈ (a,b)be another sequence that converges to b as n→ ∞. Given ǫ >0, chooseδ >0 such that (9) holds for E = (a,b). Since xnyn →0, choose NN so that nN implies|xnyn| < δ. By (9), then,|f(xn) −f(yn)| < ǫfor all nN. Taking the limit of this inequality as n → ∞,we obtain

| lim

n→∞f(xn) − lim

n→∞f(yn)| ≤ ǫ

for all ǫ >0. It follows from Theorem 1.9 that

n→∞lim f(x)= lim

n→∞f(yn).

(44)

Indeed, let yn ∈ (a,b)be another sequence that converges to b as n→ ∞. Given ǫ >0, chooseδ >0 such that (9) holds for E = (a,b). Since xnyn →0, choose NN so that nN implies|xnyn| < δ. By (9), then,|f(xn) −f(yn)| < ǫfor all nN. Taking the limit of this inequality as n → ∞, we obtain

| lim

n→∞f(xn) − lim

n→∞f(yn)| ≤ ǫ

for all ǫ >0. It follows from Theorem 1.9 that

n→∞lim f(x)= lim

n→∞f(yn).

(45)

Indeed, let yn ∈ (a,b)be another sequence that converges to b as n→ ∞. Given ǫ >0, chooseδ >0 such that (9) holds for E = (a,b). Since xnyn →0, choose NN so that nN implies|xnyn| < δ. By (9), then,|f(xn) −f(yn)| < ǫfor all nN. Taking the limit of this inequality as n → ∞,we obtain

| lim

n→∞f(xn) − lim

n→∞f(yn)| ≤ ǫ

for all ǫ >0. It follows from Theorem 1.9 that

n→∞lim f(x)= lim

n→∞f(yn).

(46)

Indeed, let yn ∈ (a,b)be another sequence that converges to b as n→ ∞. Given ǫ >0, chooseδ >0 such that (9) holds for E = (a,b). Since xnyn →0, choose NN so that nN implies|xnyn| < δ. By (9), then,|f(xn) −f(yn)| < ǫfor all nN. Taking the limit of this inequality as n → ∞, we obtain

| lim

n→∞f(xn) − lim

n→∞f(yn)| ≤ ǫ

for all ǫ >0. It follows from Theorem 1.9 that

n→∞lim f(x)= lim

n→∞f(yn).

(47)

Indeed, let yn ∈ (a,b)be another sequence that converges to b as n→ ∞. Given ǫ >0, chooseδ >0 such that (9) holds for E = (a,b). Since xnyn →0, choose NN so that nN implies|xnyn| < δ. By (9), then,|f(xn) −f(yn)| < ǫfor all nN. Taking the limit of this inequality as n → ∞, we obtain

| lim

n→∞f(xn) − lim

n→∞f(yn)| ≤ ǫ

for all ǫ >0. It follows from Theorem 1.9 that

n→∞lim f(x)= lim

n→∞f(yn).

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Thus, g(b)is well defined. A similar argument defines g(a). Set g(x) =f(x)for x ∈ (a,b). Then g is defined on [a,b], satifies (11), and is continuous on[a,b]by the Sequential Characterization of Limits. Thus, f can be

“continuously extended” to g as required.

Conversely, suppose that there is a function g

continuouos on [a,b] that satifies (11). By theorem 3.39, g is uniformly continuous on [a,b]; hence, g is uniformly continuous on (a,b). We conclude that f is uniformly continuous on (a,b).2

(49)

Thus, g(b)is well defined. A similar argument defines g(a). Set g(x) =f(x)for x ∈ (a,b). Then g is defined on [a,b], satifies (11), and is continuous on[a,b]by the Sequential Characterization of Limits. Thus, f can be

“continuously extended” to g as required.

Conversely, suppose that there is a function g

continuouos on [a,b] that satifies (11). By theorem 3.39, g is uniformly continuous on [a,b]; hence, g is uniformly continuous on (a,b). We conclude that f is uniformly continuous on (a,b).2

(50)

Thus, g(b)is well defined. A similar argument defines g(a). Set g(x) =f(x)for x ∈ (a,b). Then g is defined on [a,b], satifies (11), and is continuous on[a,b]by the Sequential Characterization of Limits. Thus,f can be

“continuously extended” to g as required.

Conversely, suppose that there is a function g

continuouos on [a,b] that satifies (11). By theorem 3.39, g is uniformly continuous on [a,b]; hence, g is uniformly continuous on (a,b). We conclude that f is uniformly continuous on (a,b).2

(51)

Thus, g(b)is well defined. A similar argument defines g(a). Set g(x) =f(x)for x ∈ (a,b). Then g is defined on [a,b], satifies (11), and is continuous on[a,b]by the Sequential Characterization of Limits. Thus, f can be

“continuously extended” to g as required.

Conversely, suppose that there is a function g

continuouos on [a,b] that satifies (11). By theorem 3.39, g is uniformly continuous on [a,b]; hence, g is uniformly continuous on (a,b). We conclude that f is uniformly continuous on (a,b).2

(52)

Thus, g(b)is well defined. A similar argument defines g(a). Set g(x) =f(x)for x ∈ (a,b). Then g is defined on [a,b], satifies (11), and is continuous on[a,b]by the Sequential Characterization of Limits. Thus,f can be

“continuously extended” to g as required.

Conversely, suppose that there is a function g

continuouos on [a,b] that satifies (11). By theorem 3.39, g is uniformly continuous on [a,b]; hence, g is uniformly continuous on (a,b). We conclude that f is uniformly continuous on (a,b).2

(53)

Thus, g(b)is well defined. A similar argument defines g(a). Set g(x) =f(x)for x ∈ (a,b). Then g is defined on [a,b], satifies (11), and is continuous on[a,b]by the Sequential Characterization of Limits. Thus, f can be

“continuously extended” to g as required.

Conversely, suppose that there is a function g

continuouos on [a,b] that satifies (11). By theorem 3.39, g is uniformly continuous on [a,b]; hence, g is uniformly continuous on (a,b). We conclude that f is uniformly continuous on (a,b).2

(54)

Thus, g(b)is well defined. A similar argument defines g(a). Set g(x) =f(x)for x ∈ (a,b). Then g is defined on [a,b], satifies (11), and is continuous on[a,b]by the Sequential Characterization of Limits. Thus, f can be

“continuously extended” to g as required.

Conversely, suppose that there is a function g

continuouos on [a,b] that satifies (11). By theorem 3.39, g is uniformly continuous on [a,b]; hence, g is uniformly continuous on (a,b). We conclude that f is uniformly continuous on (a,b).2

(55)

Thus, g(b)is well defined. A similar argument defines g(a). Set g(x) =f(x)for x ∈ (a,b). Then g is defined on [a,b], satifies (11), and is continuous on[a,b]by the Sequential Characterization of Limits. Thus, f can be

“continuously extended” to g as required.

Conversely, suppose that there is a function g

continuouos on [a,b] that satifies (11). By theorem 3.39, g is uniformly continuous on [a,b];hence, g is uniformly continuous on (a,b). We conclude that f is uniformly continuous on (a,b).2

(56)

Thus, g(b)is well defined. A similar argument defines g(a). Set g(x) =f(x)for x ∈ (a,b). Then g is defined on [a,b], satifies (11), and is continuous on[a,b]by the Sequential Characterization of Limits. Thus, f can be

“continuously extended” to g as required.

Conversely, suppose that there is a function g

continuouos on [a,b] that satifies (11). By theorem 3.39, g is uniformly continuous on [a,b]; hence, g is uniformly continuous on (a,b). We conclude that f is uniformly continuous on (a,b).2

(57)

Thus, g(b)is well defined. A similar argument defines g(a). Set g(x) =f(x)for x ∈ (a,b). Then g is defined on [a,b], satifies (11), and is continuous on[a,b]by the Sequential Characterization of Limits. Thus, f can be

“continuously extended” to g as required.

Conversely, suppose that there is a function g

continuouos on [a,b] that satifies (11). By theorem 3.39, g is uniformly continuous on [a,b];hence, g is uniformly continuous on (a,b). We conclude that f is uniformly continuous on (a,b).2

(58)

Thus, g(b)is well defined. A similar argument defines g(a). Set g(x) =f(x)for x ∈ (a,b). Then g is defined on [a,b], satifies (11), and is continuous on[a,b]by the Sequential Characterization of Limits. Thus, f can be

“continuously extended” to g as required.

Conversely, suppose that there is a function g

continuouos on [a,b] that satifies (11). By theorem 3.39, g is uniformly continuous on [a,b]; hence, g is uniformly continuous on (a,b). We conclude that f is uniformly continuous on (a,b).2

(59)

Thus, g(b)is well defined. A similar argument defines g(a). Set g(x) =f(x)for x ∈ (a,b). Then g is defined on [a,b], satifies (11), and is continuous on[a,b]by the Sequential Characterization of Limits. Thus, f can be

“continuously extended” to g as required.

Conversely, suppose that there is a function g

continuouos on [a,b] that satifies (11). By theorem 3.39, g is uniformly continuous on [a,b]; hence, g is uniformly continuous on (a,b). We conclude that f is uniformly continuous on (a,b).2

(60)

Thus, g(b)is well defined. A similar argument defines g(a). Set g(x) =f(x)for x ∈ (a,b). Then g is defined on [a,b], satifies (11), and is continuous on[a,b]by the Sequential Characterization of Limits. Thus, f can be

“continuously extended” to g as required.

Conversely, suppose that there is a function g

continuouos on [a,b] that satifies (11). By theorem 3.39, g is uniformly continuous on [a,b]; hence, g is uniformly continuous on (a,b). We conclude that f is uniformly continuous on (a,b).2

(61)

### Example:

Prove that f(x) = (x −1)

log x is uniformly continuous on (0,1).

(62)

### Example:

Prove that f(x) = (x −1)

log x is uniformly continuous on (0,1).

(63)

## Thank you.

Outline

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