## GR9768 Q 64

Forum for the GRE subject test in mathematics.
mathQ
Posts: 41
Joined: Thu Mar 25, 2010 12:14 am

### GR9768 Q 64

Ques: Suppose that f is a continuous real-valued function defined on the closed interval [0,1]. Which of the following must be true ?

I. There is a constant C>0 such that |f(x) -f(y)| <= C for all x and y in [0,1]
II. There is a constant D > 0 such that |f(x) -f(y)| <= 1 for all x and y in [0,1] that satisfy |x-y| <= D
III. There is a constant E > 0 such that |f(x) -f(y)| <= E |x-y| for all x and y in [0,1]

A) I only
B) III only
C) I & II only
D) II & III only
E) I, II * III only

Any general method of solving such problems ?

Thanks.

speedychaos4
Posts: 47
Joined: Mon Mar 22, 2010 2:42 am

### Re: GR9768 Q 64

Since f is continuous on a closed interval, then f is uniformly continuous and bounded on it.
Therefore it is easy to evaluate that I II are true.
For III, you can consider a obvious counter-example, which is $y=\sqrt{1-x^2}$

mathQ
Posts: 41
Joined: Thu Mar 25, 2010 12:14 am

### Re: GR9768 Q 64

Thanks Speedy.

Can you please elaborate more ?

As per continuous functions definition at point x0

For every e > 0 there is E >0 such that

|x-x0| < E , |f(x) - f(x0)| < e

How does this definition fit into I and II. Am I missing something related to bounded functions ?

speedychaos4
Posts: 47
Joined: Mon Mar 22, 2010 2:42 am

### Re: GR9768 Q 64

As per continuous functions definition at point x0

For every e > 0 there is E >0 such that

|x-x0| < E , |f(x) - f(x0)| < e

As I could recall, the most significant property about uniformly continuous func. is that the e and E defined above are independent.

by substituting x0 into y, e into 1 and E into D you can see II is correct.

And for I the boundedness is quite intuitively, since it is continuous and defined on [0,1], no matter how big the abs. of f goes, there will be certain endpoints, there for you could define a constant which is larger than that.

p.s. please forgive me if I make any mistakes, I don't have strong calculus basis since I skipped lots of those lectures..

EugeneKudashev
Posts: 27
Joined: Tue Apr 06, 2010 8:22 am

### Re: GR9768 Q 64

I. continuity implies that function is bounded. since f is continuous, it attains min and max value (there is a respective theorem, although it is understandable intuitively). thus, we can write |f(x)-f(y)| <= |f(x)|+|f(y)| <= 2*M, where M is max of f on the interval. therefore, (I) is true.

II. there is a nice theorem by Heine-Cantor (full version here http://en.wikipedia.org/wiki/Heine%E2%8 ... or_theorem ), it states that continuous function on a compact set (in particular, segment [a,b]) is uniformly continuous. if you read definition of uniform continuity carefully, you should notice that (II) is simply the definition of un. cont.

III. actually, this condition is named "Lipschitz condition", and it can be shown that a lipschitz-cont. func is uniformly cont., but the converse is not true. anyway, it is plain that an arbitrary continuous function does not satisfy this condition, thus it is false (and a counter has been given above).

mathQ
Posts: 41
Joined: Thu Mar 25, 2010 12:14 am

### Re: GR9768 Q 64

EugeneKudashev wrote:I. continuity implies that function is bounded. since f is continuous, it attains min and max value (there is a respective theorem, although it is understandable intuitively). thus, we can write |f(x)-f(y)| <= |f(x)|+|f(y)| <= 2*M, where M is max of f on the interval. therefore, (I) is true.

II. there is a nice theorem by Heine-Cantor (full version here http://en.wikipedia.org/wiki/Heine%E2%8 ... or_theorem ), it states that continuous function on a compact set (in particular, segment [a,b]) is uniformly continuous. if you read definition of uniform continuity carefully, you should notice that (II) is simply the definition of un. cont.

III. actually, this condition is named "Lipschitz condition", and it can be shown that a lipschitz-cont. func is uniformly cont., but the converse is not true. anyway, it is plain that an arbitrary continuous function does not satisfy this condition, thus it is false (and a counter has been given above).

Thanks Eugene.

I got why we have I as correct

For II, we know that for f(x) is uniformly continuous from the theorem, but how do we prove that |f(x) - f(y) <=1 , which I believe is the length of the closed interval .

Thanks

speedychaos4
Posts: 47
Joined: Mon Mar 22, 2010 2:42 am

### Re: GR9768 Q 64

EugeneKudashev wrote:I. continuity implies that function is bounded. since f is continuous, it attains min and max value (there is a respective theorem, although it is understandable intuitively). thus, we can write |f(x)-f(y)| <= |f(x)|+|f(y)| <= 2*M, where M is max of f on the interval. therefore, (I) is true.

II. there is a nice theorem by Heine-Cantor (full version here http://en.wikipedia.org/wiki/Heine%E2%8 ... or_theorem ), it states that continuous function on a compact set (in particular, segment [a,b]) is uniformly continuous. if you read definition of uniform continuity carefully, you should notice that (II) is simply the definition of un. cont.

III. actually, this condition is named "Lipschitz condition", and it can be shown that a lipschitz-cont. func is uniformly cont., but the converse is not true. anyway, it is plain that an arbitrary continuous function does not satisfy this condition, thus it is false (and a counter has been given above).

Beautiful ans.

EugeneKudashev
Posts: 27
Joined: Tue Apr 06, 2010 8:22 am

### Re: GR9768 Q 64

mathQ wrote:
EugeneKudashev wrote:I. continuity implies that function is bounded. since f is continuous, it attains min and max value (there is a respective theorem, although it is understandable intuitively). thus, we can write |f(x)-f(y)| <= |f(x)|+|f(y)| <= 2*M, where M is max of f on the interval. therefore, (I) is true.

II. there is a nice theorem by Heine-Cantor (full version here http://en.wikipedia.org/wiki/Heine%E2%8 ... or_theorem ), it states that continuous function on a compact set (in particular, segment [a,b]) is uniformly continuous. if you read definition of uniform continuity carefully, you should notice that (II) is simply the definition of un. cont.

III. actually, this condition is named "Lipschitz condition", and it can be shown that a lipschitz-cont. func is uniformly cont., but the converse is not true. anyway, it is plain that an arbitrary continuous function does not satisfy this condition, thus it is false (and a counter has been given above).

Thanks Eugene.

I got why we have I as correct

For II, we know that for f(x) is uniformly continuous from the theorem, but how do we prove that |f(x) - f(y) <=1 , which I believe is the length of the closed interval .

Thanks

nope, it has nothing to do with the length. look, again, at the def. of uniform continuity. it says that for every positive (no matter how small) eps we can find positive delta such that as long as |x1-x2|<delta, |f(x1)-f(x2)|<eps. if you take delta=D and eps=1, you obtain just the statement (II).

mathQ
Posts: 41
Joined: Thu Mar 25, 2010 12:14 am

### Re: GR9768 Q 64

EugeneKudashev wrote:
mathQ wrote:
EugeneKudashev wrote:I. continuity implies that function is bounded. since f is continuous, it attains min and max value (there is a respective theorem, although it is understandable intuitively). thus, we can write |f(x)-f(y)| <= |f(x)|+|f(y)| <= 2*M, where M is max of f on the interval. therefore, (I) is true.

II. there is a nice theorem by Heine-Cantor (full version here http://en.wikipedia.org/wiki/Heine%E2%8 ... or_theorem ), it states that continuous function on a compact set (in particular, segment [a,b]) is uniformly continuous. if you read definition of uniform continuity carefully, you should notice that (II) is simply the definition of un. cont.

III. actually, this condition is named "Lipschitz condition", and it can be shown that a lipschitz-cont. func is uniformly cont., but the converse is not true. anyway, it is plain that an arbitrary continuous function does not satisfy this condition, thus it is false (and a counter has been given above).

Thanks Eugene.

I got why we have I as correct

For II, we know that for f(x) is uniformly continuous from the theorem, but how do we prove that |f(x) - f(y) <=1 , which I believe is the length of the closed interval .

Thanks

nope, it has nothing to do with the length. look, again, at the def. of uniform continuity. it says that for every positive (no matter how small) eps we can find positive delta such that as long as |x1-x2|<delta, |f(x1)-f(x2)|<eps. if you take delta=D and eps=1, you obtain just the statement (II).

True...that proves it

Thanks Eugene