$f(x)g(x)$ is uniformly continuous if $f,g$ are uniformly continuous and $lim_{xto infty}g(x)=0$?
up vote
2
down vote
favorite
on $[0,infty)$, can we prove $f(x)g(x)$ is uniformly continuous if $f,g$ are uniformly continuous and $lim_{xto infty}g(x)=0$?
If $lim_{xto infty}g(x)=0$ is droped, then it is easy to see that it is wrong with the example $f(x)=g(x)=x$.
uniform-continuity
asked Nov 16 at 1:37
xldd
1,304510
add a comment |
up vote
2
down vote
favorite
on $[0,infty)$, can we prove $f(x)g(x)$ is uniformly continuous if $f,g$ are uniformly continuous and $lim_{xto infty}g(x)=0$?
If $lim_{xto infty}g(x)=0$ is droped, then it is easy to see that it is wrong with the example $f(x)=g(x)=x$.
uniform-continuity
asked Nov 16 at 1:37
xldd
1,304510
Actually the question should be, Prove that if f and g are bounded and uniformly continuous on an interval. Then the product fg is also uniformly continuous at that interval. And also bounded is not necessary for uniform continuity as you do in your first assumption. But in your second assumption boundedness is necessary.
– John Nash
Nov 16 at 1:54
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
on $[0,infty)$, can we prove $f(x)g(x)$ is uniformly continuous if $f,g$ are uniformly continuous and $lim_{xto infty}g(x)=0$?
If $lim_{xto infty}g(x)=0$ is droped, then it is easy to see that it is wrong with the example $f(x)=g(x)=x$.
uniform-continuity
asked Nov 16 at 1:37
xldd
1,304510
on $[0,infty)$, can we prove $f(x)g(x)$ is uniformly continuous if $f,g$ are uniformly continuous and $lim_{xto infty}g(x)=0$?
If $lim_{xto infty}g(x)=0$ is droped, then it is easy to see that it is wrong with the example $f(x)=g(x)=x$.
uniform-continuity
uniform-continuity
asked Nov 16 at 1:37
xldd
1,304510
asked Nov 16 at 1:37
xldd
1,304510
asked Nov 16 at 1:37
xldd
1,304510
asked Nov 16 at 1:37
xldd
1,304510
asked Nov 16 at 1:37
xldd
1,304510
1,304510
Actually the question should be, Prove that if f and g are bounded and uniformly continuous on an interval. Then the product fg is also uniformly continuous at that interval. And also bounded is not necessary for uniform continuity as you do in your first assumption. But in your second assumption boundedness is necessary.
– John Nash
Nov 16 at 1:54
add a comment |
Actually the question should be, Prove that if f and g are bounded and uniformly continuous on an interval. Then the product fg is also uniformly continuous at that interval. And also bounded is not necessary for uniform continuity as you do in your first assumption. But in your second assumption boundedness is necessary.
– John Nash
Nov 16 at 1:54
Actually the question should be, Prove that if f and g are bounded and uniformly continuous on an interval. Then the product fg is also uniformly continuous at that interval. And also bounded is not necessary for uniform continuity as you do in your first assumption. But in your second assumption boundedness is necessary.
– John Nash
Nov 16 at 1:54
Actually the question should be, Prove that if f and g are bounded and uniformly continuous on an interval. Then the product fg is also uniformly continuous at that interval. And also bounded is not necessary for uniform continuity as you do in your first assumption. But in your second assumption boundedness is necessary.
– John Nash
Nov 16 at 1:54
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
Let $f(x)=x$ and $g(x)=frac {sin(x^{2})} x$ so that $f(x)g(x)=sin(x^{2})$. $g$ is uniformly continuous because $g'$ is bounded. Also $g(x) to 0$ as $x to infty$. Now let $x_n=sqrt {npi}$ and $y_n=sqrt {(n+frac 12 ) pi}$. Then $sin (x_n^{2})-sin (y_n)^{2}=0-1=-1$ for all $n$ even though $|x_n-y_n|=frac {frac 1 2 pi} {x_n+y_n} to 0$ as $n to infty$. Hence $fg$ is not uniformly continuous.
answered Nov 16 at 5:55
Kavi Rama Murthy
39.8k31749
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Let $f(x)=x$ and $g(x)=frac {sin(x^{2})} x$ so that $f(x)g(x)=sin(x^{2})$. $g$ is uniformly continuous because $g'$ is bounded. Also $g(x) to 0$ as $x to infty$. Now let $x_n=sqrt {npi}$ and $y_n=sqrt {(n+frac 12 ) pi}$. Then $sin (x_n^{2})-sin (y_n)^{2}=0-1=-1$ for all $n$ even though $|x_n-y_n|=frac {frac 1 2 pi} {x_n+y_n} to 0$ as $n to infty$. Hence $fg$ is not uniformly continuous.
answered Nov 16 at 5:55
Kavi Rama Murthy
39.8k31749
add a comment |
up vote
1
down vote
accepted
Let $f(x)=x$ and $g(x)=frac {sin(x^{2})} x$ so that $f(x)g(x)=sin(x^{2})$. $g$ is uniformly continuous because $g'$ is bounded. Also $g(x) to 0$ as $x to infty$. Now let $x_n=sqrt {npi}$ and $y_n=sqrt {(n+frac 12 ) pi}$. Then $sin (x_n^{2})-sin (y_n)^{2}=0-1=-1$ for all $n$ even though $|x_n-y_n|=frac {frac 1 2 pi} {x_n+y_n} to 0$ as $n to infty$. Hence $fg$ is not uniformly continuous.
answered Nov 16 at 5:55
Kavi Rama Murthy
39.8k31749
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Let $f(x)=x$ and $g(x)=frac {sin(x^{2})} x$ so that $f(x)g(x)=sin(x^{2})$. $g$ is uniformly continuous because $g'$ is bounded. Also $g(x) to 0$ as $x to infty$. Now let $x_n=sqrt {npi}$ and $y_n=sqrt {(n+frac 12 ) pi}$. Then $sin (x_n^{2})-sin (y_n)^{2}=0-1=-1$ for all $n$ even though $|x_n-y_n|=frac {frac 1 2 pi} {x_n+y_n} to 0$ as $n to infty$. Hence $fg$ is not uniformly continuous.
answered Nov 16 at 5:55
Kavi Rama Murthy
39.8k31749
Let $f(x)=x$ and $g(x)=frac {sin(x^{2})} x$ so that $f(x)g(x)=sin(x^{2})$. $g$ is uniformly continuous because $g'$ is bounded. Also $g(x) to 0$ as $x to infty$. Now let $x_n=sqrt {npi}$ and $y_n=sqrt {(n+frac 12 ) pi}$. Then $sin (x_n^{2})-sin (y_n)^{2}=0-1=-1$ for all $n$ even though $|x_n-y_n|=frac {frac 1 2 pi} {x_n+y_n} to 0$ as $n to infty$. Hence $fg$ is not uniformly continuous.
answered Nov 16 at 5:55
Kavi Rama Murthy
39.8k31749
answered Nov 16 at 5:55
Kavi Rama Murthy
39.8k31749
answered Nov 16 at 5:55
Kavi Rama Murthy
39.8k31749
answered Nov 16 at 5:55
Kavi Rama Murthy
39.8k31749
39.8k31749
add a comment |
add a comment |
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3000594%2ffxgx-is-uniformly-continuous-if-f-g-are-uniformly-continuous-and-lim%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Actually the question should be, Prove that if f and g are bounded and uniformly continuous on an interval. Then the product fg is also uniformly continuous at that interval. And also bounded is not necessary for uniform continuity as you do in your first assumption. But in your second assumption boundedness is necessary.
– John Nash
Nov 16 at 1:54