Pointwise product of uniformly continuous functions
True or false: Let $f(x)$ and $g(x)$ be uniformly continuous functions from $mathbbR$ to $mathbbR$.
Then their pointwise product $f(x)g(x)$ is uniformly continuous.
I think it will be true; take $f(x)= g(x) = sqrt x$.
Am I right or wrong?
real-analysis uniform-continuity
edited Nov 11 at 19:04
jwodder
1,262918
asked Nov 11 at 12:50
santosh
969
add a comment |
True or false: Let $f(x)$ and $g(x)$ be uniformly continuous functions from $mathbbR$ to $mathbbR$.
Then their pointwise product $f(x)g(x)$ is uniformly continuous.
I think it will be true; take $f(x)= g(x) = sqrt x$.
Am I right or wrong?
real-analysis uniform-continuity
edited Nov 11 at 19:04
jwodder
1,262918
asked Nov 11 at 12:50
santosh
969
7
You don't prove a statement true by finding an example for which the statement holds.
– ab123
Nov 11 at 12:52
It would be true if you added "bounded" to your assumptions, but as stated JCSantos has the perfect counterexample in his answer.
– JonathanZ
Nov 11 at 17:57
add a comment |
True or false: Let $f(x)$ and $g(x)$ be uniformly continuous functions from $mathbbR$ to $mathbbR$.
Then their pointwise product $f(x)g(x)$ is uniformly continuous.
I think it will be true; take $f(x)= g(x) = sqrt x$.
Am I right or wrong?
real-analysis uniform-continuity
edited Nov 11 at 19:04
jwodder
1,262918
asked Nov 11 at 12:50
santosh
969
True or false: Let $f(x)$ and $g(x)$ be uniformly continuous functions from $mathbbR$ to $mathbbR$.
Then their pointwise product $f(x)g(x)$ is uniformly continuous.
I think it will be true; take $f(x)= g(x) = sqrt x$.
Am I right or wrong?
real-analysis uniform-continuity
real-analysis uniform-continuity
edited Nov 11 at 19:04
jwodder
1,262918
asked Nov 11 at 12:50
santosh
969
edited Nov 11 at 19:04
jwodder
1,262918
asked Nov 11 at 12:50
santosh
969
edited Nov 11 at 19:04
jwodder
1,262918
edited Nov 11 at 19:04
jwodder
1,262918
edited Nov 11 at 19:04
jwodder
1,262918
1,262918
asked Nov 11 at 12:50
santosh
969
asked Nov 11 at 12:50
santosh
969
asked Nov 11 at 12:50
santosh
969
969
7
You don't prove a statement true by finding an example for which the statement holds.
– ab123
Nov 11 at 12:52
It would be true if you added "bounded" to your assumptions, but as stated JCSantos has the perfect counterexample in his answer.
– JonathanZ
Nov 11 at 17:57
add a comment |
7
You don't prove a statement true by finding an example for which the statement holds.
– ab123
Nov 11 at 12:52
It would be true if you added "bounded" to your assumptions, but as stated JCSantos has the perfect counterexample in his answer.
– JonathanZ
Nov 11 at 17:57
7
7
You don't prove a statement true by finding an example for which the statement holds.
– ab123
Nov 11 at 12:52
You don't prove a statement true by finding an example for which the statement holds.
– ab123
Nov 11 at 12:52
It would be true if you added "bounded" to your assumptions, but as stated JCSantos has the perfect counterexample in his answer.
– JonathanZ
Nov 11 at 17:57
It would be true if you added "bounded" to your assumptions, but as stated JCSantos has the perfect counterexample in his answer.
– JonathanZ
Nov 11 at 17:57
add a comment |
2 Answers
2
active
oldest
votes
It is false. Just take $f(x)=g(x)=x$.
answered Nov 11 at 12:53
José Carlos Santos
149k22117219
add a comment |
The uniform continuity property refers to a spatially homogeneous behavior of continuity. Look at the graph of the following functions.
- $f(x)=x$
- $f(x)=sqrt$
- $f(x)=x^2$
- $$ f(x)= begincases
xsin(frac1x) & xne 0 \
0 & x=0
endcases$$
Note that (2) has a "vertiginous behavior" around $0$, since that $$f'(x)=frac12sqrtxto infty$$ when $xdownarrow 0$.
Also for (3) this behavior occurs when $|x|to infty$.
(4) is continuous but is even "less uniform" than previous cases.
answered Nov 11 at 13:38
Daniel Camarena Perez
57538
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
It is false. Just take $f(x)=g(x)=x$.
answered Nov 11 at 12:53
José Carlos Santos
149k22117219
add a comment |
It is false. Just take $f(x)=g(x)=x$.
answered Nov 11 at 12:53
José Carlos Santos
149k22117219
add a comment |
It is false. Just take $f(x)=g(x)=x$.
answered Nov 11 at 12:53
José Carlos Santos
149k22117219
It is false. Just take $f(x)=g(x)=x$.
answered Nov 11 at 12:53
José Carlos Santos
149k22117219
answered Nov 11 at 12:53
José Carlos Santos
149k22117219
answered Nov 11 at 12:53
José Carlos Santos
149k22117219
answered Nov 11 at 12:53
José Carlos Santos
149k22117219
149k22117219
add a comment |
add a comment |
The uniform continuity property refers to a spatially homogeneous behavior of continuity. Look at the graph of the following functions.
- $f(x)=x$
- $f(x)=sqrt$
- $f(x)=x^2$
- $$ f(x)= begincases
xsin(frac1x) & xne 0 \
0 & x=0
endcases$$
Note that (2) has a "vertiginous behavior" around $0$, since that $$f'(x)=frac12sqrtxto infty$$ when $xdownarrow 0$.
Also for (3) this behavior occurs when $|x|to infty$.
(4) is continuous but is even "less uniform" than previous cases.
answered Nov 11 at 13:38
Daniel Camarena Perez
57538
add a comment |
The uniform continuity property refers to a spatially homogeneous behavior of continuity. Look at the graph of the following functions.
- $f(x)=x$
- $f(x)=sqrt$
- $f(x)=x^2$
- $$ f(x)= begincases
xsin(frac1x) & xne 0 \
0 & x=0
endcases$$
Note that (2) has a "vertiginous behavior" around $0$, since that $$f'(x)=frac12sqrtxto infty$$ when $xdownarrow 0$.
Also for (3) this behavior occurs when $|x|to infty$.
(4) is continuous but is even "less uniform" than previous cases.
answered Nov 11 at 13:38
Daniel Camarena Perez
57538
add a comment |
The uniform continuity property refers to a spatially homogeneous behavior of continuity. Look at the graph of the following functions.
- $f(x)=x$
- $f(x)=sqrt$
- $f(x)=x^2$
- $$ f(x)= begincases
xsin(frac1x) & xne 0 \
0 & x=0
endcases$$
Note that (2) has a "vertiginous behavior" around $0$, since that $$f'(x)=frac12sqrtxto infty$$ when $xdownarrow 0$.
Also for (3) this behavior occurs when $|x|to infty$.
(4) is continuous but is even "less uniform" than previous cases.
answered Nov 11 at 13:38
Daniel Camarena Perez
57538
The uniform continuity property refers to a spatially homogeneous behavior of continuity. Look at the graph of the following functions.
- $f(x)=x$
- $f(x)=sqrt$
- $f(x)=x^2$
- $$ f(x)= begincases
xsin(frac1x) & xne 0 \
0 & x=0
endcases$$
Note that (2) has a "vertiginous behavior" around $0$, since that $$f'(x)=frac12sqrtxto infty$$ when $xdownarrow 0$.
Also for (3) this behavior occurs when $|x|to infty$.
(4) is continuous but is even "less uniform" than previous cases.
answered Nov 11 at 13:38
Daniel Camarena Perez
57538
answered Nov 11 at 13:38
Daniel Camarena Perez
57538
answered Nov 11 at 13:38
Daniel Camarena Perez
57538
answered Nov 11 at 13:38
Daniel Camarena Perez
57538
57538
add a comment |
add a comment |
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7
You don't prove a statement true by finding an example for which the statement holds.
– ab123
Nov 11 at 12:52
It would be true if you added "bounded" to your assumptions, but as stated JCSantos has the perfect counterexample in his answer.
– JonathanZ
Nov 11 at 17:57