Checking uniform continuity
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I want to show that $x^2sin frac {1}{x}$ is not uniformly continuous on $(0,infty)$. I tried to find sequences $(x_n),(y_n)$ in $(0,infty)$ such that $|x_n-y_n|to 0$ but $|f(x_n)-f(y_n)|$ does not go to $0$. Please help.
real-analysis continuity uniform-continuity
edited Nov 16 at 1:58
RRL
46.7k42366
asked Jan 21 '17 at 6:59
Anupam
2,3361823
add a comment |
up vote
1
down vote
favorite
I want to show that $x^2sin frac {1}{x}$ is not uniformly continuous on $(0,infty)$. I tried to find sequences $(x_n),(y_n)$ in $(0,infty)$ such that $|x_n-y_n|to 0$ but $|f(x_n)-f(y_n)|$ does not go to $0$. Please help.
real-analysis continuity uniform-continuity
edited Nov 16 at 1:58
RRL
46.7k42366
asked Jan 21 '17 at 6:59
Anupam
2,3361823
Why do you think it's not uniformly continuous?
– RRL
Jan 21 '17 at 9:10
@RRL yes you are correct. It is uniformly continuous.
– Anupam
Jan 21 '17 at 11:21
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I want to show that $x^2sin frac {1}{x}$ is not uniformly continuous on $(0,infty)$. I tried to find sequences $(x_n),(y_n)$ in $(0,infty)$ such that $|x_n-y_n|to 0$ but $|f(x_n)-f(y_n)|$ does not go to $0$. Please help.
real-analysis continuity uniform-continuity
edited Nov 16 at 1:58
RRL
46.7k42366
asked Jan 21 '17 at 6:59
Anupam
2,3361823
I want to show that $x^2sin frac {1}{x}$ is not uniformly continuous on $(0,infty)$. I tried to find sequences $(x_n),(y_n)$ in $(0,infty)$ such that $|x_n-y_n|to 0$ but $|f(x_n)-f(y_n)|$ does not go to $0$. Please help.
real-analysis continuity uniform-continuity
real-analysis continuity uniform-continuity
edited Nov 16 at 1:58
RRL
46.7k42366
asked Jan 21 '17 at 6:59
Anupam
2,3361823
edited Nov 16 at 1:58
RRL
46.7k42366
asked Jan 21 '17 at 6:59
Anupam
2,3361823
edited Nov 16 at 1:58
RRL
46.7k42366
edited Nov 16 at 1:58
RRL
46.7k42366
edited Nov 16 at 1:58
RRL
46.7k42366
46.7k42366
asked Jan 21 '17 at 6:59
Anupam
2,3361823
asked Jan 21 '17 at 6:59
Anupam
2,3361823
asked Jan 21 '17 at 6:59
Anupam
2,3361823
2,3361823
Why do you think it's not uniformly continuous?
– RRL
Jan 21 '17 at 9:10
@RRL yes you are correct. It is uniformly continuous.
– Anupam
Jan 21 '17 at 11:21
add a comment |
Why do you think it's not uniformly continuous?
– RRL
Jan 21 '17 at 9:10
@RRL yes you are correct. It is uniformly continuous.
– Anupam
Jan 21 '17 at 11:21
Why do you think it's not uniformly continuous?
– RRL
Jan 21 '17 at 9:10
Why do you think it's not uniformly continuous?
– RRL
Jan 21 '17 at 9:10
@RRL yes you are correct. It is uniformly continuous.
– Anupam
Jan 21 '17 at 11:21
@RRL yes you are correct. It is uniformly continuous.
– Anupam
Jan 21 '17 at 11:21
add a comment |
1 Answer
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For $f(x) = x^2 sin(1/x)$ we have
$$f'(x) = 2x sin(1/x) - cos(1/x).$$
For $ x in (0,infty)$ we have
$$|f'(x)| leqslant 2|xsin(1/x)| + |cos(1/x)| leqslant 3.$$
Since the derivative is bounded the function is uniformly continuous on $(0,infty).$
Note that $|xsin(1/x)| leqslant 1$ for $x in (0,infty)$. On $[1,infty)$ we have $0 < 1/x leqslant 1$ and $|x sin(1/x)| = frac{sin(1/x)}{1/x} leqslant 1$ since $frac{sin y}{y} uparrow 1$ as $y to 0+$. On $(0,1)$ we have $|x sin(1/x)| leqslant 1$ since $|x sin(1/x)| to 0$ as $x to 0$.
answered Jan 21 '17 at 7:45
RRL
46.7k42366
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
For $f(x) = x^2 sin(1/x)$ we have
$$f'(x) = 2x sin(1/x) - cos(1/x).$$
For $ x in (0,infty)$ we have
$$|f'(x)| leqslant 2|xsin(1/x)| + |cos(1/x)| leqslant 3.$$
Since the derivative is bounded the function is uniformly continuous on $(0,infty).$
Note that $|xsin(1/x)| leqslant 1$ for $x in (0,infty)$. On $[1,infty)$ we have $0 < 1/x leqslant 1$ and $|x sin(1/x)| = frac{sin(1/x)}{1/x} leqslant 1$ since $frac{sin y}{y} uparrow 1$ as $y to 0+$. On $(0,1)$ we have $|x sin(1/x)| leqslant 1$ since $|x sin(1/x)| to 0$ as $x to 0$.
answered Jan 21 '17 at 7:45
RRL
46.7k42366
add a comment |
up vote
3
down vote
accepted
For $f(x) = x^2 sin(1/x)$ we have
$$f'(x) = 2x sin(1/x) - cos(1/x).$$
For $ x in (0,infty)$ we have
$$|f'(x)| leqslant 2|xsin(1/x)| + |cos(1/x)| leqslant 3.$$
Since the derivative is bounded the function is uniformly continuous on $(0,infty).$
Note that $|xsin(1/x)| leqslant 1$ for $x in (0,infty)$. On $[1,infty)$ we have $0 < 1/x leqslant 1$ and $|x sin(1/x)| = frac{sin(1/x)}{1/x} leqslant 1$ since $frac{sin y}{y} uparrow 1$ as $y to 0+$. On $(0,1)$ we have $|x sin(1/x)| leqslant 1$ since $|x sin(1/x)| to 0$ as $x to 0$.
answered Jan 21 '17 at 7:45
RRL
46.7k42366
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
For $f(x) = x^2 sin(1/x)$ we have
$$f'(x) = 2x sin(1/x) - cos(1/x).$$
For $ x in (0,infty)$ we have
$$|f'(x)| leqslant 2|xsin(1/x)| + |cos(1/x)| leqslant 3.$$
Since the derivative is bounded the function is uniformly continuous on $(0,infty).$
Note that $|xsin(1/x)| leqslant 1$ for $x in (0,infty)$. On $[1,infty)$ we have $0 < 1/x leqslant 1$ and $|x sin(1/x)| = frac{sin(1/x)}{1/x} leqslant 1$ since $frac{sin y}{y} uparrow 1$ as $y to 0+$. On $(0,1)$ we have $|x sin(1/x)| leqslant 1$ since $|x sin(1/x)| to 0$ as $x to 0$.
answered Jan 21 '17 at 7:45
RRL
46.7k42366
For $f(x) = x^2 sin(1/x)$ we have
$$f'(x) = 2x sin(1/x) - cos(1/x).$$
For $ x in (0,infty)$ we have
$$|f'(x)| leqslant 2|xsin(1/x)| + |cos(1/x)| leqslant 3.$$
Since the derivative is bounded the function is uniformly continuous on $(0,infty).$
Note that $|xsin(1/x)| leqslant 1$ for $x in (0,infty)$. On $[1,infty)$ we have $0 < 1/x leqslant 1$ and $|x sin(1/x)| = frac{sin(1/x)}{1/x} leqslant 1$ since $frac{sin y}{y} uparrow 1$ as $y to 0+$. On $(0,1)$ we have $|x sin(1/x)| leqslant 1$ since $|x sin(1/x)| to 0$ as $x to 0$.
answered Jan 21 '17 at 7:45
RRL
46.7k42366
edited Jan 21 '17 at 8:11
answered Jan 21 '17 at 7:45
RRL
46.7k42366
answered Jan 21 '17 at 7:45
RRL
46.7k42366
answered Jan 21 '17 at 7:45
RRL
46.7k42366
46.7k42366
add a comment |
add a comment |
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Why do you think it's not uniformly continuous?
– RRL
Jan 21 '17 at 9:10
@RRL yes you are correct. It is uniformly continuous.
– Anupam
Jan 21 '17 at 11:21