Mixing
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Proving a function of a bounded variation
For $a$, $b$ > 0, define the function $f$ on $[0, 1]$ by $f(x) = x^{a}sin(1/x^{b})$ for $0 < x leq 1$ and $f(x) = 0$ for $x=0$. Show that if $a> b$, then $f$ is of bounded variation on $[0, 1]$, by showing that $f'$ is integrable over $[0, 1]$. Then show that if $a leq b$, then $f$ is not of bounded variation on $[0, 1]$.
First, we see clearly that $f$ is continuous on $(0,1]$. Also, $f$ is continuous at $x=0$ because $lim_{x rightarrow 0}f(x)=0=f(0).$ Moreover, $f$ is differentiable over $(a,b)$ with $f'=ax^{a-1}sin(1/x^b)-bx^{a-(b+1)}cos(1/x^b)$ on $(0,1)$, and we see that $f'$ is continuous on $(0,1]$. I can not see the continuity for $f$ at x=0 becouse when $1>a>0$ we have a problem.
Any thoughts or ideas about that would be appreciated.
measure-theory lebesgue-measure
asked Nov 20 '18 at 6:03
Ahmed
29019
add a comment |
For $a$, $b$ > 0, define the function $f$ on $[0, 1]$ by $f(x) = x^{a}sin(1/x^{b})$ for $0 < x leq 1$ and $f(x) = 0$ for $x=0$. Show that if $a> b$, then $f$ is of bounded variation on $[0, 1]$, by showing that $f'$ is integrable over $[0, 1]$. Then show that if $a leq b$, then $f$ is not of bounded variation on $[0, 1]$.
First, we see clearly that $f$ is continuous on $(0,1]$. Also, $f$ is continuous at $x=0$ because $lim_{x rightarrow 0}f(x)=0=f(0).$ Moreover, $f$ is differentiable over $(a,b)$ with $f'=ax^{a-1}sin(1/x^b)-bx^{a-(b+1)}cos(1/x^b)$ on $(0,1)$, and we see that $f'$ is continuous on $(0,1]$. I can not see the continuity for $f$ at x=0 becouse when $1>a>0$ we have a problem.
Any thoughts or ideas about that would be appreciated.
measure-theory lebesgue-measure
asked Nov 20 '18 at 6:03
Ahmed
29019
add a comment |
For $a$, $b$ > 0, define the function $f$ on $[0, 1]$ by $f(x) = x^{a}sin(1/x^{b})$ for $0 < x leq 1$ and $f(x) = 0$ for $x=0$. Show that if $a> b$, then $f$ is of bounded variation on $[0, 1]$, by showing that $f'$ is integrable over $[0, 1]$. Then show that if $a leq b$, then $f$ is not of bounded variation on $[0, 1]$.
First, we see clearly that $f$ is continuous on $(0,1]$. Also, $f$ is continuous at $x=0$ because $lim_{x rightarrow 0}f(x)=0=f(0).$ Moreover, $f$ is differentiable over $(a,b)$ with $f'=ax^{a-1}sin(1/x^b)-bx^{a-(b+1)}cos(1/x^b)$ on $(0,1)$, and we see that $f'$ is continuous on $(0,1]$. I can not see the continuity for $f$ at x=0 becouse when $1>a>0$ we have a problem.
Any thoughts or ideas about that would be appreciated.
measure-theory lebesgue-measure
asked Nov 20 '18 at 6:03
Ahmed
29019
For $a$, $b$ > 0, define the function $f$ on $[0, 1]$ by $f(x) = x^{a}sin(1/x^{b})$ for $0 < x leq 1$ and $f(x) = 0$ for $x=0$. Show that if $a> b$, then $f$ is of bounded variation on $[0, 1]$, by showing that $f'$ is integrable over $[0, 1]$. Then show that if $a leq b$, then $f$ is not of bounded variation on $[0, 1]$.
First, we see clearly that $f$ is continuous on $(0,1]$. Also, $f$ is continuous at $x=0$ because $lim_{x rightarrow 0}f(x)=0=f(0).$ Moreover, $f$ is differentiable over $(a,b)$ with $f'=ax^{a-1}sin(1/x^b)-bx^{a-(b+1)}cos(1/x^b)$ on $(0,1)$, and we see that $f'$ is continuous on $(0,1]$. I can not see the continuity for $f$ at x=0 becouse when $1>a>0$ we have a problem.
Any thoughts or ideas about that would be appreciated.
measure-theory lebesgue-measure
measure-theory lebesgue-measure
asked Nov 20 '18 at 6:03
Ahmed
29019
asked Nov 20 '18 at 6:03
Ahmed
29019
asked Nov 20 '18 at 6:03
Ahmed
29019
asked Nov 20 '18 at 6:03
Ahmed
29019
asked Nov 20 '18 at 6:03
Ahmed
29019
29019
add a comment |
add a comment |
1 Answer
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It is enough to show that $f'$ is integrable; continuity at $0$ is not required. Can you verify that each of the two terms in $f'$ is intergrable? Once you do that use the fact that $f(y)-f(x)=int_x^{y} f'(t) , dt$ to show that $f$ is of bounded variation. [Total variation of $f$ is less than or equal to $int_0^{1}|f'(t)|, dt$].
answered Nov 20 '18 at 6:13
Kavi Rama Murthy
50.4k31854
I could not veify that for the first term because of $a-1$ if its negative!!
– Ahmed
Nov 20 '18 at 6:18
@Ahmed $sin (1/x^{b})$ is bounded. $x^{a-1}$ is integrable: $int_0^{1}x^{a-1} , dx=frac 1 a x^{a}|_0^{1}=frac 1 a$.
– Kavi Rama Murthy
Nov 20 '18 at 6:21
Yeah yeah I got it than you
– Ahmed
Nov 20 '18 at 6:23
add a comment |
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1 Answer
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1 Answer
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It is enough to show that $f'$ is integrable; continuity at $0$ is not required. Can you verify that each of the two terms in $f'$ is intergrable? Once you do that use the fact that $f(y)-f(x)=int_x^{y} f'(t) , dt$ to show that $f$ is of bounded variation. [Total variation of $f$ is less than or equal to $int_0^{1}|f'(t)|, dt$].
answered Nov 20 '18 at 6:13
Kavi Rama Murthy
50.4k31854
I could not veify that for the first term because of $a-1$ if its negative!!
– Ahmed
Nov 20 '18 at 6:18
@Ahmed $sin (1/x^{b})$ is bounded. $x^{a-1}$ is integrable: $int_0^{1}x^{a-1} , dx=frac 1 a x^{a}|_0^{1}=frac 1 a$.
– Kavi Rama Murthy
Nov 20 '18 at 6:21
Yeah yeah I got it than you
– Ahmed
Nov 20 '18 at 6:23
add a comment |
It is enough to show that $f'$ is integrable; continuity at $0$ is not required. Can you verify that each of the two terms in $f'$ is intergrable? Once you do that use the fact that $f(y)-f(x)=int_x^{y} f'(t) , dt$ to show that $f$ is of bounded variation. [Total variation of $f$ is less than or equal to $int_0^{1}|f'(t)|, dt$].
answered Nov 20 '18 at 6:13
Kavi Rama Murthy
50.4k31854
I could not veify that for the first term because of $a-1$ if its negative!!
– Ahmed
Nov 20 '18 at 6:18
@Ahmed $sin (1/x^{b})$ is bounded. $x^{a-1}$ is integrable: $int_0^{1}x^{a-1} , dx=frac 1 a x^{a}|_0^{1}=frac 1 a$.
– Kavi Rama Murthy
Nov 20 '18 at 6:21
Yeah yeah I got it than you
– Ahmed
Nov 20 '18 at 6:23
add a comment |
It is enough to show that $f'$ is integrable; continuity at $0$ is not required. Can you verify that each of the two terms in $f'$ is intergrable? Once you do that use the fact that $f(y)-f(x)=int_x^{y} f'(t) , dt$ to show that $f$ is of bounded variation. [Total variation of $f$ is less than or equal to $int_0^{1}|f'(t)|, dt$].
answered Nov 20 '18 at 6:13
Kavi Rama Murthy
50.4k31854
It is enough to show that $f'$ is integrable; continuity at $0$ is not required. Can you verify that each of the two terms in $f'$ is intergrable? Once you do that use the fact that $f(y)-f(x)=int_x^{y} f'(t) , dt$ to show that $f$ is of bounded variation. [Total variation of $f$ is less than or equal to $int_0^{1}|f'(t)|, dt$].
answered Nov 20 '18 at 6:13
Kavi Rama Murthy
50.4k31854
answered Nov 20 '18 at 6:13
Kavi Rama Murthy
50.4k31854
answered Nov 20 '18 at 6:13
Kavi Rama Murthy
50.4k31854
answered Nov 20 '18 at 6:13
Kavi Rama Murthy
50.4k31854
50.4k31854
I could not veify that for the first term because of $a-1$ if its negative!!
– Ahmed
Nov 20 '18 at 6:18
@Ahmed $sin (1/x^{b})$ is bounded. $x^{a-1}$ is integrable: $int_0^{1}x^{a-1} , dx=frac 1 a x^{a}|_0^{1}=frac 1 a$.
– Kavi Rama Murthy
Nov 20 '18 at 6:21
Yeah yeah I got it than you
– Ahmed
Nov 20 '18 at 6:23
add a comment |
I could not veify that for the first term because of $a-1$ if its negative!!
– Ahmed
Nov 20 '18 at 6:18
@Ahmed $sin (1/x^{b})$ is bounded. $x^{a-1}$ is integrable: $int_0^{1}x^{a-1} , dx=frac 1 a x^{a}|_0^{1}=frac 1 a$.
– Kavi Rama Murthy
Nov 20 '18 at 6:21
Yeah yeah I got it than you
– Ahmed
Nov 20 '18 at 6:23
I could not veify that for the first term because of $a-1$ if its negative!!
– Ahmed
Nov 20 '18 at 6:18
I could not veify that for the first term because of $a-1$ if its negative!!
– Ahmed
Nov 20 '18 at 6:18
@Ahmed $sin (1/x^{b})$ is bounded. $x^{a-1}$ is integrable: $int_0^{1}x^{a-1} , dx=frac 1 a x^{a}|_0^{1}=frac 1 a$.
– Kavi Rama Murthy
Nov 20 '18 at 6:21
@Ahmed $sin (1/x^{b})$ is bounded. $x^{a-1}$ is integrable: $int_0^{1}x^{a-1} , dx=frac 1 a x^{a}|_0^{1}=frac 1 a$.
– Kavi Rama Murthy
Nov 20 '18 at 6:21
Yeah yeah I got it than you
– Ahmed
Nov 20 '18 at 6:23
Yeah yeah I got it than you
– Ahmed
Nov 20 '18 at 6:23
add a comment |
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