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Show that the function $f(x)= arcsin(x)$ is Lipschitz on $[-1,1]$
Clash Royale CLAN TAG#URR8PPP
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Actually , it is easy to see that$ f(x)$ is uniformly continuous on $[-1,1]$ , but via it's graph , I am sure that it's Lipschitz but I am unable to prove it . To show for any x in [-1,1] , $$|arcsin(x+c) - arcsin(x)| < K.c$$ where $K >0$ for any $0< c < 2$
real-analysis
asked Jul 26 at 17:38
RABI KUMAR CHAKRABORTY
244
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up vote
3
down vote
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Actually , it is easy to see that$ f(x)$ is uniformly continuous on $[-1,1]$ , but via it's graph , I am sure that it's Lipschitz but I am unable to prove it . To show for any x in [-1,1] , $$|arcsin(x+c) - arcsin(x)| < K.c$$ where $K >0$ for any $0< c < 2$
real-analysis
asked Jul 26 at 17:38
RABI KUMAR CHAKRABORTY
244
1
"I am sure that it's Lipschitz but I am disable to prove it" It is odd that you are sure it is and very comforting that you are unable to prove it is... What is the behaviour of $f(x)$ near $x=1$ already? Of course $f(1)=pi/2$, but $f(1-epsilon)$ is roughly... what?
– Did
Jul 26 at 17:47
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Actually , it is easy to see that$ f(x)$ is uniformly continuous on $[-1,1]$ , but via it's graph , I am sure that it's Lipschitz but I am unable to prove it . To show for any x in [-1,1] , $$|arcsin(x+c) - arcsin(x)| < K.c$$ where $K >0$ for any $0< c < 2$
real-analysis
asked Jul 26 at 17:38
RABI KUMAR CHAKRABORTY
244
Actually , it is easy to see that$ f(x)$ is uniformly continuous on $[-1,1]$ , but via it's graph , I am sure that it's Lipschitz but I am unable to prove it . To show for any x in [-1,1] , $$|arcsin(x+c) - arcsin(x)| < K.c$$ where $K >0$ for any $0< c < 2$
real-analysis
asked Jul 26 at 17:38
RABI KUMAR CHAKRABORTY
244
edited Jul 26 at 18:17
asked Jul 26 at 17:38
RABI KUMAR CHAKRABORTY
244
asked Jul 26 at 17:38
RABI KUMAR CHAKRABORTY
244
asked Jul 26 at 17:38
RABI KUMAR CHAKRABORTY
244
244
1
"I am sure that it's Lipschitz but I am disable to prove it" It is odd that you are sure it is and very comforting that you are unable to prove it is... What is the behaviour of $f(x)$ near $x=1$ already? Of course $f(1)=pi/2$, but $f(1-epsilon)$ is roughly... what?
– Did
Jul 26 at 17:47
add a comment |Â
1
"I am sure that it's Lipschitz but I am disable to prove it" It is odd that you are sure it is and very comforting that you are unable to prove it is... What is the behaviour of $f(x)$ near $x=1$ already? Of course $f(1)=pi/2$, but $f(1-epsilon)$ is roughly... what?
– Did
Jul 26 at 17:47
1
1
"I am sure that it's Lipschitz but I am disable to prove it" It is odd that you are sure it is and very comforting that you are unable to prove it is... What is the behaviour of $f(x)$ near $x=1$ already? Of course $f(1)=pi/2$, but $f(1-epsilon)$ is roughly... what?
– Did
Jul 26 at 17:47
"I am sure that it's Lipschitz but I am disable to prove it" It is odd that you are sure it is and very comforting that you are unable to prove it is... What is the behaviour of $f(x)$ near $x=1$ already? Of course $f(1)=pi/2$, but $f(1-epsilon)$ is roughly... what?
– Did
Jul 26 at 17:47
add a comment |Â
2 Answers
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3
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$fracddxmboxarcsin(x)=frac1sqrt1-x^2$ blows up as $x$ gets close to $pm 1$, which means it could not possibly be Lipschitz on $[-1,1]$, as Lipschitz functions must have bounded derivatives whenever they exist. As an analogous example, consider $f(x)=sqrtx$ near $x=0$, which is again not Lipschitz.
answered Jul 26 at 18:37
Alex R.
23.7k12352
2
@GonzaloBenavides: A Lipschitz function must have a bounded derivative wherever it exists. This follows straight from definitions: $lim_xrightarrow y left|fracf(x)-f(y)x-yright|leq L$.
– Alex R.
Jul 26 at 18:54
add a comment |Â
up vote
2
down vote
Note that for $0<varepsilon<1$, we have
$$beginalign
left|arcsin(1)-arcsin(1-varepsilon)right|&=left|int_1-varepsilon^1 frac1sqrt1-t^2,dtright|\\
&gefrac12int_1-varepsilon^1frac1sqrt1-t,dt\\
&=sqrtvarepsilon
endalign$$
Can you conclude now?
answered Jul 26 at 20:09
Mark Viola
126k1172167
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
$fracddxmboxarcsin(x)=frac1sqrt1-x^2$ blows up as $x$ gets close to $pm 1$, which means it could not possibly be Lipschitz on $[-1,1]$, as Lipschitz functions must have bounded derivatives whenever they exist. As an analogous example, consider $f(x)=sqrtx$ near $x=0$, which is again not Lipschitz.
answered Jul 26 at 18:37
Alex R.
23.7k12352
2
@GonzaloBenavides: A Lipschitz function must have a bounded derivative wherever it exists. This follows straight from definitions: $lim_xrightarrow y left|fracf(x)-f(y)x-yright|leq L$.
– Alex R.
Jul 26 at 18:54
add a comment |Â
up vote
3
down vote
$fracddxmboxarcsin(x)=frac1sqrt1-x^2$ blows up as $x$ gets close to $pm 1$, which means it could not possibly be Lipschitz on $[-1,1]$, as Lipschitz functions must have bounded derivatives whenever they exist. As an analogous example, consider $f(x)=sqrtx$ near $x=0$, which is again not Lipschitz.
answered Jul 26 at 18:37
Alex R.
23.7k12352
2
@GonzaloBenavides: A Lipschitz function must have a bounded derivative wherever it exists. This follows straight from definitions: $lim_xrightarrow y left|fracf(x)-f(y)x-yright|leq L$.
– Alex R.
Jul 26 at 18:54
add a comment |Â
up vote
3
down vote
up vote
3
down vote
$fracddxmboxarcsin(x)=frac1sqrt1-x^2$ blows up as $x$ gets close to $pm 1$, which means it could not possibly be Lipschitz on $[-1,1]$, as Lipschitz functions must have bounded derivatives whenever they exist. As an analogous example, consider $f(x)=sqrtx$ near $x=0$, which is again not Lipschitz.
answered Jul 26 at 18:37
Alex R.
23.7k12352
$fracddxmboxarcsin(x)=frac1sqrt1-x^2$ blows up as $x$ gets close to $pm 1$, which means it could not possibly be Lipschitz on $[-1,1]$, as Lipschitz functions must have bounded derivatives whenever they exist. As an analogous example, consider $f(x)=sqrtx$ near $x=0$, which is again not Lipschitz.
answered Jul 26 at 18:37
Alex R.
23.7k12352
edited Jul 26 at 18:54
answered Jul 26 at 18:37
Alex R.
23.7k12352
answered Jul 26 at 18:37
Alex R.
23.7k12352
answered Jul 26 at 18:37
Alex R.
23.7k12352
23.7k12352
2
@GonzaloBenavides: A Lipschitz function must have a bounded derivative wherever it exists. This follows straight from definitions: $lim_xrightarrow y left|fracf(x)-f(y)x-yright|leq L$.
– Alex R.
Jul 26 at 18:54
add a comment |Â
2
@GonzaloBenavides: A Lipschitz function must have a bounded derivative wherever it exists. This follows straight from definitions: $lim_xrightarrow y left|fracf(x)-f(y)x-yright|leq L$.
– Alex R.
Jul 26 at 18:54
2
2
@GonzaloBenavides: A Lipschitz function must have a bounded derivative wherever it exists. This follows straight from definitions: $lim_xrightarrow y left|fracf(x)-f(y)x-yright|leq L$.
– Alex R.
Jul 26 at 18:54
@GonzaloBenavides: A Lipschitz function must have a bounded derivative wherever it exists. This follows straight from definitions: $lim_xrightarrow y left|fracf(x)-f(y)x-yright|leq L$.
– Alex R.
Jul 26 at 18:54
add a comment |Â
up vote
2
down vote
Note that for $0<varepsilon<1$, we have
$$beginalign
left|arcsin(1)-arcsin(1-varepsilon)right|&=left|int_1-varepsilon^1 frac1sqrt1-t^2,dtright|\\
&gefrac12int_1-varepsilon^1frac1sqrt1-t,dt\\
&=sqrtvarepsilon
endalign$$
Can you conclude now?
answered Jul 26 at 20:09
Mark Viola
126k1172167
add a comment |Â
up vote
2
down vote
Note that for $0<varepsilon<1$, we have
$$beginalign
left|arcsin(1)-arcsin(1-varepsilon)right|&=left|int_1-varepsilon^1 frac1sqrt1-t^2,dtright|\\
&gefrac12int_1-varepsilon^1frac1sqrt1-t,dt\\
&=sqrtvarepsilon
endalign$$
Can you conclude now?
answered Jul 26 at 20:09
Mark Viola
126k1172167
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Note that for $0<varepsilon<1$, we have
$$beginalign
left|arcsin(1)-arcsin(1-varepsilon)right|&=left|int_1-varepsilon^1 frac1sqrt1-t^2,dtright|\\
&gefrac12int_1-varepsilon^1frac1sqrt1-t,dt\\
&=sqrtvarepsilon
endalign$$
Can you conclude now?
answered Jul 26 at 20:09
Mark Viola
126k1172167
Note that for $0<varepsilon<1$, we have
$$beginalign
left|arcsin(1)-arcsin(1-varepsilon)right|&=left|int_1-varepsilon^1 frac1sqrt1-t^2,dtright|\\
&gefrac12int_1-varepsilon^1frac1sqrt1-t,dt\\
&=sqrtvarepsilon
endalign$$
Can you conclude now?
answered Jul 26 at 20:09
Mark Viola
126k1172167
answered Jul 26 at 20:09
Mark Viola
126k1172167
answered Jul 26 at 20:09
Mark Viola
126k1172167
answered Jul 26 at 20:09
Mark Viola
126k1172167
126k1172167
add a comment |Â
add a comment |Â
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1
"I am sure that it's Lipschitz but I am disable to prove it" It is odd that you are sure it is and very comforting that you are unable to prove it is... What is the behaviour of $f(x)$ near $x=1$ already? Of course $f(1)=pi/2$, but $f(1-epsilon)$ is roughly... what?
– Did
Jul 26 at 17:47