Mixing
Finding the value of function
$begingroup$
$f$ is a continuous function from $R$ to $R$ such that for any two real numbers $x$ and $y$
$$|f(x)-f(y)| leq 7{|x-y|}^{201}$$
Then will this function be constant?
My attempt
I divide both sides by $|x-y|$ and then take limit as $x$ goes to $y$ on both sides which will give the derivative as 0 which means the function will be constant, but I am not sure if this is the correct answer or not, I am getting confused because of the seven and the power. If my approach is correct, is there a general form for which this kind of equation will always give a constant function?
real-analysis calculus limits
asked Jan 14 at 23:28
user601297user601297
37119
$endgroup$
add a comment |
$begingroup$
$f$ is a continuous function from $R$ to $R$ such that for any two real numbers $x$ and $y$
$$|f(x)-f(y)| leq 7{|x-y|}^{201}$$
Then will this function be constant?
My attempt
I divide both sides by $|x-y|$ and then take limit as $x$ goes to $y$ on both sides which will give the derivative as 0 which means the function will be constant, but I am not sure if this is the correct answer or not, I am getting confused because of the seven and the power. If my approach is correct, is there a general form for which this kind of equation will always give a constant function?
real-analysis calculus limits
asked Jan 14 at 23:28
user601297user601297
37119
$endgroup$
add a comment |
$begingroup$
$f$ is a continuous function from $R$ to $R$ such that for any two real numbers $x$ and $y$
$$|f(x)-f(y)| leq 7{|x-y|}^{201}$$
Then will this function be constant?
My attempt
I divide both sides by $|x-y|$ and then take limit as $x$ goes to $y$ on both sides which will give the derivative as 0 which means the function will be constant, but I am not sure if this is the correct answer or not, I am getting confused because of the seven and the power. If my approach is correct, is there a general form for which this kind of equation will always give a constant function?
real-analysis calculus limits
asked Jan 14 at 23:28
user601297user601297
37119
$endgroup$
$f$ is a continuous function from $R$ to $R$ such that for any two real numbers $x$ and $y$
$$|f(x)-f(y)| leq 7{|x-y|}^{201}$$
Then will this function be constant?
My attempt
I divide both sides by $|x-y|$ and then take limit as $x$ goes to $y$ on both sides which will give the derivative as 0 which means the function will be constant, but I am not sure if this is the correct answer or not, I am getting confused because of the seven and the power. If my approach is correct, is there a general form for which this kind of equation will always give a constant function?
real-analysis calculus limits
real-analysis calculus limits
asked Jan 14 at 23:28
user601297user601297
37119
asked Jan 14 at 23:28
user601297user601297
37119
asked Jan 14 at 23:28
user601297user601297
37119
asked Jan 14 at 23:28
user601297user601297
37119
asked Jan 14 at 23:28
user601297user601297
37119
37119
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
Yes, your approach is correct. More generally, if $a,b>0$ and$$(forall x,yinmathbb{R}):bigllvert f(x)-f(y)bigrrvertleqslant alvert x-yrvert^{b+1},$$then $f$ is constant. That's so because$$xneq yimpliesleftlvertfrac{f(x)-f(y)}{x-y}rightrvertleqslant alvert x-yrvert^b$$and therefore $f'(x)=0$ for each real $x$
answered Jan 14 at 23:49
José Carlos SantosJosé Carlos Santos
162k22128232
$endgroup$
$begingroup$
In the second to last step, we have to take the limit as well , right?
$endgroup$
– user601297
Jan 14 at 23:50
$begingroup$
Yes, of course.
$endgroup$
– José Carlos Santos
Jan 15 at 6:27
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Yes, your approach is correct. More generally, if $a,b>0$ and$$(forall x,yinmathbb{R}):bigllvert f(x)-f(y)bigrrvertleqslant alvert x-yrvert^{b+1},$$then $f$ is constant. That's so because$$xneq yimpliesleftlvertfrac{f(x)-f(y)}{x-y}rightrvertleqslant alvert x-yrvert^b$$and therefore $f'(x)=0$ for each real $x$
answered Jan 14 at 23:49
José Carlos SantosJosé Carlos Santos
162k22128232
$endgroup$
$begingroup$
In the second to last step, we have to take the limit as well , right?
$endgroup$
– user601297
Jan 14 at 23:50
$begingroup$
Yes, of course.
$endgroup$
– José Carlos Santos
Jan 15 at 6:27
add a comment |
$begingroup$
Yes, your approach is correct. More generally, if $a,b>0$ and$$(forall x,yinmathbb{R}):bigllvert f(x)-f(y)bigrrvertleqslant alvert x-yrvert^{b+1},$$then $f$ is constant. That's so because$$xneq yimpliesleftlvertfrac{f(x)-f(y)}{x-y}rightrvertleqslant alvert x-yrvert^b$$and therefore $f'(x)=0$ for each real $x$
answered Jan 14 at 23:49
José Carlos SantosJosé Carlos Santos
162k22128232
$endgroup$
$begingroup$
In the second to last step, we have to take the limit as well , right?
$endgroup$
– user601297
Jan 14 at 23:50
$begingroup$
Yes, of course.
$endgroup$
– José Carlos Santos
Jan 15 at 6:27
add a comment |
$begingroup$
Yes, your approach is correct. More generally, if $a,b>0$ and$$(forall x,yinmathbb{R}):bigllvert f(x)-f(y)bigrrvertleqslant alvert x-yrvert^{b+1},$$then $f$ is constant. That's so because$$xneq yimpliesleftlvertfrac{f(x)-f(y)}{x-y}rightrvertleqslant alvert x-yrvert^b$$and therefore $f'(x)=0$ for each real $x$
answered Jan 14 at 23:49
José Carlos SantosJosé Carlos Santos
162k22128232
$endgroup$
Yes, your approach is correct. More generally, if $a,b>0$ and$$(forall x,yinmathbb{R}):bigllvert f(x)-f(y)bigrrvertleqslant alvert x-yrvert^{b+1},$$then $f$ is constant. That's so because$$xneq yimpliesleftlvertfrac{f(x)-f(y)}{x-y}rightrvertleqslant alvert x-yrvert^b$$and therefore $f'(x)=0$ for each real $x$
answered Jan 14 at 23:49
José Carlos SantosJosé Carlos Santos
162k22128232
answered Jan 14 at 23:49
José Carlos SantosJosé Carlos Santos
162k22128232
answered Jan 14 at 23:49
José Carlos SantosJosé Carlos Santos
162k22128232
answered Jan 14 at 23:49
José Carlos SantosJosé Carlos Santos
162k22128232
162k22128232
$begingroup$
In the second to last step, we have to take the limit as well , right?
$endgroup$
– user601297
Jan 14 at 23:50
$begingroup$
Yes, of course.
$endgroup$
– José Carlos Santos
Jan 15 at 6:27
add a comment |
$begingroup$
In the second to last step, we have to take the limit as well , right?
$endgroup$
– user601297
Jan 14 at 23:50
$begingroup$
Yes, of course.
$endgroup$
– José Carlos Santos
Jan 15 at 6:27
$begingroup$
In the second to last step, we have to take the limit as well , right?
$endgroup$
– user601297
Jan 14 at 23:50
$begingroup$
In the second to last step, we have to take the limit as well , right?
$endgroup$
– user601297
Jan 14 at 23:50
$begingroup$
Yes, of course.
$endgroup$
– José Carlos Santos
Jan 15 at 6:27
$begingroup$
Yes, of course.
$endgroup$
– José Carlos Santos
Jan 15 at 6:27
add a comment |
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