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FInd a function $f$ which satisfies $af(kx)-bf(x)$ is a function of $k$ and irrevelent to $x$ [on hold]

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1

1

$begingroup$

I want to find a function $f$ that satisfies
$af(kx) - bf(x)=c$ where $a,b,c,k$ are constant. In other words, a function that makes the left part irrelevent to $x$.

For example, when $a=b$, we can find a function $f(x)=log(x)$, then $af(kx) - bf(x) = alog(kx)-alog(x)=alog(k)$ which is a constant.
Although it's true that an identity function satisfies the condition, but it's too trivial and not what I want.

functional-analysis

share|cite|improve this question

edited 2 days ago

troy.tang

asked Mar 11 at 8:50

troy.tangtroy.tang

112

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troy.tang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$

put on hold as too broad by 5xum, Eevee Trainer, Claude Leibovici, uniquesolution, Riccardo.Alestra Mar 11 at 10:21

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
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  $begingroup$
  If you just want to find examples there are plenty: $f$ can be a constant, it can be the identity function with $b=ak$ etc.
  $endgroup$
  – Kavi Rama Murthy
  Mar 11 at 8:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Are $a,b,c,k$ known constants? Are there other limitations to $f$? Right now, this question is waaaay to broad.
  $endgroup$
  – 5xum
  Mar 11 at 8:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $f(x)$ should be a monotone function with respect to $x$, and $a,b,c$ are known while $k$ can be solved.
  $endgroup$
  – troy.tang
  Mar 11 at 8:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For example, when $a=b$, we can find a $f(x) = log(x)$ which satisfy the equation.
  $endgroup$
  – troy.tang
  Mar 11 at 8:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The identity function satisfies the condition, but it's kind of trivial and not what I'm looking for, I'm looking for a function form like $log(x)$
  $endgroup$
  – troy.tang
  Mar 11 at 9:08
  

  
  
  
  

 | 
show 2 more comments

1

1

$begingroup$

I want to find a function $f$ that satisfies
$af(kx) - bf(x)=c$ where $a,b,c,k$ are constant. In other words, a function that makes the left part irrelevent to $x$.

For example, when $a=b$, we can find a function $f(x)=log(x)$, then $af(kx) - bf(x) = alog(kx)-alog(x)=alog(k)$ which is a constant.
Although it's true that an identity function satisfies the condition, but it's too trivial and not what I want.

functional-analysis

share|cite|improve this question

edited 2 days ago

troy.tang

asked Mar 11 at 8:50

troy.tangtroy.tang

112

New contributor

troy.tang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

put on hold as too broad by 5xum, Eevee Trainer, Claude Leibovici, uniquesolution, Riccardo.Alestra Mar 11 at 10:21

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  If you just want to find examples there are plenty: $f$ can be a constant, it can be the identity function with $b=ak$ etc.
  $endgroup$
  – Kavi Rama Murthy
  Mar 11 at 8:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Are $a,b,c,k$ known constants? Are there other limitations to $f$? Right now, this question is waaaay to broad.
  $endgroup$
  – 5xum
  Mar 11 at 8:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $f(x)$ should be a monotone function with respect to $x$, and $a,b,c$ are known while $k$ can be solved.
  $endgroup$
  – troy.tang
  Mar 11 at 8:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For example, when $a=b$, we can find a $f(x) = log(x)$ which satisfy the equation.
  $endgroup$
  – troy.tang
  Mar 11 at 8:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The identity function satisfies the condition, but it's kind of trivial and not what I'm looking for, I'm looking for a function form like $log(x)$
  $endgroup$
  – troy.tang
  Mar 11 at 9:08
  

  
  
  
  

 | 
show 2 more comments

$begingroup$

I want to find a function $f$ that satisfies
$af(kx) - bf(x)=c$ where $a,b,c,k$ are constant. In other words, a function that makes the left part irrelevent to $x$.

For example, when $a=b$, we can find a function $f(x)=log(x)$, then $af(kx) - bf(x) = alog(kx)-alog(x)=alog(k)$ which is a constant.
Although it's true that an identity function satisfies the condition, but it's too trivial and not what I want.

functional-analysis

share|cite|improve this question

edited 2 days ago

troy.tang

asked Mar 11 at 8:50

troy.tangtroy.tang

112

New contributor

troy.tang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited 2 days ago

troy.tang

asked Mar 11 at 8:50

troy.tangtroy.tang

112

New contributor

troy.tang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited 2 days ago

troy.tang

asked Mar 11 at 8:50

troy.tangtroy.tang

112

New contributor

troy.tang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked Mar 11 at 8:50

troy.tangtroy.tang

112

New contributor

troy.tang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked Mar 11 at 8:50

troy.tangtroy.tang

112

1 Answer
1

active

oldest

votes

1

$begingroup$

In general, $f(x)=fracca-b$, the constant function is the only solution defined for all $x$.

Plug in $x=1$, then you can rearrange the equation to $f(k)=frac1a(c+bcdot f(1))$, so $f$ has to be constant. You can compute the constant by plugging in $k=1$.

There can be more solutions for specific values of $a,b,c,k$ and maybe also other solutions that are only defined for some values of $x$.

share|cite|improve this answer

answered Mar 11 at 9:20

quaraguequarague

436210

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Given that the OP accepts $f(x) = log(x)$ as a solution, they may permit $c$ to depend on $k$
  $endgroup$
  – preferred_anon
  Mar 11 at 9:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, in fact. I want to find a non-trivial function that satisfies $af(kx)-bf(x)$ is irrelevant to $x$. When $a=b$, given a $f(x)=log(x)$, then it becomes $a log(kx)-a log(x)=a log(k)$ which is a constant. The problem is what's the function will be when $a neq b$?
  $endgroup$
  – troy.tang
  Mar 11 at 11:32
  
  

  
  
  
  

add a comment | 

1

$begingroup$

In general, $f(x)=fracca-b$, the constant function is the only solution defined for all $x$.

Plug in $x=1$, then you can rearrange the equation to $f(k)=frac1a(c+bcdot f(1))$, so $f$ has to be constant. You can compute the constant by plugging in $k=1$.

There can be more solutions for specific values of $a,b,c,k$ and maybe also other solutions that are only defined for some values of $x$.

share|cite|improve this answer

answered Mar 11 at 9:20

quaraguequarague

436210

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Given that the OP accepts $f(x) = log(x)$ as a solution, they may permit $c$ to depend on $k$
  $endgroup$
  – preferred_anon
  Mar 11 at 9:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, in fact. I want to find a non-trivial function that satisfies $af(kx)-bf(x)$ is irrelevant to $x$. When $a=b$, given a $f(x)=log(x)$, then it becomes $a log(kx)-a log(x)=a log(k)$ which is a constant. The problem is what's the function will be when $a neq b$?
  $endgroup$
  – troy.tang
  Mar 11 at 11:32
  
  

  
  
  
  

add a comment | 

1

$begingroup$

In general, $f(x)=fracca-b$, the constant function is the only solution defined for all $x$.

Plug in $x=1$, then you can rearrange the equation to $f(k)=frac1a(c+bcdot f(1))$, so $f$ has to be constant. You can compute the constant by plugging in $k=1$.

There can be more solutions for specific values of $a,b,c,k$ and maybe also other solutions that are only defined for some values of $x$.

share|cite|improve this answer

answered Mar 11 at 9:20

quaraguequarague

436210

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Given that the OP accepts $f(x) = log(x)$ as a solution, they may permit $c$ to depend on $k$
  $endgroup$
  – preferred_anon
  Mar 11 at 9:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, in fact. I want to find a non-trivial function that satisfies $af(kx)-bf(x)$ is irrelevant to $x$. When $a=b$, given a $f(x)=log(x)$, then it becomes $a log(kx)-a log(x)=a log(k)$ which is a constant. The problem is what's the function will be when $a neq b$?
  $endgroup$
  – troy.tang
  Mar 11 at 11:32
  
  

  
  
  
  

add a comment | 

$begingroup$

In general, $f(x)=fracca-b$, the constant function is the only solution defined for all $x$.

Plug in $x=1$, then you can rearrange the equation to $f(k)=frac1a(c+bcdot f(1))$, so $f$ has to be constant. You can compute the constant by plugging in $k=1$.

There can be more solutions for specific values of $a,b,c,k$ and maybe also other solutions that are only defined for some values of $x$.

share|cite|improve this answer

answered Mar 11 at 9:20

quaraguequarague

436210

$endgroup$

share|cite|improve this answer

answered Mar 11 at 9:20

quaraguequarague

436210

answered Mar 11 at 9:20

quaraguequarague

436210

answered Mar 11 at 9:20

quaraguequarague

436210