Mixing
Find all continuous functions in $0$ that $2f(2x) = f(x) + x $
$begingroup$
I need to find all functions that they are continuous in zero and
$$ 2f(2x) = f(x) + x $$
About
I know that there are many examples and that forum but I don't understand one thing in it and I need additional explanation. (Nowhere I see similar problem :( )
My try
I take $ y= 2x$ then
$$f(y) = frac{1}{2}fleft(frac{1}{2}yright) + frac{1}{4}$$
after induction I get:
$$f(y) = frac{1}{2^n}fleft(frac{1}{2^n}yright) + yleft(frac{1}{2^2} + frac{1}{2^4} + ... + frac{1}{2^{2n}} right)$$
I take $lim_{nrightarrow infty} $
$$ lim_{nrightarrow infty}f(y) = f(y) = lim_{nrightarrow infty} frac{1}{2^n}fleft(frac{1}{2^n}yright) + ycdot lim_{nrightarrow infty} left(frac{1}{2^2} + frac{1}{2^4} + ... + frac{1}{2^{2n}} right)$$
$$f(y) = lim_{nrightarrow infty} frac{1}{2^n} cdot fleft( lim_{nrightarrow infty} frac{1}{2^n}y right) + frac{1}{3}y$$
Ok, there I have question - what I should there after? How do I know that $$f(0) = 0 $$?
I think that it can be related with " continuous functions in $0$ " but
function is continous in $0$ when
$$ lim_{yrightarrow 0^+}f(y)=f(0)=lim_{yrightarrow 0^-}f(y)$$
And I don't see a reason why $f(0)=0$
edit
- Ok, I know why $f(0) =0$ but why I need informations about "Continuity at a point $0$ " ? It comes to
$$lim_{nrightarrow infty}fleft(frac{1}{2^n}yright) = fleft( lim_{nrightarrow infty} frac{1}{2^n}y right)$$ ?
real-analysis functional-equations
edited Jan 23 at 7:04
daw
24.7k1645
asked Jan 22 at 19:55
VirtualUserVirtualUser
1,094115
$endgroup$
|
show 4 more comments
$begingroup$
I need to find all functions that they are continuous in zero and
$$ 2f(2x) = f(x) + x $$
About
I know that there are many examples and that forum but I don't understand one thing in it and I need additional explanation. (Nowhere I see similar problem :( )
My try
I take $ y= 2x$ then
$$f(y) = frac{1}{2}fleft(frac{1}{2}yright) + frac{1}{4}$$
after induction I get:
$$f(y) = frac{1}{2^n}fleft(frac{1}{2^n}yright) + yleft(frac{1}{2^2} + frac{1}{2^4} + ... + frac{1}{2^{2n}} right)$$
I take $lim_{nrightarrow infty} $
$$ lim_{nrightarrow infty}f(y) = f(y) = lim_{nrightarrow infty} frac{1}{2^n}fleft(frac{1}{2^n}yright) + ycdot lim_{nrightarrow infty} left(frac{1}{2^2} + frac{1}{2^4} + ... + frac{1}{2^{2n}} right)$$
$$f(y) = lim_{nrightarrow infty} frac{1}{2^n} cdot fleft( lim_{nrightarrow infty} frac{1}{2^n}y right) + frac{1}{3}y$$
Ok, there I have question - what I should there after? How do I know that $$f(0) = 0 $$?
I think that it can be related with " continuous functions in $0$ " but
function is continous in $0$ when
$$ lim_{yrightarrow 0^+}f(y)=f(0)=lim_{yrightarrow 0^-}f(y)$$
And I don't see a reason why $f(0)=0$
edit
- Ok, I know why $f(0) =0$ but why I need informations about "Continuity at a point $0$ " ? It comes to
$$lim_{nrightarrow infty}fleft(frac{1}{2^n}yright) = fleft( lim_{nrightarrow infty} frac{1}{2^n}y right)$$ ?
real-analysis functional-equations
edited Jan 23 at 7:04
daw
24.7k1645
asked Jan 22 at 19:55
VirtualUserVirtualUser
1,094115
$endgroup$
$begingroup$
Are you sure you meant what you wrote? $2 f(x) = f(x) + x$ just says $f(x) = x$.
$endgroup$
– Robert Israel
Jan 22 at 19:57
$begingroup$
Are you sure of your functional equation?
$endgroup$
– lulu
Jan 22 at 19:57
3
$begingroup$
To get $f(0)=0$ just plug in $x=0$ into your equation
$endgroup$
– GReyes
Jan 22 at 20:00
1
$begingroup$
$f(0)=0$ because when you set $x=0$ in your (corrected) functional equation you get $2f(0)=f(0)$.
$endgroup$
– Dog_69
Jan 22 at 20:01
1
$begingroup$
You need continuity to conclude that, since $y/2^nto 0$ as $ntoinfty$, then $f(y/2^n)to 0$ as well (for any $y$). The only solution is $f(x)=x/3$
$endgroup$
– GReyes
Jan 22 at 20:35
|
show 4 more comments
$begingroup$
I need to find all functions that they are continuous in zero and
$$ 2f(2x) = f(x) + x $$
About
I know that there are many examples and that forum but I don't understand one thing in it and I need additional explanation. (Nowhere I see similar problem :( )
My try
I take $ y= 2x$ then
$$f(y) = frac{1}{2}fleft(frac{1}{2}yright) + frac{1}{4}$$
after induction I get:
$$f(y) = frac{1}{2^n}fleft(frac{1}{2^n}yright) + yleft(frac{1}{2^2} + frac{1}{2^4} + ... + frac{1}{2^{2n}} right)$$
I take $lim_{nrightarrow infty} $
$$ lim_{nrightarrow infty}f(y) = f(y) = lim_{nrightarrow infty} frac{1}{2^n}fleft(frac{1}{2^n}yright) + ycdot lim_{nrightarrow infty} left(frac{1}{2^2} + frac{1}{2^4} + ... + frac{1}{2^{2n}} right)$$
$$f(y) = lim_{nrightarrow infty} frac{1}{2^n} cdot fleft( lim_{nrightarrow infty} frac{1}{2^n}y right) + frac{1}{3}y$$
Ok, there I have question - what I should there after? How do I know that $$f(0) = 0 $$?
I think that it can be related with " continuous functions in $0$ " but
function is continous in $0$ when
$$ lim_{yrightarrow 0^+}f(y)=f(0)=lim_{yrightarrow 0^-}f(y)$$
And I don't see a reason why $f(0)=0$
edit
- Ok, I know why $f(0) =0$ but why I need informations about "Continuity at a point $0$ " ? It comes to
$$lim_{nrightarrow infty}fleft(frac{1}{2^n}yright) = fleft( lim_{nrightarrow infty} frac{1}{2^n}y right)$$ ?
real-analysis functional-equations
edited Jan 23 at 7:04
daw
24.7k1645
asked Jan 22 at 19:55
VirtualUserVirtualUser
1,094115
$endgroup$
I need to find all functions that they are continuous in zero and
$$ 2f(2x) = f(x) + x $$
About
I know that there are many examples and that forum but I don't understand one thing in it and I need additional explanation. (Nowhere I see similar problem :( )
My try
I take $ y= 2x$ then
$$f(y) = frac{1}{2}fleft(frac{1}{2}yright) + frac{1}{4}$$
after induction I get:
$$f(y) = frac{1}{2^n}fleft(frac{1}{2^n}yright) + yleft(frac{1}{2^2} + frac{1}{2^4} + ... + frac{1}{2^{2n}} right)$$
I take $lim_{nrightarrow infty} $
$$ lim_{nrightarrow infty}f(y) = f(y) = lim_{nrightarrow infty} frac{1}{2^n}fleft(frac{1}{2^n}yright) + ycdot lim_{nrightarrow infty} left(frac{1}{2^2} + frac{1}{2^4} + ... + frac{1}{2^{2n}} right)$$
$$f(y) = lim_{nrightarrow infty} frac{1}{2^n} cdot fleft( lim_{nrightarrow infty} frac{1}{2^n}y right) + frac{1}{3}y$$
Ok, there I have question - what I should there after? How do I know that $$f(0) = 0 $$?
I think that it can be related with " continuous functions in $0$ " but
function is continous in $0$ when
$$ lim_{yrightarrow 0^+}f(y)=f(0)=lim_{yrightarrow 0^-}f(y)$$
And I don't see a reason why $f(0)=0$
edit
- Ok, I know why $f(0) =0$ but why I need informations about "Continuity at a point $0$ " ? It comes to
$$lim_{nrightarrow infty}fleft(frac{1}{2^n}yright) = fleft( lim_{nrightarrow infty} frac{1}{2^n}y right)$$ ?
real-analysis functional-equations
real-analysis functional-equations
edited Jan 23 at 7:04
daw
24.7k1645
asked Jan 22 at 19:55
VirtualUserVirtualUser
1,094115
edited Jan 23 at 7:04
daw
24.7k1645
asked Jan 22 at 19:55
VirtualUserVirtualUser
1,094115
edited Jan 23 at 7:04
daw
24.7k1645
edited Jan 23 at 7:04
daw
24.7k1645
edited Jan 23 at 7:04
daw
24.7k1645
24.7k1645
asked Jan 22 at 19:55
VirtualUserVirtualUser
1,094115
asked Jan 22 at 19:55
VirtualUserVirtualUser
1,094115
asked Jan 22 at 19:55
VirtualUserVirtualUser
1,094115
1,094115
$begingroup$
Are you sure you meant what you wrote? $2 f(x) = f(x) + x$ just says $f(x) = x$.
$endgroup$
– Robert Israel
Jan 22 at 19:57
$begingroup$
Are you sure of your functional equation?
$endgroup$
– lulu
Jan 22 at 19:57
3
$begingroup$
To get $f(0)=0$ just plug in $x=0$ into your equation
$endgroup$
– GReyes
Jan 22 at 20:00
1
$begingroup$
$f(0)=0$ because when you set $x=0$ in your (corrected) functional equation you get $2f(0)=f(0)$.
$endgroup$
– Dog_69
Jan 22 at 20:01
1
$begingroup$
You need continuity to conclude that, since $y/2^nto 0$ as $ntoinfty$, then $f(y/2^n)to 0$ as well (for any $y$). The only solution is $f(x)=x/3$
$endgroup$
– GReyes
Jan 22 at 20:35
|
show 4 more comments
$begingroup$
Are you sure you meant what you wrote? $2 f(x) = f(x) + x$ just says $f(x) = x$.
$endgroup$
– Robert Israel
Jan 22 at 19:57
$begingroup$
Are you sure of your functional equation?
$endgroup$
– lulu
Jan 22 at 19:57
3
$begingroup$
To get $f(0)=0$ just plug in $x=0$ into your equation
$endgroup$
– GReyes
Jan 22 at 20:00
1
$begingroup$
$f(0)=0$ because when you set $x=0$ in your (corrected) functional equation you get $2f(0)=f(0)$.
$endgroup$
– Dog_69
Jan 22 at 20:01
1
$begingroup$
You need continuity to conclude that, since $y/2^nto 0$ as $ntoinfty$, then $f(y/2^n)to 0$ as well (for any $y$). The only solution is $f(x)=x/3$
$endgroup$
– GReyes
Jan 22 at 20:35
$begingroup$
Are you sure you meant what you wrote? $2 f(x) = f(x) + x$ just says $f(x) = x$.
$endgroup$
– Robert Israel
Jan 22 at 19:57
$begingroup$
Are you sure you meant what you wrote? $2 f(x) = f(x) + x$ just says $f(x) = x$.
$endgroup$
– Robert Israel
Jan 22 at 19:57
$begingroup$
Are you sure of your functional equation?
$endgroup$
– lulu
Jan 22 at 19:57
$begingroup$
Are you sure of your functional equation?
$endgroup$
– lulu
Jan 22 at 19:57
3
3
$begingroup$
To get $f(0)=0$ just plug in $x=0$ into your equation
$endgroup$
– GReyes
Jan 22 at 20:00
$begingroup$
To get $f(0)=0$ just plug in $x=0$ into your equation
$endgroup$
– GReyes
Jan 22 at 20:00
1
1
$begingroup$
$f(0)=0$ because when you set $x=0$ in your (corrected) functional equation you get $2f(0)=f(0)$.
$endgroup$
– Dog_69
Jan 22 at 20:01
$begingroup$
$f(0)=0$ because when you set $x=0$ in your (corrected) functional equation you get $2f(0)=f(0)$.
$endgroup$
– Dog_69
Jan 22 at 20:01
1
1
$begingroup$
You need continuity to conclude that, since $y/2^nto 0$ as $ntoinfty$, then $f(y/2^n)to 0$ as well (for any $y$). The only solution is $f(x)=x/3$
$endgroup$
– GReyes
Jan 22 at 20:35
$begingroup$
You need continuity to conclude that, since $y/2^nto 0$ as $ntoinfty$, then $f(y/2^n)to 0$ as well (for any $y$). The only solution is $f(x)=x/3$
$endgroup$
– GReyes
Jan 22 at 20:35
|
show 4 more comments
4 Answers
4
active
oldest
votes
$begingroup$
A powerful method to solve these kinds of problems is to reduce to a simpler equation. In this case we want to eliminate the $x$ in the right hand side. Set $g(x)=f(x)+ax$, with $a$ to be found later. Note that $f$ is continuous if and only if $g$ is. Then the equality becomes
$$2(g(2x)-a(2x))=g(x)-ax+x$$
$$2g(2x)=g(x)+x(1+3a)$$
Therefore setting $a=-frac13$ the equality simplifies to
$$g(2x)=frac12g(x).$$
Now plugging zero gives $g(0)=0$. You can now prove by induction that for every $x$
$$
gleft(frac{x}{2^n}right)=2^ng(x).tag{1}
$$
If $g$ is not identically zero, say $g(x_0)neq 0$, then we find a contradiction. Indeed by continuity in zero (which is still true for $g$) $g(frac{x_0}{2^n})$ should converge to zero, while by $(1)$ it does not.
Therefore we conclude that $g$ must be identically zero, or equivalently $f(x)=frac13 x$.
answered Jan 22 at 21:06
DelDel
3,1221524
$endgroup$
add a comment |
$begingroup$
Let $g(x) = xf(x)$. We obtain
$$
g(2x) = g(x) +x^2.
$$ Since $limlimits_{xto 0}g(x)=g(0)=0$,
$$begin{eqnarray}
g(x)=g(x) -lim_{ntoinfty}g(2^{-n-1}x) &=&sum_{j=0}^infty g(2^{-j}x)-g(2^{-j-1}x)\
&=&sum_{j=0}^infty 2^{-2j-2}cdot x^2=frac{x^2}{3}.
end{eqnarray}$$ This gives $$f(x) =frac{g(x)}{x}=frac{x}{3}.$$
answered Jan 22 at 20:18
SongSong
17k21145
$endgroup$
add a comment |
$begingroup$
Let assume any required regularity for the moment.
Plugging $x=0$ give $2f(0)=f(0)$ thus $f(0)=0$.
Derivating gives $4f'(2x)=f'(x)+1$ then $8f''(2x)=f''(x)$
So let solve first $g(2x)=frac 18 g(x)$.
$g(2^n)=a/8^n$ let assume $g(x)=dfrac a{x^3}$
This would give $f(x)=dfrac ax+bx+c$ continuity in $0$ implies $a=0$ so $f(x)=bx+c$.
$f(0)=c=0$ so $f(x)=bx$.
$2f(2x)=4bx=f(x)+x=(b+1)xiff b=frac 13$ so $f(x)=dfrac x3$
Now that we have found what $f$ should look like, lets work by substitution.
Set $f(x)=dfrac x3g(x)$ then $2f(2x)=dfrac{2x}3g(2x)=dfrac x3g(x)+xiff 2g(2x)=g(x)+1$
Set $h(x)=g(x)-1$ then $2h(2x)+2=h(x)+2iff 2h(2x)=h(x)$
In particular $h(x)=frac 12h(frac x2)=cdotsfrac 1{2^n}h(frac x{2^n})to 0$ assuming $h$ is bounded in $0$.
But we need a little more than continuity of $f$ here to conclude $h=0$, we need info on $dfrac{f(x)}{x}$ at $0$, so since $f(0)=0$ we need derivability of $f$ in $0$.
Assuming this condition then $f(x)=dfrac x3$.
answered Jan 22 at 20:47
zwimzwim
12.5k831
$endgroup$
1
$begingroup$
Too all. Am I too restrictive ? Can we conclude assuming only continuity in $0$ and not derivability ?
$endgroup$
– zwim
Jan 22 at 20:48
$begingroup$
Doesn't the original post already give an argument using only continuity (once the point, which you have proved, about $f$ mapping $0$ to $0$ is resolved)?
$endgroup$
– LSpice
Jan 22 at 21:07
$begingroup$
In fact I should have substituted directly like Del did, it avoids the point f(x)/x. But the method is basically the same. I created an artificial problem.
$endgroup$
– zwim
Jan 22 at 22:07
add a comment |
$begingroup$
This recurrence equation is linear then
$$
f(x) = f_h(x)+f_p(x)
$$
such that
$$
a f_h(a x)-f_h(x) = 0\
a f_p(a x)-f_p(x) = x
$$
for the homogeneous equation we assume
$$
f_h(x) = frac Cx
$$
and then for the paticular we assume
$$
f_p(x) = frac{C(x)}{x}
$$
then
$$
afrac{C(a x)}{a x}-frac{C(x)}{x} = x
$$
or
$$
C(a x)-C(x) = x^2
$$
for this last recurrence equation we choose
$$
C(x) = frac{x^2}{a^2-1}
$$
so the final solution is
$$
f(x) = frac{C}{x}+frac{x}{a^2-1}
$$
in our case $a = 2$ then
$$
f(x) = frac Cx+frac x3
$$
and to assure continuity at $x=0$ we choose $C = 0$ so the final result is
$$
f(x) = frac x3
$$
answered Jan 22 at 22:10
CesareoCesareo
9,3063517
$endgroup$
$begingroup$
The function is required to be continuous at $x = 0$, so you'd better have $C = 0$.
$endgroup$
– LSpice
Jan 22 at 23:58
$begingroup$
@LSpice Yes. I will fix that accordingly. Thanks.
$endgroup$
– Cesareo
Jan 23 at 0:01
add a comment |
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4 Answers
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active
oldest
votes
4 Answers
4
active
oldest
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active
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active
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$begingroup$
A powerful method to solve these kinds of problems is to reduce to a simpler equation. In this case we want to eliminate the $x$ in the right hand side. Set $g(x)=f(x)+ax$, with $a$ to be found later. Note that $f$ is continuous if and only if $g$ is. Then the equality becomes
$$2(g(2x)-a(2x))=g(x)-ax+x$$
$$2g(2x)=g(x)+x(1+3a)$$
Therefore setting $a=-frac13$ the equality simplifies to
$$g(2x)=frac12g(x).$$
Now plugging zero gives $g(0)=0$. You can now prove by induction that for every $x$
$$
gleft(frac{x}{2^n}right)=2^ng(x).tag{1}
$$
If $g$ is not identically zero, say $g(x_0)neq 0$, then we find a contradiction. Indeed by continuity in zero (which is still true for $g$) $g(frac{x_0}{2^n})$ should converge to zero, while by $(1)$ it does not.
Therefore we conclude that $g$ must be identically zero, or equivalently $f(x)=frac13 x$.
answered Jan 22 at 21:06
DelDel
3,1221524
$endgroup$
add a comment |
$begingroup$
A powerful method to solve these kinds of problems is to reduce to a simpler equation. In this case we want to eliminate the $x$ in the right hand side. Set $g(x)=f(x)+ax$, with $a$ to be found later. Note that $f$ is continuous if and only if $g$ is. Then the equality becomes
$$2(g(2x)-a(2x))=g(x)-ax+x$$
$$2g(2x)=g(x)+x(1+3a)$$
Therefore setting $a=-frac13$ the equality simplifies to
$$g(2x)=frac12g(x).$$
Now plugging zero gives $g(0)=0$. You can now prove by induction that for every $x$
$$
gleft(frac{x}{2^n}right)=2^ng(x).tag{1}
$$
If $g$ is not identically zero, say $g(x_0)neq 0$, then we find a contradiction. Indeed by continuity in zero (which is still true for $g$) $g(frac{x_0}{2^n})$ should converge to zero, while by $(1)$ it does not.
Therefore we conclude that $g$ must be identically zero, or equivalently $f(x)=frac13 x$.
answered Jan 22 at 21:06
DelDel
3,1221524
$endgroup$
add a comment |
$begingroup$
A powerful method to solve these kinds of problems is to reduce to a simpler equation. In this case we want to eliminate the $x$ in the right hand side. Set $g(x)=f(x)+ax$, with $a$ to be found later. Note that $f$ is continuous if and only if $g$ is. Then the equality becomes
$$2(g(2x)-a(2x))=g(x)-ax+x$$
$$2g(2x)=g(x)+x(1+3a)$$
Therefore setting $a=-frac13$ the equality simplifies to
$$g(2x)=frac12g(x).$$
Now plugging zero gives $g(0)=0$. You can now prove by induction that for every $x$
$$
gleft(frac{x}{2^n}right)=2^ng(x).tag{1}
$$
If $g$ is not identically zero, say $g(x_0)neq 0$, then we find a contradiction. Indeed by continuity in zero (which is still true for $g$) $g(frac{x_0}{2^n})$ should converge to zero, while by $(1)$ it does not.
Therefore we conclude that $g$ must be identically zero, or equivalently $f(x)=frac13 x$.
answered Jan 22 at 21:06
DelDel
3,1221524
$endgroup$
A powerful method to solve these kinds of problems is to reduce to a simpler equation. In this case we want to eliminate the $x$ in the right hand side. Set $g(x)=f(x)+ax$, with $a$ to be found later. Note that $f$ is continuous if and only if $g$ is. Then the equality becomes
$$2(g(2x)-a(2x))=g(x)-ax+x$$
$$2g(2x)=g(x)+x(1+3a)$$
Therefore setting $a=-frac13$ the equality simplifies to
$$g(2x)=frac12g(x).$$
Now plugging zero gives $g(0)=0$. You can now prove by induction that for every $x$
$$
gleft(frac{x}{2^n}right)=2^ng(x).tag{1}
$$
If $g$ is not identically zero, say $g(x_0)neq 0$, then we find a contradiction. Indeed by continuity in zero (which is still true for $g$) $g(frac{x_0}{2^n})$ should converge to zero, while by $(1)$ it does not.
Therefore we conclude that $g$ must be identically zero, or equivalently $f(x)=frac13 x$.
answered Jan 22 at 21:06
DelDel
3,1221524
answered Jan 22 at 21:06
DelDel
3,1221524
answered Jan 22 at 21:06
DelDel
3,1221524
answered Jan 22 at 21:06
DelDel
3,1221524
3,1221524
add a comment |
add a comment |
$begingroup$
Let $g(x) = xf(x)$. We obtain
$$
g(2x) = g(x) +x^2.
$$ Since $limlimits_{xto 0}g(x)=g(0)=0$,
$$begin{eqnarray}
g(x)=g(x) -lim_{ntoinfty}g(2^{-n-1}x) &=&sum_{j=0}^infty g(2^{-j}x)-g(2^{-j-1}x)\
&=&sum_{j=0}^infty 2^{-2j-2}cdot x^2=frac{x^2}{3}.
end{eqnarray}$$ This gives $$f(x) =frac{g(x)}{x}=frac{x}{3}.$$
answered Jan 22 at 20:18
SongSong
17k21145
$endgroup$
add a comment |
$begingroup$
Let $g(x) = xf(x)$. We obtain
$$
g(2x) = g(x) +x^2.
$$ Since $limlimits_{xto 0}g(x)=g(0)=0$,
$$begin{eqnarray}
g(x)=g(x) -lim_{ntoinfty}g(2^{-n-1}x) &=&sum_{j=0}^infty g(2^{-j}x)-g(2^{-j-1}x)\
&=&sum_{j=0}^infty 2^{-2j-2}cdot x^2=frac{x^2}{3}.
end{eqnarray}$$ This gives $$f(x) =frac{g(x)}{x}=frac{x}{3}.$$
answered Jan 22 at 20:18
SongSong
17k21145
$endgroup$
add a comment |
$begingroup$
Let $g(x) = xf(x)$. We obtain
$$
g(2x) = g(x) +x^2.
$$ Since $limlimits_{xto 0}g(x)=g(0)=0$,
$$begin{eqnarray}
g(x)=g(x) -lim_{ntoinfty}g(2^{-n-1}x) &=&sum_{j=0}^infty g(2^{-j}x)-g(2^{-j-1}x)\
&=&sum_{j=0}^infty 2^{-2j-2}cdot x^2=frac{x^2}{3}.
end{eqnarray}$$ This gives $$f(x) =frac{g(x)}{x}=frac{x}{3}.$$
answered Jan 22 at 20:18
SongSong
17k21145
$endgroup$
Let $g(x) = xf(x)$. We obtain
$$
g(2x) = g(x) +x^2.
$$ Since $limlimits_{xto 0}g(x)=g(0)=0$,
$$begin{eqnarray}
g(x)=g(x) -lim_{ntoinfty}g(2^{-n-1}x) &=&sum_{j=0}^infty g(2^{-j}x)-g(2^{-j-1}x)\
&=&sum_{j=0}^infty 2^{-2j-2}cdot x^2=frac{x^2}{3}.
end{eqnarray}$$ This gives $$f(x) =frac{g(x)}{x}=frac{x}{3}.$$
answered Jan 22 at 20:18
SongSong
17k21145
answered Jan 22 at 20:18
SongSong
17k21145
answered Jan 22 at 20:18
SongSong
17k21145
answered Jan 22 at 20:18
SongSong
17k21145
17k21145
add a comment |
add a comment |
$begingroup$
Let assume any required regularity for the moment.
Plugging $x=0$ give $2f(0)=f(0)$ thus $f(0)=0$.
Derivating gives $4f'(2x)=f'(x)+1$ then $8f''(2x)=f''(x)$
So let solve first $g(2x)=frac 18 g(x)$.
$g(2^n)=a/8^n$ let assume $g(x)=dfrac a{x^3}$
This would give $f(x)=dfrac ax+bx+c$ continuity in $0$ implies $a=0$ so $f(x)=bx+c$.
$f(0)=c=0$ so $f(x)=bx$.
$2f(2x)=4bx=f(x)+x=(b+1)xiff b=frac 13$ so $f(x)=dfrac x3$
Now that we have found what $f$ should look like, lets work by substitution.
Set $f(x)=dfrac x3g(x)$ then $2f(2x)=dfrac{2x}3g(2x)=dfrac x3g(x)+xiff 2g(2x)=g(x)+1$
Set $h(x)=g(x)-1$ then $2h(2x)+2=h(x)+2iff 2h(2x)=h(x)$
In particular $h(x)=frac 12h(frac x2)=cdotsfrac 1{2^n}h(frac x{2^n})to 0$ assuming $h$ is bounded in $0$.
But we need a little more than continuity of $f$ here to conclude $h=0$, we need info on $dfrac{f(x)}{x}$ at $0$, so since $f(0)=0$ we need derivability of $f$ in $0$.
Assuming this condition then $f(x)=dfrac x3$.
answered Jan 22 at 20:47
zwimzwim
12.5k831
$endgroup$
1
$begingroup$
Too all. Am I too restrictive ? Can we conclude assuming only continuity in $0$ and not derivability ?
$endgroup$
– zwim
Jan 22 at 20:48
$begingroup$
Doesn't the original post already give an argument using only continuity (once the point, which you have proved, about $f$ mapping $0$ to $0$ is resolved)?
$endgroup$
– LSpice
Jan 22 at 21:07
$begingroup$
In fact I should have substituted directly like Del did, it avoids the point f(x)/x. But the method is basically the same. I created an artificial problem.
$endgroup$
– zwim
Jan 22 at 22:07
add a comment |
$begingroup$
Let assume any required regularity for the moment.
Plugging $x=0$ give $2f(0)=f(0)$ thus $f(0)=0$.
Derivating gives $4f'(2x)=f'(x)+1$ then $8f''(2x)=f''(x)$
So let solve first $g(2x)=frac 18 g(x)$.
$g(2^n)=a/8^n$ let assume $g(x)=dfrac a{x^3}$
This would give $f(x)=dfrac ax+bx+c$ continuity in $0$ implies $a=0$ so $f(x)=bx+c$.
$f(0)=c=0$ so $f(x)=bx$.
$2f(2x)=4bx=f(x)+x=(b+1)xiff b=frac 13$ so $f(x)=dfrac x3$
Now that we have found what $f$ should look like, lets work by substitution.
Set $f(x)=dfrac x3g(x)$ then $2f(2x)=dfrac{2x}3g(2x)=dfrac x3g(x)+xiff 2g(2x)=g(x)+1$
Set $h(x)=g(x)-1$ then $2h(2x)+2=h(x)+2iff 2h(2x)=h(x)$
In particular $h(x)=frac 12h(frac x2)=cdotsfrac 1{2^n}h(frac x{2^n})to 0$ assuming $h$ is bounded in $0$.
But we need a little more than continuity of $f$ here to conclude $h=0$, we need info on $dfrac{f(x)}{x}$ at $0$, so since $f(0)=0$ we need derivability of $f$ in $0$.
Assuming this condition then $f(x)=dfrac x3$.
answered Jan 22 at 20:47
zwimzwim
12.5k831
$endgroup$
1
$begingroup$
Too all. Am I too restrictive ? Can we conclude assuming only continuity in $0$ and not derivability ?
$endgroup$
– zwim
Jan 22 at 20:48
$begingroup$
Doesn't the original post already give an argument using only continuity (once the point, which you have proved, about $f$ mapping $0$ to $0$ is resolved)?
$endgroup$
– LSpice
Jan 22 at 21:07
$begingroup$
In fact I should have substituted directly like Del did, it avoids the point f(x)/x. But the method is basically the same. I created an artificial problem.
$endgroup$
– zwim
Jan 22 at 22:07
add a comment |
$begingroup$
Let assume any required regularity for the moment.
Plugging $x=0$ give $2f(0)=f(0)$ thus $f(0)=0$.
Derivating gives $4f'(2x)=f'(x)+1$ then $8f''(2x)=f''(x)$
So let solve first $g(2x)=frac 18 g(x)$.
$g(2^n)=a/8^n$ let assume $g(x)=dfrac a{x^3}$
This would give $f(x)=dfrac ax+bx+c$ continuity in $0$ implies $a=0$ so $f(x)=bx+c$.
$f(0)=c=0$ so $f(x)=bx$.
$2f(2x)=4bx=f(x)+x=(b+1)xiff b=frac 13$ so $f(x)=dfrac x3$
Now that we have found what $f$ should look like, lets work by substitution.
Set $f(x)=dfrac x3g(x)$ then $2f(2x)=dfrac{2x}3g(2x)=dfrac x3g(x)+xiff 2g(2x)=g(x)+1$
Set $h(x)=g(x)-1$ then $2h(2x)+2=h(x)+2iff 2h(2x)=h(x)$
In particular $h(x)=frac 12h(frac x2)=cdotsfrac 1{2^n}h(frac x{2^n})to 0$ assuming $h$ is bounded in $0$.
But we need a little more than continuity of $f$ here to conclude $h=0$, we need info on $dfrac{f(x)}{x}$ at $0$, so since $f(0)=0$ we need derivability of $f$ in $0$.
Assuming this condition then $f(x)=dfrac x3$.
answered Jan 22 at 20:47
zwimzwim
12.5k831
$endgroup$
Let assume any required regularity for the moment.
Plugging $x=0$ give $2f(0)=f(0)$ thus $f(0)=0$.
Derivating gives $4f'(2x)=f'(x)+1$ then $8f''(2x)=f''(x)$
So let solve first $g(2x)=frac 18 g(x)$.
$g(2^n)=a/8^n$ let assume $g(x)=dfrac a{x^3}$
This would give $f(x)=dfrac ax+bx+c$ continuity in $0$ implies $a=0$ so $f(x)=bx+c$.
$f(0)=c=0$ so $f(x)=bx$.
$2f(2x)=4bx=f(x)+x=(b+1)xiff b=frac 13$ so $f(x)=dfrac x3$
Now that we have found what $f$ should look like, lets work by substitution.
Set $f(x)=dfrac x3g(x)$ then $2f(2x)=dfrac{2x}3g(2x)=dfrac x3g(x)+xiff 2g(2x)=g(x)+1$
Set $h(x)=g(x)-1$ then $2h(2x)+2=h(x)+2iff 2h(2x)=h(x)$
In particular $h(x)=frac 12h(frac x2)=cdotsfrac 1{2^n}h(frac x{2^n})to 0$ assuming $h$ is bounded in $0$.
But we need a little more than continuity of $f$ here to conclude $h=0$, we need info on $dfrac{f(x)}{x}$ at $0$, so since $f(0)=0$ we need derivability of $f$ in $0$.
Assuming this condition then $f(x)=dfrac x3$.
answered Jan 22 at 20:47
zwimzwim
12.5k831
answered Jan 22 at 20:47
zwimzwim
12.5k831
answered Jan 22 at 20:47
zwimzwim
12.5k831
answered Jan 22 at 20:47
zwimzwim
12.5k831
12.5k831
1
$begingroup$
Too all. Am I too restrictive ? Can we conclude assuming only continuity in $0$ and not derivability ?
$endgroup$
– zwim
Jan 22 at 20:48
$begingroup$
Doesn't the original post already give an argument using only continuity (once the point, which you have proved, about $f$ mapping $0$ to $0$ is resolved)?
$endgroup$
– LSpice
Jan 22 at 21:07
$begingroup$
In fact I should have substituted directly like Del did, it avoids the point f(x)/x. But the method is basically the same. I created an artificial problem.
$endgroup$
– zwim
Jan 22 at 22:07
add a comment |
1
$begingroup$
Too all. Am I too restrictive ? Can we conclude assuming only continuity in $0$ and not derivability ?
$endgroup$
– zwim
Jan 22 at 20:48
$begingroup$
Doesn't the original post already give an argument using only continuity (once the point, which you have proved, about $f$ mapping $0$ to $0$ is resolved)?
$endgroup$
– LSpice
Jan 22 at 21:07
$begingroup$
In fact I should have substituted directly like Del did, it avoids the point f(x)/x. But the method is basically the same. I created an artificial problem.
$endgroup$
– zwim
Jan 22 at 22:07
1
1
$begingroup$
Too all. Am I too restrictive ? Can we conclude assuming only continuity in $0$ and not derivability ?
$endgroup$
– zwim
Jan 22 at 20:48
$begingroup$
Too all. Am I too restrictive ? Can we conclude assuming only continuity in $0$ and not derivability ?
$endgroup$
– zwim
Jan 22 at 20:48
$begingroup$
Doesn't the original post already give an argument using only continuity (once the point, which you have proved, about $f$ mapping $0$ to $0$ is resolved)?
$endgroup$
– LSpice
Jan 22 at 21:07
$begingroup$
Doesn't the original post already give an argument using only continuity (once the point, which you have proved, about $f$ mapping $0$ to $0$ is resolved)?
$endgroup$
– LSpice
Jan 22 at 21:07
$begingroup$
In fact I should have substituted directly like Del did, it avoids the point f(x)/x. But the method is basically the same. I created an artificial problem.
$endgroup$
– zwim
Jan 22 at 22:07
$begingroup$
In fact I should have substituted directly like Del did, it avoids the point f(x)/x. But the method is basically the same. I created an artificial problem.
$endgroup$
– zwim
Jan 22 at 22:07
add a comment |
$begingroup$
This recurrence equation is linear then
$$
f(x) = f_h(x)+f_p(x)
$$
such that
$$
a f_h(a x)-f_h(x) = 0\
a f_p(a x)-f_p(x) = x
$$
for the homogeneous equation we assume
$$
f_h(x) = frac Cx
$$
and then for the paticular we assume
$$
f_p(x) = frac{C(x)}{x}
$$
then
$$
afrac{C(a x)}{a x}-frac{C(x)}{x} = x
$$
or
$$
C(a x)-C(x) = x^2
$$
for this last recurrence equation we choose
$$
C(x) = frac{x^2}{a^2-1}
$$
so the final solution is
$$
f(x) = frac{C}{x}+frac{x}{a^2-1}
$$
in our case $a = 2$ then
$$
f(x) = frac Cx+frac x3
$$
and to assure continuity at $x=0$ we choose $C = 0$ so the final result is
$$
f(x) = frac x3
$$
answered Jan 22 at 22:10
CesareoCesareo
9,3063517
$endgroup$
$begingroup$
The function is required to be continuous at $x = 0$, so you'd better have $C = 0$.
$endgroup$
– LSpice
Jan 22 at 23:58
$begingroup$
@LSpice Yes. I will fix that accordingly. Thanks.
$endgroup$
– Cesareo
Jan 23 at 0:01
add a comment |
$begingroup$
This recurrence equation is linear then
$$
f(x) = f_h(x)+f_p(x)
$$
such that
$$
a f_h(a x)-f_h(x) = 0\
a f_p(a x)-f_p(x) = x
$$
for the homogeneous equation we assume
$$
f_h(x) = frac Cx
$$
and then for the paticular we assume
$$
f_p(x) = frac{C(x)}{x}
$$
then
$$
afrac{C(a x)}{a x}-frac{C(x)}{x} = x
$$
or
$$
C(a x)-C(x) = x^2
$$
for this last recurrence equation we choose
$$
C(x) = frac{x^2}{a^2-1}
$$
so the final solution is
$$
f(x) = frac{C}{x}+frac{x}{a^2-1}
$$
in our case $a = 2$ then
$$
f(x) = frac Cx+frac x3
$$
and to assure continuity at $x=0$ we choose $C = 0$ so the final result is
$$
f(x) = frac x3
$$
answered Jan 22 at 22:10
CesareoCesareo
9,3063517
$endgroup$
$begingroup$
The function is required to be continuous at $x = 0$, so you'd better have $C = 0$.
$endgroup$
– LSpice
Jan 22 at 23:58
$begingroup$
@LSpice Yes. I will fix that accordingly. Thanks.
$endgroup$
– Cesareo
Jan 23 at 0:01
add a comment |
$begingroup$
This recurrence equation is linear then
$$
f(x) = f_h(x)+f_p(x)
$$
such that
$$
a f_h(a x)-f_h(x) = 0\
a f_p(a x)-f_p(x) = x
$$
for the homogeneous equation we assume
$$
f_h(x) = frac Cx
$$
and then for the paticular we assume
$$
f_p(x) = frac{C(x)}{x}
$$
then
$$
afrac{C(a x)}{a x}-frac{C(x)}{x} = x
$$
or
$$
C(a x)-C(x) = x^2
$$
for this last recurrence equation we choose
$$
C(x) = frac{x^2}{a^2-1}
$$
so the final solution is
$$
f(x) = frac{C}{x}+frac{x}{a^2-1}
$$
in our case $a = 2$ then
$$
f(x) = frac Cx+frac x3
$$
and to assure continuity at $x=0$ we choose $C = 0$ so the final result is
$$
f(x) = frac x3
$$
answered Jan 22 at 22:10
CesareoCesareo
9,3063517
$endgroup$
This recurrence equation is linear then
$$
f(x) = f_h(x)+f_p(x)
$$
such that
$$
a f_h(a x)-f_h(x) = 0\
a f_p(a x)-f_p(x) = x
$$
for the homogeneous equation we assume
$$
f_h(x) = frac Cx
$$
and then for the paticular we assume
$$
f_p(x) = frac{C(x)}{x}
$$
then
$$
afrac{C(a x)}{a x}-frac{C(x)}{x} = x
$$
or
$$
C(a x)-C(x) = x^2
$$
for this last recurrence equation we choose
$$
C(x) = frac{x^2}{a^2-1}
$$
so the final solution is
$$
f(x) = frac{C}{x}+frac{x}{a^2-1}
$$
in our case $a = 2$ then
$$
f(x) = frac Cx+frac x3
$$
and to assure continuity at $x=0$ we choose $C = 0$ so the final result is
$$
f(x) = frac x3
$$
answered Jan 22 at 22:10
CesareoCesareo
9,3063517
edited Jan 23 at 0:02
answered Jan 22 at 22:10
CesareoCesareo
9,3063517
answered Jan 22 at 22:10
CesareoCesareo
9,3063517
answered Jan 22 at 22:10
CesareoCesareo
9,3063517
9,3063517
$begingroup$
The function is required to be continuous at $x = 0$, so you'd better have $C = 0$.
$endgroup$
– LSpice
Jan 22 at 23:58
$begingroup$
@LSpice Yes. I will fix that accordingly. Thanks.
$endgroup$
– Cesareo
Jan 23 at 0:01
add a comment |
$begingroup$
The function is required to be continuous at $x = 0$, so you'd better have $C = 0$.
$endgroup$
– LSpice
Jan 22 at 23:58
$begingroup$
@LSpice Yes. I will fix that accordingly. Thanks.
$endgroup$
– Cesareo
Jan 23 at 0:01
$begingroup$
The function is required to be continuous at $x = 0$, so you'd better have $C = 0$.
$endgroup$
– LSpice
Jan 22 at 23:58
$begingroup$
The function is required to be continuous at $x = 0$, so you'd better have $C = 0$.
$endgroup$
– LSpice
Jan 22 at 23:58
$begingroup$
@LSpice Yes. I will fix that accordingly. Thanks.
$endgroup$
– Cesareo
Jan 23 at 0:01
$begingroup$
@LSpice Yes. I will fix that accordingly. Thanks.
$endgroup$
– Cesareo
Jan 23 at 0:01
add a comment |
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$begingroup$
Are you sure you meant what you wrote? $2 f(x) = f(x) + x$ just says $f(x) = x$.
$endgroup$
– Robert Israel
Jan 22 at 19:57
$begingroup$
Are you sure of your functional equation?
$endgroup$
– lulu
Jan 22 at 19:57
3
$begingroup$
To get $f(0)=0$ just plug in $x=0$ into your equation
$endgroup$
– GReyes
Jan 22 at 20:00
1
$begingroup$
$f(0)=0$ because when you set $x=0$ in your (corrected) functional equation you get $2f(0)=f(0)$.
$endgroup$
– Dog_69
Jan 22 at 20:01
1
$begingroup$
You need continuity to conclude that, since $y/2^nto 0$ as $ntoinfty$, then $f(y/2^n)to 0$ as well (for any $y$). The only solution is $f(x)=x/3$
$endgroup$
– GReyes
Jan 22 at 20:35