Mixing
The functional equation $f(x)+f(y)+f(z)+f(x+y+z)=f(x+y)+f(y+z)+f(z+x)$
$begingroup$
How to find the all continuous functions $fcolon mathbb{R}tomathbb{R}$ such that $f(x)+f(y)+f(z)+f(x+y+z)=f(x+y)+f(y+z)+f(z+x)$
calculus functional-equations
edited Aug 23 '12 at 20:55
MJD
47.4k29214396
asked Aug 23 '12 at 19:49
bibbib
562
$endgroup$
add a comment |
$begingroup$
How to find the all continuous functions $fcolon mathbb{R}tomathbb{R}$ such that $f(x)+f(y)+f(z)+f(x+y+z)=f(x+y)+f(y+z)+f(z+x)$
calculus functional-equations
edited Aug 23 '12 at 20:55
MJD
47.4k29214396
asked Aug 23 '12 at 19:49
bibbib
562
$endgroup$
3
$begingroup$
It's easy to see that $f(0)=0$. The solutions form a linear space, and include $f(x)=x$ and $f(x) = x^2$ but no other powers. I would guess that $f(x) = ax + bx^2$ is the general solution.
$endgroup$
– Robert Israel
Aug 23 '12 at 20:12
3
$begingroup$
A more descriptive title would be useful: as it stands, it is almost unconnected to the question!
$endgroup$
– Mariano Suárez-Álvarez
Aug 23 '12 at 20:25
3
$begingroup$
Setting $x=y=z$ and differentiating the equality many times and evaluating at zero shows Robert's solutions are all the entire ones.
$endgroup$
– Mariano Suárez-Álvarez
Aug 23 '12 at 20:31
$begingroup$
@MarianoSuárez-Alvarez Is it obvious then any continuous function satisfying that equality is differentiable?
$endgroup$
– JSchlather
Aug 23 '12 at 20:38
$begingroup$
If it were obvious I would have said «Robert's solutions are all solutions» and not what I wrote!
$endgroup$
– Mariano Suárez-Álvarez
Aug 23 '12 at 20:39
add a comment |
$begingroup$
How to find the all continuous functions $fcolon mathbb{R}tomathbb{R}$ such that $f(x)+f(y)+f(z)+f(x+y+z)=f(x+y)+f(y+z)+f(z+x)$
calculus functional-equations
edited Aug 23 '12 at 20:55
MJD
47.4k29214396
asked Aug 23 '12 at 19:49
bibbib
562
$endgroup$
How to find the all continuous functions $fcolon mathbb{R}tomathbb{R}$ such that $f(x)+f(y)+f(z)+f(x+y+z)=f(x+y)+f(y+z)+f(z+x)$
calculus functional-equations
calculus functional-equations
edited Aug 23 '12 at 20:55
MJD
47.4k29214396
asked Aug 23 '12 at 19:49
bibbib
562
edited Aug 23 '12 at 20:55
MJD
47.4k29214396
asked Aug 23 '12 at 19:49
bibbib
562
edited Aug 23 '12 at 20:55
MJD
47.4k29214396
edited Aug 23 '12 at 20:55
MJD
47.4k29214396
edited Aug 23 '12 at 20:55
MJD
47.4k29214396
47.4k29214396
asked Aug 23 '12 at 19:49
bibbib
562
asked Aug 23 '12 at 19:49
bibbib
562
asked Aug 23 '12 at 19:49
bibbib
562
562
3
$begingroup$
It's easy to see that $f(0)=0$. The solutions form a linear space, and include $f(x)=x$ and $f(x) = x^2$ but no other powers. I would guess that $f(x) = ax + bx^2$ is the general solution.
$endgroup$
– Robert Israel
Aug 23 '12 at 20:12
3
$begingroup$
A more descriptive title would be useful: as it stands, it is almost unconnected to the question!
$endgroup$
– Mariano Suárez-Álvarez
Aug 23 '12 at 20:25
3
$begingroup$
Setting $x=y=z$ and differentiating the equality many times and evaluating at zero shows Robert's solutions are all the entire ones.
$endgroup$
– Mariano Suárez-Álvarez
Aug 23 '12 at 20:31
$begingroup$
@MarianoSuárez-Alvarez Is it obvious then any continuous function satisfying that equality is differentiable?
$endgroup$
– JSchlather
Aug 23 '12 at 20:38
$begingroup$
If it were obvious I would have said «Robert's solutions are all solutions» and not what I wrote!
$endgroup$
– Mariano Suárez-Álvarez
Aug 23 '12 at 20:39
add a comment |
3
$begingroup$
It's easy to see that $f(0)=0$. The solutions form a linear space, and include $f(x)=x$ and $f(x) = x^2$ but no other powers. I would guess that $f(x) = ax + bx^2$ is the general solution.
$endgroup$
– Robert Israel
Aug 23 '12 at 20:12
3
$begingroup$
A more descriptive title would be useful: as it stands, it is almost unconnected to the question!
$endgroup$
– Mariano Suárez-Álvarez
Aug 23 '12 at 20:25
3
$begingroup$
Setting $x=y=z$ and differentiating the equality many times and evaluating at zero shows Robert's solutions are all the entire ones.
$endgroup$
– Mariano Suárez-Álvarez
Aug 23 '12 at 20:31
$begingroup$
@MarianoSuárez-Alvarez Is it obvious then any continuous function satisfying that equality is differentiable?
$endgroup$
– JSchlather
Aug 23 '12 at 20:38
$begingroup$
If it were obvious I would have said «Robert's solutions are all solutions» and not what I wrote!
$endgroup$
– Mariano Suárez-Álvarez
Aug 23 '12 at 20:39
3
3
$begingroup$
It's easy to see that $f(0)=0$. The solutions form a linear space, and include $f(x)=x$ and $f(x) = x^2$ but no other powers. I would guess that $f(x) = ax + bx^2$ is the general solution.
$endgroup$
– Robert Israel
Aug 23 '12 at 20:12
$begingroup$
It's easy to see that $f(0)=0$. The solutions form a linear space, and include $f(x)=x$ and $f(x) = x^2$ but no other powers. I would guess that $f(x) = ax + bx^2$ is the general solution.
$endgroup$
– Robert Israel
Aug 23 '12 at 20:12
3
3
$begingroup$
A more descriptive title would be useful: as it stands, it is almost unconnected to the question!
$endgroup$
– Mariano Suárez-Álvarez
Aug 23 '12 at 20:25
$begingroup$
A more descriptive title would be useful: as it stands, it is almost unconnected to the question!
$endgroup$
– Mariano Suárez-Álvarez
Aug 23 '12 at 20:25
3
3
$begingroup$
Setting $x=y=z$ and differentiating the equality many times and evaluating at zero shows Robert's solutions are all the entire ones.
$endgroup$
– Mariano Suárez-Álvarez
Aug 23 '12 at 20:31
$begingroup$
Setting $x=y=z$ and differentiating the equality many times and evaluating at zero shows Robert's solutions are all the entire ones.
$endgroup$
– Mariano Suárez-Álvarez
Aug 23 '12 at 20:31
$begingroup$
@MarianoSuárez-Alvarez Is it obvious then any continuous function satisfying that equality is differentiable?
$endgroup$
– JSchlather
Aug 23 '12 at 20:38
$begingroup$
@MarianoSuárez-Alvarez Is it obvious then any continuous function satisfying that equality is differentiable?
$endgroup$
– JSchlather
Aug 23 '12 at 20:38
$begingroup$
If it were obvious I would have said «Robert's solutions are all solutions» and not what I wrote!
$endgroup$
– Mariano Suárez-Álvarez
Aug 23 '12 at 20:39
$begingroup$
If it were obvious I would have said «Robert's solutions are all solutions» and not what I wrote!
$endgroup$
– Mariano Suárez-Álvarez
Aug 23 '12 at 20:39
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
We first use divided differences to show that $f$ is three times differentiable. In fact, continuity is not needed. Let
$$
delta_h f(x)=f(x+h)-f(x).
$$
Then writing the functional equation with $x$ replaced by $x+h$ and subtracting the original equation from it, we derive
$$
delta_hf(x)+delta_hf(x+y+z)=delta_hf(x+y)+delta_hf(z+x).
$$
Now doing the same with $y$, and subsequently with $z$, we get
$$
delta_elldelta_kdelta_hf(x)=0,
$$
for all $h,k,ellinmathbb{R}$ and $xinmathbb{R}$. In particular, the following limit exists
$$
delta_kdelta_hf'(x)=lim_{ellto0}frac{delta_elldelta_kdelta_hf(x)}{ell}=0.
$$
Proceeding similarly, we conclude that $f'''(x)$ exists and equal to $0$ everywhere.
This means that $f$ is of the form
$$
f(x)=ax^2+bx+c.
$$
We then simply check if each of $1,x$, and $x^2$ satisfies the functional equation, and conclude that the general solution is
$$
f(x)=ax^2+bx.
$$
answered Aug 23 '12 at 21:07
timurtimur
12k2143
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add a comment |
$begingroup$
All continuous solutions are of the form $f(x) = a x + b x^2$.
For given $z$, let $g(x) = f(x+z) - f(x) - f(z)$. The functional equation becomes
$g(x+y) = g(x) + g(y)$. This is the Cauchy functional equation, and it is known that its
only continuous solutions are $g(x) = c x$. Of course $c$ can depend on $z$. Now we need to solve
$f(x+z) - f(x) - f(z) = c(z) x$. Since the left side is symmetric in $x$ and $z$, $c(z) x = c(x) z$, so $c(z) = k z$ for some constant $k$.
Taking $z=-x$, and using $f(0)=0$, we get $- f(x) - f(-x) = - k x^2$, or $f(x) + f(-x) = k x^2$.
Note that if $f(x)$ is a solution of our equation, so is $f(-x)$, and by linearity so are the even and odd parts $ (f(x) + f(-x))/2$ and $(f(x) - f(-x))/2$. Thus it suffices to consider the two cases $f$ even and $f$ odd.
If $f$ is even,
$f(x) + f(-x) = 2 f(x) = k x^2$, so $f(x) = (k/2) x^2$.
If $f$ is odd, $f(x) + f(-x) = 0$ so $k=0$. now we have $f(x+z) - f(x) - f(z) = 0$, which is again Cauchy's functional equation, and so $f(x) = a x$ for some constant $a$.
answered Aug 23 '12 at 21:17
Robert IsraelRobert Israel
324k23214468
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add a comment |
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Consider the left and right as functions of three variables.
- Take $partial / partial x$ of both sides (keeping track of the chain rule when necessary):
$$ f'(x) + f'(x+y+z) = f'(x+y) + f'(z+x) $$
- Take $partial / partial y$ of both sides of the result:
$$ f''(x+y+z) = f''(x + y) $$
- Take $partial / partial z$ of both sides of that:
$$ f'''(x+y+z) = 0 $$
Now substitute $y=z=0$ to find that $f'''(x)=0$. This shows that the only possible functions would be of the form $f(x) = ax^2 + bx + c$ for constants $a, b, c$. Then apply Robert's observation (see comments above) that $f(0) = 0$ (hence $c=0$), and that the set of functions form a linear space (hence all choices of $a, b in mathbb{R}$ are valid).
Hope this helps!
answered Aug 23 '12 at 20:33
Shaun AultShaun Ault
7,9021828
$endgroup$
$begingroup$
I just saw Mariano's comment. :)
$endgroup$
– Shaun Ault
Aug 23 '12 at 20:34
3
$begingroup$
And what if $f$ is not differentiable?
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– Chris Eagle
Aug 23 '12 at 20:37
add a comment |
$begingroup$
Let $f: mathbb R to mathbb R$ be a continous solution of the functional equation and let $g(x) := ax^2+bx$ where $a$ and $b$ are chosen in such a way that $f(-1) = g(-1)$ and $f(1) = g(1)$. Then $tilde f := f-g$ is also a solution and we have $tilde f(-1) = tilde f(1) = 0$. We will show $tilde f equiv 0$, so that $f = g$.
By replacing $f$ with $tilde f$ way may assume that $f$ is a solution with $f(-1) = f(1) = 0$, and we will show $f equiv 0$. Let $Z = {x in mathbb R: f(x) = 0 }$. A priori we have ${-1,+1} subseteq Z$. Plugging $(1,-1,x)$ into the functional equation yields
$$f(x) = frac{f(x+1) + f(x-1)}{2},$$
which implies $mathbb Z subseteq Z$. If we can show
$$2x,x in Z Rightarrow frac{x}{2} in Z$$
we are finished since then every dyadic rational $a/2^n$ is in $Z$ and these are dense in $mathbb R$ (here the continuity of $f$ is needed). So let $2x in Z$ and $x in Z$. Plug $(frac{x}{2},frac{x}{2},x)$ into the functional equation to get
$$2fleft(frac{x}{2}right) + f(x) + f(2x) = f(x) + 2fleft(frac{3}{2}xright),$$
hence $fleft(frac{x}{2}right) = fleft(frac{3}{2}xright)$. Then plugging in $(frac{x}{2},frac{x}{2},frac{x}{2})$ yields
$$3fleft(frac{x}{2}right) + fleft(frac{3}{2}xright) = 3f(x),$$
hence $f(frac{x}{2}) = 0$ and $frac{x}{2} in Z$.
answered Aug 23 '12 at 22:43
marlumarlu
11k2041
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add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
We first use divided differences to show that $f$ is three times differentiable. In fact, continuity is not needed. Let
$$
delta_h f(x)=f(x+h)-f(x).
$$
Then writing the functional equation with $x$ replaced by $x+h$ and subtracting the original equation from it, we derive
$$
delta_hf(x)+delta_hf(x+y+z)=delta_hf(x+y)+delta_hf(z+x).
$$
Now doing the same with $y$, and subsequently with $z$, we get
$$
delta_elldelta_kdelta_hf(x)=0,
$$
for all $h,k,ellinmathbb{R}$ and $xinmathbb{R}$. In particular, the following limit exists
$$
delta_kdelta_hf'(x)=lim_{ellto0}frac{delta_elldelta_kdelta_hf(x)}{ell}=0.
$$
Proceeding similarly, we conclude that $f'''(x)$ exists and equal to $0$ everywhere.
This means that $f$ is of the form
$$
f(x)=ax^2+bx+c.
$$
We then simply check if each of $1,x$, and $x^2$ satisfies the functional equation, and conclude that the general solution is
$$
f(x)=ax^2+bx.
$$
answered Aug 23 '12 at 21:07
timurtimur
12k2143
$endgroup$
add a comment |
$begingroup$
We first use divided differences to show that $f$ is three times differentiable. In fact, continuity is not needed. Let
$$
delta_h f(x)=f(x+h)-f(x).
$$
Then writing the functional equation with $x$ replaced by $x+h$ and subtracting the original equation from it, we derive
$$
delta_hf(x)+delta_hf(x+y+z)=delta_hf(x+y)+delta_hf(z+x).
$$
Now doing the same with $y$, and subsequently with $z$, we get
$$
delta_elldelta_kdelta_hf(x)=0,
$$
for all $h,k,ellinmathbb{R}$ and $xinmathbb{R}$. In particular, the following limit exists
$$
delta_kdelta_hf'(x)=lim_{ellto0}frac{delta_elldelta_kdelta_hf(x)}{ell}=0.
$$
Proceeding similarly, we conclude that $f'''(x)$ exists and equal to $0$ everywhere.
This means that $f$ is of the form
$$
f(x)=ax^2+bx+c.
$$
We then simply check if each of $1,x$, and $x^2$ satisfies the functional equation, and conclude that the general solution is
$$
f(x)=ax^2+bx.
$$
answered Aug 23 '12 at 21:07
timurtimur
12k2143
$endgroup$
add a comment |
$begingroup$
We first use divided differences to show that $f$ is three times differentiable. In fact, continuity is not needed. Let
$$
delta_h f(x)=f(x+h)-f(x).
$$
Then writing the functional equation with $x$ replaced by $x+h$ and subtracting the original equation from it, we derive
$$
delta_hf(x)+delta_hf(x+y+z)=delta_hf(x+y)+delta_hf(z+x).
$$
Now doing the same with $y$, and subsequently with $z$, we get
$$
delta_elldelta_kdelta_hf(x)=0,
$$
for all $h,k,ellinmathbb{R}$ and $xinmathbb{R}$. In particular, the following limit exists
$$
delta_kdelta_hf'(x)=lim_{ellto0}frac{delta_elldelta_kdelta_hf(x)}{ell}=0.
$$
Proceeding similarly, we conclude that $f'''(x)$ exists and equal to $0$ everywhere.
This means that $f$ is of the form
$$
f(x)=ax^2+bx+c.
$$
We then simply check if each of $1,x$, and $x^2$ satisfies the functional equation, and conclude that the general solution is
$$
f(x)=ax^2+bx.
$$
answered Aug 23 '12 at 21:07
timurtimur
12k2143
$endgroup$
We first use divided differences to show that $f$ is three times differentiable. In fact, continuity is not needed. Let
$$
delta_h f(x)=f(x+h)-f(x).
$$
Then writing the functional equation with $x$ replaced by $x+h$ and subtracting the original equation from it, we derive
$$
delta_hf(x)+delta_hf(x+y+z)=delta_hf(x+y)+delta_hf(z+x).
$$
Now doing the same with $y$, and subsequently with $z$, we get
$$
delta_elldelta_kdelta_hf(x)=0,
$$
for all $h,k,ellinmathbb{R}$ and $xinmathbb{R}$. In particular, the following limit exists
$$
delta_kdelta_hf'(x)=lim_{ellto0}frac{delta_elldelta_kdelta_hf(x)}{ell}=0.
$$
Proceeding similarly, we conclude that $f'''(x)$ exists and equal to $0$ everywhere.
This means that $f$ is of the form
$$
f(x)=ax^2+bx+c.
$$
We then simply check if each of $1,x$, and $x^2$ satisfies the functional equation, and conclude that the general solution is
$$
f(x)=ax^2+bx.
$$
answered Aug 23 '12 at 21:07
timurtimur
12k2143
edited Jan 17 at 6:03
answered Aug 23 '12 at 21:07
timurtimur
12k2143
answered Aug 23 '12 at 21:07
timurtimur
12k2143
answered Aug 23 '12 at 21:07
timurtimur
12k2143
12k2143
add a comment |
add a comment |
$begingroup$
All continuous solutions are of the form $f(x) = a x + b x^2$.
For given $z$, let $g(x) = f(x+z) - f(x) - f(z)$. The functional equation becomes
$g(x+y) = g(x) + g(y)$. This is the Cauchy functional equation, and it is known that its
only continuous solutions are $g(x) = c x$. Of course $c$ can depend on $z$. Now we need to solve
$f(x+z) - f(x) - f(z) = c(z) x$. Since the left side is symmetric in $x$ and $z$, $c(z) x = c(x) z$, so $c(z) = k z$ for some constant $k$.
Taking $z=-x$, and using $f(0)=0$, we get $- f(x) - f(-x) = - k x^2$, or $f(x) + f(-x) = k x^2$.
Note that if $f(x)$ is a solution of our equation, so is $f(-x)$, and by linearity so are the even and odd parts $ (f(x) + f(-x))/2$ and $(f(x) - f(-x))/2$. Thus it suffices to consider the two cases $f$ even and $f$ odd.
If $f$ is even,
$f(x) + f(-x) = 2 f(x) = k x^2$, so $f(x) = (k/2) x^2$.
If $f$ is odd, $f(x) + f(-x) = 0$ so $k=0$. now we have $f(x+z) - f(x) - f(z) = 0$, which is again Cauchy's functional equation, and so $f(x) = a x$ for some constant $a$.
answered Aug 23 '12 at 21:17
Robert IsraelRobert Israel
324k23214468
$endgroup$
add a comment |
$begingroup$
All continuous solutions are of the form $f(x) = a x + b x^2$.
For given $z$, let $g(x) = f(x+z) - f(x) - f(z)$. The functional equation becomes
$g(x+y) = g(x) + g(y)$. This is the Cauchy functional equation, and it is known that its
only continuous solutions are $g(x) = c x$. Of course $c$ can depend on $z$. Now we need to solve
$f(x+z) - f(x) - f(z) = c(z) x$. Since the left side is symmetric in $x$ and $z$, $c(z) x = c(x) z$, so $c(z) = k z$ for some constant $k$.
Taking $z=-x$, and using $f(0)=0$, we get $- f(x) - f(-x) = - k x^2$, or $f(x) + f(-x) = k x^2$.
Note that if $f(x)$ is a solution of our equation, so is $f(-x)$, and by linearity so are the even and odd parts $ (f(x) + f(-x))/2$ and $(f(x) - f(-x))/2$. Thus it suffices to consider the two cases $f$ even and $f$ odd.
If $f$ is even,
$f(x) + f(-x) = 2 f(x) = k x^2$, so $f(x) = (k/2) x^2$.
If $f$ is odd, $f(x) + f(-x) = 0$ so $k=0$. now we have $f(x+z) - f(x) - f(z) = 0$, which is again Cauchy's functional equation, and so $f(x) = a x$ for some constant $a$.
answered Aug 23 '12 at 21:17
Robert IsraelRobert Israel
324k23214468
$endgroup$
add a comment |
$begingroup$
All continuous solutions are of the form $f(x) = a x + b x^2$.
For given $z$, let $g(x) = f(x+z) - f(x) - f(z)$. The functional equation becomes
$g(x+y) = g(x) + g(y)$. This is the Cauchy functional equation, and it is known that its
only continuous solutions are $g(x) = c x$. Of course $c$ can depend on $z$. Now we need to solve
$f(x+z) - f(x) - f(z) = c(z) x$. Since the left side is symmetric in $x$ and $z$, $c(z) x = c(x) z$, so $c(z) = k z$ for some constant $k$.
Taking $z=-x$, and using $f(0)=0$, we get $- f(x) - f(-x) = - k x^2$, or $f(x) + f(-x) = k x^2$.
Note that if $f(x)$ is a solution of our equation, so is $f(-x)$, and by linearity so are the even and odd parts $ (f(x) + f(-x))/2$ and $(f(x) - f(-x))/2$. Thus it suffices to consider the two cases $f$ even and $f$ odd.
If $f$ is even,
$f(x) + f(-x) = 2 f(x) = k x^2$, so $f(x) = (k/2) x^2$.
If $f$ is odd, $f(x) + f(-x) = 0$ so $k=0$. now we have $f(x+z) - f(x) - f(z) = 0$, which is again Cauchy's functional equation, and so $f(x) = a x$ for some constant $a$.
answered Aug 23 '12 at 21:17
Robert IsraelRobert Israel
324k23214468
$endgroup$
All continuous solutions are of the form $f(x) = a x + b x^2$.
For given $z$, let $g(x) = f(x+z) - f(x) - f(z)$. The functional equation becomes
$g(x+y) = g(x) + g(y)$. This is the Cauchy functional equation, and it is known that its
only continuous solutions are $g(x) = c x$. Of course $c$ can depend on $z$. Now we need to solve
$f(x+z) - f(x) - f(z) = c(z) x$. Since the left side is symmetric in $x$ and $z$, $c(z) x = c(x) z$, so $c(z) = k z$ for some constant $k$.
Taking $z=-x$, and using $f(0)=0$, we get $- f(x) - f(-x) = - k x^2$, or $f(x) + f(-x) = k x^2$.
Note that if $f(x)$ is a solution of our equation, so is $f(-x)$, and by linearity so are the even and odd parts $ (f(x) + f(-x))/2$ and $(f(x) - f(-x))/2$. Thus it suffices to consider the two cases $f$ even and $f$ odd.
If $f$ is even,
$f(x) + f(-x) = 2 f(x) = k x^2$, so $f(x) = (k/2) x^2$.
If $f$ is odd, $f(x) + f(-x) = 0$ so $k=0$. now we have $f(x+z) - f(x) - f(z) = 0$, which is again Cauchy's functional equation, and so $f(x) = a x$ for some constant $a$.
answered Aug 23 '12 at 21:17
Robert IsraelRobert Israel
324k23214468
answered Aug 23 '12 at 21:17
Robert IsraelRobert Israel
324k23214468
answered Aug 23 '12 at 21:17
Robert IsraelRobert Israel
324k23214468
answered Aug 23 '12 at 21:17
Robert IsraelRobert Israel
324k23214468
324k23214468
add a comment |
add a comment |
$begingroup$
Consider the left and right as functions of three variables.
- Take $partial / partial x$ of both sides (keeping track of the chain rule when necessary):
$$ f'(x) + f'(x+y+z) = f'(x+y) + f'(z+x) $$
- Take $partial / partial y$ of both sides of the result:
$$ f''(x+y+z) = f''(x + y) $$
- Take $partial / partial z$ of both sides of that:
$$ f'''(x+y+z) = 0 $$
Now substitute $y=z=0$ to find that $f'''(x)=0$. This shows that the only possible functions would be of the form $f(x) = ax^2 + bx + c$ for constants $a, b, c$. Then apply Robert's observation (see comments above) that $f(0) = 0$ (hence $c=0$), and that the set of functions form a linear space (hence all choices of $a, b in mathbb{R}$ are valid).
Hope this helps!
answered Aug 23 '12 at 20:33
Shaun AultShaun Ault
7,9021828
$endgroup$
$begingroup$
I just saw Mariano's comment. :)
$endgroup$
– Shaun Ault
Aug 23 '12 at 20:34
3
$begingroup$
And what if $f$ is not differentiable?
$endgroup$
– Chris Eagle
Aug 23 '12 at 20:37
add a comment |
$begingroup$
Consider the left and right as functions of three variables.
- Take $partial / partial x$ of both sides (keeping track of the chain rule when necessary):
$$ f'(x) + f'(x+y+z) = f'(x+y) + f'(z+x) $$
- Take $partial / partial y$ of both sides of the result:
$$ f''(x+y+z) = f''(x + y) $$
- Take $partial / partial z$ of both sides of that:
$$ f'''(x+y+z) = 0 $$
Now substitute $y=z=0$ to find that $f'''(x)=0$. This shows that the only possible functions would be of the form $f(x) = ax^2 + bx + c$ for constants $a, b, c$. Then apply Robert's observation (see comments above) that $f(0) = 0$ (hence $c=0$), and that the set of functions form a linear space (hence all choices of $a, b in mathbb{R}$ are valid).
Hope this helps!
answered Aug 23 '12 at 20:33
Shaun AultShaun Ault
7,9021828
$endgroup$
$begingroup$
I just saw Mariano's comment. :)
$endgroup$
– Shaun Ault
Aug 23 '12 at 20:34
3
$begingroup$
And what if $f$ is not differentiable?
$endgroup$
– Chris Eagle
Aug 23 '12 at 20:37
add a comment |
$begingroup$
Consider the left and right as functions of three variables.
- Take $partial / partial x$ of both sides (keeping track of the chain rule when necessary):
$$ f'(x) + f'(x+y+z) = f'(x+y) + f'(z+x) $$
- Take $partial / partial y$ of both sides of the result:
$$ f''(x+y+z) = f''(x + y) $$
- Take $partial / partial z$ of both sides of that:
$$ f'''(x+y+z) = 0 $$
Now substitute $y=z=0$ to find that $f'''(x)=0$. This shows that the only possible functions would be of the form $f(x) = ax^2 + bx + c$ for constants $a, b, c$. Then apply Robert's observation (see comments above) that $f(0) = 0$ (hence $c=0$), and that the set of functions form a linear space (hence all choices of $a, b in mathbb{R}$ are valid).
Hope this helps!
answered Aug 23 '12 at 20:33
Shaun AultShaun Ault
7,9021828
$endgroup$
Consider the left and right as functions of three variables.
- Take $partial / partial x$ of both sides (keeping track of the chain rule when necessary):
$$ f'(x) + f'(x+y+z) = f'(x+y) + f'(z+x) $$
- Take $partial / partial y$ of both sides of the result:
$$ f''(x+y+z) = f''(x + y) $$
- Take $partial / partial z$ of both sides of that:
$$ f'''(x+y+z) = 0 $$
Now substitute $y=z=0$ to find that $f'''(x)=0$. This shows that the only possible functions would be of the form $f(x) = ax^2 + bx + c$ for constants $a, b, c$. Then apply Robert's observation (see comments above) that $f(0) = 0$ (hence $c=0$), and that the set of functions form a linear space (hence all choices of $a, b in mathbb{R}$ are valid).
Hope this helps!
answered Aug 23 '12 at 20:33
Shaun AultShaun Ault
7,9021828
answered Aug 23 '12 at 20:33
Shaun AultShaun Ault
7,9021828
answered Aug 23 '12 at 20:33
Shaun AultShaun Ault
7,9021828
answered Aug 23 '12 at 20:33
Shaun AultShaun Ault
7,9021828
7,9021828
$begingroup$
I just saw Mariano's comment. :)
$endgroup$
– Shaun Ault
Aug 23 '12 at 20:34
3
$begingroup$
And what if $f$ is not differentiable?
$endgroup$
– Chris Eagle
Aug 23 '12 at 20:37
add a comment |
$begingroup$
I just saw Mariano's comment. :)
$endgroup$
– Shaun Ault
Aug 23 '12 at 20:34
3
$begingroup$
And what if $f$ is not differentiable?
$endgroup$
– Chris Eagle
Aug 23 '12 at 20:37
$begingroup$
I just saw Mariano's comment. :)
$endgroup$
– Shaun Ault
Aug 23 '12 at 20:34
$begingroup$
I just saw Mariano's comment. :)
$endgroup$
– Shaun Ault
Aug 23 '12 at 20:34
3
3
$begingroup$
And what if $f$ is not differentiable?
$endgroup$
– Chris Eagle
Aug 23 '12 at 20:37
$begingroup$
And what if $f$ is not differentiable?
$endgroup$
– Chris Eagle
Aug 23 '12 at 20:37
add a comment |
$begingroup$
Let $f: mathbb R to mathbb R$ be a continous solution of the functional equation and let $g(x) := ax^2+bx$ where $a$ and $b$ are chosen in such a way that $f(-1) = g(-1)$ and $f(1) = g(1)$. Then $tilde f := f-g$ is also a solution and we have $tilde f(-1) = tilde f(1) = 0$. We will show $tilde f equiv 0$, so that $f = g$.
By replacing $f$ with $tilde f$ way may assume that $f$ is a solution with $f(-1) = f(1) = 0$, and we will show $f equiv 0$. Let $Z = {x in mathbb R: f(x) = 0 }$. A priori we have ${-1,+1} subseteq Z$. Plugging $(1,-1,x)$ into the functional equation yields
$$f(x) = frac{f(x+1) + f(x-1)}{2},$$
which implies $mathbb Z subseteq Z$. If we can show
$$2x,x in Z Rightarrow frac{x}{2} in Z$$
we are finished since then every dyadic rational $a/2^n$ is in $Z$ and these are dense in $mathbb R$ (here the continuity of $f$ is needed). So let $2x in Z$ and $x in Z$. Plug $(frac{x}{2},frac{x}{2},x)$ into the functional equation to get
$$2fleft(frac{x}{2}right) + f(x) + f(2x) = f(x) + 2fleft(frac{3}{2}xright),$$
hence $fleft(frac{x}{2}right) = fleft(frac{3}{2}xright)$. Then plugging in $(frac{x}{2},frac{x}{2},frac{x}{2})$ yields
$$3fleft(frac{x}{2}right) + fleft(frac{3}{2}xright) = 3f(x),$$
hence $f(frac{x}{2}) = 0$ and $frac{x}{2} in Z$.
answered Aug 23 '12 at 22:43
marlumarlu
11k2041
$endgroup$
add a comment |
$begingroup$
Let $f: mathbb R to mathbb R$ be a continous solution of the functional equation and let $g(x) := ax^2+bx$ where $a$ and $b$ are chosen in such a way that $f(-1) = g(-1)$ and $f(1) = g(1)$. Then $tilde f := f-g$ is also a solution and we have $tilde f(-1) = tilde f(1) = 0$. We will show $tilde f equiv 0$, so that $f = g$.
By replacing $f$ with $tilde f$ way may assume that $f$ is a solution with $f(-1) = f(1) = 0$, and we will show $f equiv 0$. Let $Z = {x in mathbb R: f(x) = 0 }$. A priori we have ${-1,+1} subseteq Z$. Plugging $(1,-1,x)$ into the functional equation yields
$$f(x) = frac{f(x+1) + f(x-1)}{2},$$
which implies $mathbb Z subseteq Z$. If we can show
$$2x,x in Z Rightarrow frac{x}{2} in Z$$
we are finished since then every dyadic rational $a/2^n$ is in $Z$ and these are dense in $mathbb R$ (here the continuity of $f$ is needed). So let $2x in Z$ and $x in Z$. Plug $(frac{x}{2},frac{x}{2},x)$ into the functional equation to get
$$2fleft(frac{x}{2}right) + f(x) + f(2x) = f(x) + 2fleft(frac{3}{2}xright),$$
hence $fleft(frac{x}{2}right) = fleft(frac{3}{2}xright)$. Then plugging in $(frac{x}{2},frac{x}{2},frac{x}{2})$ yields
$$3fleft(frac{x}{2}right) + fleft(frac{3}{2}xright) = 3f(x),$$
hence $f(frac{x}{2}) = 0$ and $frac{x}{2} in Z$.
answered Aug 23 '12 at 22:43
marlumarlu
11k2041
$endgroup$
add a comment |
$begingroup$
Let $f: mathbb R to mathbb R$ be a continous solution of the functional equation and let $g(x) := ax^2+bx$ where $a$ and $b$ are chosen in such a way that $f(-1) = g(-1)$ and $f(1) = g(1)$. Then $tilde f := f-g$ is also a solution and we have $tilde f(-1) = tilde f(1) = 0$. We will show $tilde f equiv 0$, so that $f = g$.
By replacing $f$ with $tilde f$ way may assume that $f$ is a solution with $f(-1) = f(1) = 0$, and we will show $f equiv 0$. Let $Z = {x in mathbb R: f(x) = 0 }$. A priori we have ${-1,+1} subseteq Z$. Plugging $(1,-1,x)$ into the functional equation yields
$$f(x) = frac{f(x+1) + f(x-1)}{2},$$
which implies $mathbb Z subseteq Z$. If we can show
$$2x,x in Z Rightarrow frac{x}{2} in Z$$
we are finished since then every dyadic rational $a/2^n$ is in $Z$ and these are dense in $mathbb R$ (here the continuity of $f$ is needed). So let $2x in Z$ and $x in Z$. Plug $(frac{x}{2},frac{x}{2},x)$ into the functional equation to get
$$2fleft(frac{x}{2}right) + f(x) + f(2x) = f(x) + 2fleft(frac{3}{2}xright),$$
hence $fleft(frac{x}{2}right) = fleft(frac{3}{2}xright)$. Then plugging in $(frac{x}{2},frac{x}{2},frac{x}{2})$ yields
$$3fleft(frac{x}{2}right) + fleft(frac{3}{2}xright) = 3f(x),$$
hence $f(frac{x}{2}) = 0$ and $frac{x}{2} in Z$.
answered Aug 23 '12 at 22:43
marlumarlu
11k2041
$endgroup$
Let $f: mathbb R to mathbb R$ be a continous solution of the functional equation and let $g(x) := ax^2+bx$ where $a$ and $b$ are chosen in such a way that $f(-1) = g(-1)$ and $f(1) = g(1)$. Then $tilde f := f-g$ is also a solution and we have $tilde f(-1) = tilde f(1) = 0$. We will show $tilde f equiv 0$, so that $f = g$.
By replacing $f$ with $tilde f$ way may assume that $f$ is a solution with $f(-1) = f(1) = 0$, and we will show $f equiv 0$. Let $Z = {x in mathbb R: f(x) = 0 }$. A priori we have ${-1,+1} subseteq Z$. Plugging $(1,-1,x)$ into the functional equation yields
$$f(x) = frac{f(x+1) + f(x-1)}{2},$$
which implies $mathbb Z subseteq Z$. If we can show
$$2x,x in Z Rightarrow frac{x}{2} in Z$$
we are finished since then every dyadic rational $a/2^n$ is in $Z$ and these are dense in $mathbb R$ (here the continuity of $f$ is needed). So let $2x in Z$ and $x in Z$. Plug $(frac{x}{2},frac{x}{2},x)$ into the functional equation to get
$$2fleft(frac{x}{2}right) + f(x) + f(2x) = f(x) + 2fleft(frac{3}{2}xright),$$
hence $fleft(frac{x}{2}right) = fleft(frac{3}{2}xright)$. Then plugging in $(frac{x}{2},frac{x}{2},frac{x}{2})$ yields
$$3fleft(frac{x}{2}right) + fleft(frac{3}{2}xright) = 3f(x),$$
hence $f(frac{x}{2}) = 0$ and $frac{x}{2} in Z$.
answered Aug 23 '12 at 22:43
marlumarlu
11k2041
edited Aug 23 '12 at 23:02
answered Aug 23 '12 at 22:43
marlumarlu
11k2041
answered Aug 23 '12 at 22:43
marlumarlu
11k2041
answered Aug 23 '12 at 22:43
marlumarlu
11k2041
11k2041
add a comment |
add a comment |
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3
$begingroup$
It's easy to see that $f(0)=0$. The solutions form a linear space, and include $f(x)=x$ and $f(x) = x^2$ but no other powers. I would guess that $f(x) = ax + bx^2$ is the general solution.
$endgroup$
– Robert Israel
Aug 23 '12 at 20:12
3
$begingroup$
A more descriptive title would be useful: as it stands, it is almost unconnected to the question!
$endgroup$
– Mariano Suárez-Álvarez
Aug 23 '12 at 20:25
3
$begingroup$
Setting $x=y=z$ and differentiating the equality many times and evaluating at zero shows Robert's solutions are all the entire ones.
$endgroup$
– Mariano Suárez-Álvarez
Aug 23 '12 at 20:31
$begingroup$
@MarianoSuárez-Alvarez Is it obvious then any continuous function satisfying that equality is differentiable?
$endgroup$
– JSchlather
Aug 23 '12 at 20:38
$begingroup$
If it were obvious I would have said «Robert's solutions are all solutions» and not what I wrote!
$endgroup$
– Mariano Suárez-Álvarez
Aug 23 '12 at 20:39