Mixing
What functions satisfy $f(x)+f(y-x) = g(y)$?
$begingroup$
What functions $f$ have the property that, for all $x$ and $y$:
$$
f(x) + f(y-x) = g(y)
$$
i.e., the sum does not depend on $x$?
Linear functions obviously this property: if $f(x) = ax+b$, then:
$$
f(x) + f(y-x) = ay+2b = g(y).
$$
On the other hand, if we look only at differentiable functions, then only linear functions have this property. By taking the derivative of the first equation as a function of $x$:
$$
f'(x) - f'(y-x) = 0 iff f'(x)=f'(y-x)
$$
Since this is true for every $y$, $f'(cdot)$ must be a constant function, so $f$ must be linear.
Are there non-differentiable functions with this property?
functions
asked Feb 1 at 8:48
Erel Segal-HaleviErel Segal-Halevi
4,28712061
$endgroup$
add a comment |
$begingroup$
What functions $f$ have the property that, for all $x$ and $y$:
$$
f(x) + f(y-x) = g(y)
$$
i.e., the sum does not depend on $x$?
Linear functions obviously this property: if $f(x) = ax+b$, then:
$$
f(x) + f(y-x) = ay+2b = g(y).
$$
On the other hand, if we look only at differentiable functions, then only linear functions have this property. By taking the derivative of the first equation as a function of $x$:
$$
f'(x) - f'(y-x) = 0 iff f'(x)=f'(y-x)
$$
Since this is true for every $y$, $f'(cdot)$ must be a constant function, so $f$ must be linear.
Are there non-differentiable functions with this property?
functions
asked Feb 1 at 8:48
Erel Segal-HaleviErel Segal-Halevi
4,28712061
$endgroup$
add a comment |
$begingroup$
What functions $f$ have the property that, for all $x$ and $y$:
$$
f(x) + f(y-x) = g(y)
$$
i.e., the sum does not depend on $x$?
Linear functions obviously this property: if $f(x) = ax+b$, then:
$$
f(x) + f(y-x) = ay+2b = g(y).
$$
On the other hand, if we look only at differentiable functions, then only linear functions have this property. By taking the derivative of the first equation as a function of $x$:
$$
f'(x) - f'(y-x) = 0 iff f'(x)=f'(y-x)
$$
Since this is true for every $y$, $f'(cdot)$ must be a constant function, so $f$ must be linear.
Are there non-differentiable functions with this property?
functions
asked Feb 1 at 8:48
Erel Segal-HaleviErel Segal-Halevi
4,28712061
$endgroup$
What functions $f$ have the property that, for all $x$ and $y$:
$$
f(x) + f(y-x) = g(y)
$$
i.e., the sum does not depend on $x$?
Linear functions obviously this property: if $f(x) = ax+b$, then:
$$
f(x) + f(y-x) = ay+2b = g(y).
$$
On the other hand, if we look only at differentiable functions, then only linear functions have this property. By taking the derivative of the first equation as a function of $x$:
$$
f'(x) - f'(y-x) = 0 iff f'(x)=f'(y-x)
$$
Since this is true for every $y$, $f'(cdot)$ must be a constant function, so $f$ must be linear.
Are there non-differentiable functions with this property?
functions
functions
asked Feb 1 at 8:48
Erel Segal-HaleviErel Segal-Halevi
4,28712061
asked Feb 1 at 8:48
Erel Segal-HaleviErel Segal-Halevi
4,28712061
asked Feb 1 at 8:48
Erel Segal-HaleviErel Segal-Halevi
4,28712061
asked Feb 1 at 8:48
Erel Segal-HaleviErel Segal-Halevi
4,28712061
asked Feb 1 at 8:48
Erel Segal-HaleviErel Segal-Halevi
4,28712061
4,28712061
add a comment |
add a comment |
1 Answer
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$begingroup$
Put $y=x$ to get $f(x)+f(0)=g(x)$. Now verify that $h(x)=f(x)-f(0)$ satisfies $h(x)+h(y-x)=h(y)$. This is same thing as $h(x+y)=h(x)+h(y)$. There are discontinuous functions satisfying this equation. A continuous function $f$ satisfies the given property (that is to say the property stated in the question) iff $f(x)=ax+b$ for some constants $a$ and $b$.
answered Feb 1 at 8:57
Kavi Rama MurthyKavi Rama Murthy
73.8k53170
$endgroup$
$begingroup$
The solution $f(x)=ax+b$ is for the original equation in the question, not for additive functions.
$endgroup$
– Kavi Rama Murthy
Feb 1 at 9:12
$begingroup$
Then say so.
$endgroup$
– Did
Feb 1 at 9:12
$begingroup$
This is what I was missing: en.wikipedia.org/wiki/Cauchy%27s_functional_equation so if $h$ is continuous, then $h(x) = a x$ for some constant $a$, which implies that $f(x) = h(x) + f(0) = a x + b$ for some constants $a,b$.
$endgroup$
– Erel Segal-Halevi
Feb 2 at 20:32
add a comment |
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1 Answer
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active
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votes
1 Answer
1
active
oldest
votes
active
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active
oldest
votes
$begingroup$
Put $y=x$ to get $f(x)+f(0)=g(x)$. Now verify that $h(x)=f(x)-f(0)$ satisfies $h(x)+h(y-x)=h(y)$. This is same thing as $h(x+y)=h(x)+h(y)$. There are discontinuous functions satisfying this equation. A continuous function $f$ satisfies the given property (that is to say the property stated in the question) iff $f(x)=ax+b$ for some constants $a$ and $b$.
answered Feb 1 at 8:57
Kavi Rama MurthyKavi Rama Murthy
73.8k53170
$endgroup$
$begingroup$
The solution $f(x)=ax+b$ is for the original equation in the question, not for additive functions.
$endgroup$
– Kavi Rama Murthy
Feb 1 at 9:12
$begingroup$
Then say so.
$endgroup$
– Did
Feb 1 at 9:12
$begingroup$
This is what I was missing: en.wikipedia.org/wiki/Cauchy%27s_functional_equation so if $h$ is continuous, then $h(x) = a x$ for some constant $a$, which implies that $f(x) = h(x) + f(0) = a x + b$ for some constants $a,b$.
$endgroup$
– Erel Segal-Halevi
Feb 2 at 20:32
add a comment |
$begingroup$
Put $y=x$ to get $f(x)+f(0)=g(x)$. Now verify that $h(x)=f(x)-f(0)$ satisfies $h(x)+h(y-x)=h(y)$. This is same thing as $h(x+y)=h(x)+h(y)$. There are discontinuous functions satisfying this equation. A continuous function $f$ satisfies the given property (that is to say the property stated in the question) iff $f(x)=ax+b$ for some constants $a$ and $b$.
answered Feb 1 at 8:57
Kavi Rama MurthyKavi Rama Murthy
73.8k53170
$endgroup$
$begingroup$
The solution $f(x)=ax+b$ is for the original equation in the question, not for additive functions.
$endgroup$
– Kavi Rama Murthy
Feb 1 at 9:12
$begingroup$
Then say so.
$endgroup$
– Did
Feb 1 at 9:12
$begingroup$
This is what I was missing: en.wikipedia.org/wiki/Cauchy%27s_functional_equation so if $h$ is continuous, then $h(x) = a x$ for some constant $a$, which implies that $f(x) = h(x) + f(0) = a x + b$ for some constants $a,b$.
$endgroup$
– Erel Segal-Halevi
Feb 2 at 20:32
add a comment |
$begingroup$
Put $y=x$ to get $f(x)+f(0)=g(x)$. Now verify that $h(x)=f(x)-f(0)$ satisfies $h(x)+h(y-x)=h(y)$. This is same thing as $h(x+y)=h(x)+h(y)$. There are discontinuous functions satisfying this equation. A continuous function $f$ satisfies the given property (that is to say the property stated in the question) iff $f(x)=ax+b$ for some constants $a$ and $b$.
answered Feb 1 at 8:57
Kavi Rama MurthyKavi Rama Murthy
73.8k53170
$endgroup$
Put $y=x$ to get $f(x)+f(0)=g(x)$. Now verify that $h(x)=f(x)-f(0)$ satisfies $h(x)+h(y-x)=h(y)$. This is same thing as $h(x+y)=h(x)+h(y)$. There are discontinuous functions satisfying this equation. A continuous function $f$ satisfies the given property (that is to say the property stated in the question) iff $f(x)=ax+b$ for some constants $a$ and $b$.
answered Feb 1 at 8:57
Kavi Rama MurthyKavi Rama Murthy
73.8k53170
edited Feb 1 at 9:13
answered Feb 1 at 8:57
Kavi Rama MurthyKavi Rama Murthy
73.8k53170
answered Feb 1 at 8:57
Kavi Rama MurthyKavi Rama Murthy
73.8k53170
answered Feb 1 at 8:57
Kavi Rama MurthyKavi Rama Murthy
73.8k53170
73.8k53170
$begingroup$
The solution $f(x)=ax+b$ is for the original equation in the question, not for additive functions.
$endgroup$
– Kavi Rama Murthy
Feb 1 at 9:12
$begingroup$
Then say so.
$endgroup$
– Did
Feb 1 at 9:12
$begingroup$
This is what I was missing: en.wikipedia.org/wiki/Cauchy%27s_functional_equation so if $h$ is continuous, then $h(x) = a x$ for some constant $a$, which implies that $f(x) = h(x) + f(0) = a x + b$ for some constants $a,b$.
$endgroup$
– Erel Segal-Halevi
Feb 2 at 20:32
add a comment |
$begingroup$
The solution $f(x)=ax+b$ is for the original equation in the question, not for additive functions.
$endgroup$
– Kavi Rama Murthy
Feb 1 at 9:12
$begingroup$
Then say so.
$endgroup$
– Did
Feb 1 at 9:12
$begingroup$
This is what I was missing: en.wikipedia.org/wiki/Cauchy%27s_functional_equation so if $h$ is continuous, then $h(x) = a x$ for some constant $a$, which implies that $f(x) = h(x) + f(0) = a x + b$ for some constants $a,b$.
$endgroup$
– Erel Segal-Halevi
Feb 2 at 20:32
$begingroup$
The solution $f(x)=ax+b$ is for the original equation in the question, not for additive functions.
$endgroup$
– Kavi Rama Murthy
Feb 1 at 9:12
$begingroup$
The solution $f(x)=ax+b$ is for the original equation in the question, not for additive functions.
$endgroup$
– Kavi Rama Murthy
Feb 1 at 9:12
$begingroup$
Then say so.
$endgroup$
– Did
Feb 1 at 9:12
$begingroup$
Then say so.
$endgroup$
– Did
Feb 1 at 9:12
$begingroup$
This is what I was missing: en.wikipedia.org/wiki/Cauchy%27s_functional_equation so if $h$ is continuous, then $h(x) = a x$ for some constant $a$, which implies that $f(x) = h(x) + f(0) = a x + b$ for some constants $a,b$.
$endgroup$
– Erel Segal-Halevi
Feb 2 at 20:32
$begingroup$
This is what I was missing: en.wikipedia.org/wiki/Cauchy%27s_functional_equation so if $h$ is continuous, then $h(x) = a x$ for some constant $a$, which implies that $f(x) = h(x) + f(0) = a x + b$ for some constants $a,b$.
$endgroup$
– Erel Segal-Halevi
Feb 2 at 20:32
add a comment |
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