$f(y+zf(x))=f(y)+xf(z) $ with $x,y,zin mathbb{R}$
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Find all $fcolonmathbb{R} rightarrow mathbb{R}$ such that
$f(y+zf(x))=f(y)+xf(z)$ with $x,y,zin mathbb{R}$
When I study functional equation, I have some difficult in solving above equation. Can anyone help me please?
functional-equations
edited Nov 22 at 15:02
Ivan Neretin
8,75021535
asked Nov 22 at 13:47
Trong Tuan
766
add a comment |
up vote
1
down vote
favorite
Find all $fcolonmathbb{R} rightarrow mathbb{R}$ such that
$f(y+zf(x))=f(y)+xf(z)$ with $x,y,zin mathbb{R}$
When I study functional equation, I have some difficult in solving above equation. Can anyone help me please?
functional-equations
edited Nov 22 at 15:02
Ivan Neretin
8,75021535
asked Nov 22 at 13:47
Trong Tuan
766
1
Does the function need to have all three properties at once? Or are you looking for all functions that satisfy a, and then all that satisfy b?
– 5xum
Nov 22 at 13:49
I'd narrow down the number of functions by noting that these would hold for $x=y=z$
– Rhys Hughes
Nov 22 at 14:00
For (a) can you show that $f(1) = 1$?. Hint: If $f(1) = p$ can you find $f(p^n)$ and $f(p^{-n})$ in terms of p? If you assume that f is continuous at 0 you can find $f(1)$.
– Paul
Nov 22 at 14:29
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Find all $fcolonmathbb{R} rightarrow mathbb{R}$ such that
$f(y+zf(x))=f(y)+xf(z)$ with $x,y,zin mathbb{R}$
When I study functional equation, I have some difficult in solving above equation. Can anyone help me please?
functional-equations
edited Nov 22 at 15:02
Ivan Neretin
8,75021535
asked Nov 22 at 13:47
Trong Tuan
766
Find all $fcolonmathbb{R} rightarrow mathbb{R}$ such that
$f(y+zf(x))=f(y)+xf(z)$ with $x,y,zin mathbb{R}$
When I study functional equation, I have some difficult in solving above equation. Can anyone help me please?
functional-equations
functional-equations
edited Nov 22 at 15:02
Ivan Neretin
8,75021535
asked Nov 22 at 13:47
Trong Tuan
766
edited Nov 22 at 15:02
Ivan Neretin
8,75021535
asked Nov 22 at 13:47
Trong Tuan
766
edited Nov 22 at 15:02
Ivan Neretin
8,75021535
edited Nov 22 at 15:02
Ivan Neretin
8,75021535
edited Nov 22 at 15:02
Ivan Neretin
8,75021535
8,75021535
asked Nov 22 at 13:47
Trong Tuan
766
asked Nov 22 at 13:47
Trong Tuan
766
asked Nov 22 at 13:47
Trong Tuan
766
766
1
Does the function need to have all three properties at once? Or are you looking for all functions that satisfy a, and then all that satisfy b?
– 5xum
Nov 22 at 13:49
I'd narrow down the number of functions by noting that these would hold for $x=y=z$
– Rhys Hughes
Nov 22 at 14:00
For (a) can you show that $f(1) = 1$?. Hint: If $f(1) = p$ can you find $f(p^n)$ and $f(p^{-n})$ in terms of p? If you assume that f is continuous at 0 you can find $f(1)$.
– Paul
Nov 22 at 14:29
add a comment |
1
Does the function need to have all three properties at once? Or are you looking for all functions that satisfy a, and then all that satisfy b?
– 5xum
Nov 22 at 13:49
I'd narrow down the number of functions by noting that these would hold for $x=y=z$
– Rhys Hughes
Nov 22 at 14:00
For (a) can you show that $f(1) = 1$?. Hint: If $f(1) = p$ can you find $f(p^n)$ and $f(p^{-n})$ in terms of p? If you assume that f is continuous at 0 you can find $f(1)$.
– Paul
Nov 22 at 14:29
1
1
Does the function need to have all three properties at once? Or are you looking for all functions that satisfy a, and then all that satisfy b?
– 5xum
Nov 22 at 13:49
Does the function need to have all three properties at once? Or are you looking for all functions that satisfy a, and then all that satisfy b?
– 5xum
Nov 22 at 13:49
I'd narrow down the number of functions by noting that these would hold for $x=y=z$
– Rhys Hughes
Nov 22 at 14:00
I'd narrow down the number of functions by noting that these would hold for $x=y=z$
– Rhys Hughes
Nov 22 at 14:00
For (a) can you show that $f(1) = 1$?. Hint: If $f(1) = p$ can you find $f(p^n)$ and $f(p^{-n})$ in terms of p? If you assume that f is continuous at 0 you can find $f(1)$.
– Paul
Nov 22 at 14:29
For (a) can you show that $f(1) = 1$?. Hint: If $f(1) = p$ can you find $f(p^n)$ and $f(p^{-n})$ in terms of p? If you assume that f is continuous at 0 you can find $f(1)$.
– Paul
Nov 22 at 14:29
add a comment |
1 Answer
1
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By applying $z = 0$, we have $f(0) = 0$.
$f(x) = 0$ for all $xinmathbb{R}$ is clearly a solution.
Assume that there exists $tinmathbb{R}$ such that $f(t)neq 0$. Since $f(y+zf(t)) = f(y) + tf(z)$, $f$ is surjective.
Let $m, ninmathbb{R}$ such that $f(m) = f(n)$. $f(y) + mf(t) = f(y+tf(m)) = f(y+tf(n)) = f(y) + nf(t)$, so $m = n$. Hence $f$ is bijective.
By applying $y = 0$, we have $f(zf(x)) = xf(z)$. By changing the roles of $x$ and $z$, we have $f(f(zf(x))) = f(xf(z)) = zf(x)$. So for all $linmathbb{R}$, by applying $z = frac{l}{f(t)}, x = t$, we obtain $f(f(l)) = l$.
By applying $z = 1$ in $f(zf(t)) = tf(z)$, we have $f(1) = 1$.
So $f$ satisfies $f(y + z) = f(y)+ f(z)$.
By applying $y = 0, z = f(m)$, we obtain $f(f(x)f(m)) = f(x)f(m)$ for all $x, minmathbb{R}$. Since $f$ is bijective, $f(xy) = f(x)f(y)$ for all $x, yinmathbb{R}$.
Therefore $f$ is an automorphsim of field $mathbb{R}$ ! And the identity is the only automorphism of field $mathbb{R}$.
It can be demonstrated by the following steps :
- Show that $f(x) = x$ for all $xinmathbb{Q}$.
- Show that $f(mathbb{R}_{+})subsetmathbb{R}_{+}$. (use the fact that $xgeq 0$ iff x is a square)
- Deduce that $f$ is strictly increasing.
- For all $xinmathbb{R}setminusmathbb{Q}$, if $xneq f(x)$, choose a rational number between $x$ and $f(x)$.
answered Nov 22 at 15:49
曾靖國
3868
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
By applying $z = 0$, we have $f(0) = 0$.
$f(x) = 0$ for all $xinmathbb{R}$ is clearly a solution.
Assume that there exists $tinmathbb{R}$ such that $f(t)neq 0$. Since $f(y+zf(t)) = f(y) + tf(z)$, $f$ is surjective.
Let $m, ninmathbb{R}$ such that $f(m) = f(n)$. $f(y) + mf(t) = f(y+tf(m)) = f(y+tf(n)) = f(y) + nf(t)$, so $m = n$. Hence $f$ is bijective.
By applying $y = 0$, we have $f(zf(x)) = xf(z)$. By changing the roles of $x$ and $z$, we have $f(f(zf(x))) = f(xf(z)) = zf(x)$. So for all $linmathbb{R}$, by applying $z = frac{l}{f(t)}, x = t$, we obtain $f(f(l)) = l$.
By applying $z = 1$ in $f(zf(t)) = tf(z)$, we have $f(1) = 1$.
So $f$ satisfies $f(y + z) = f(y)+ f(z)$.
By applying $y = 0, z = f(m)$, we obtain $f(f(x)f(m)) = f(x)f(m)$ for all $x, minmathbb{R}$. Since $f$ is bijective, $f(xy) = f(x)f(y)$ for all $x, yinmathbb{R}$.
Therefore $f$ is an automorphsim of field $mathbb{R}$ ! And the identity is the only automorphism of field $mathbb{R}$.
It can be demonstrated by the following steps :
- Show that $f(x) = x$ for all $xinmathbb{Q}$.
- Show that $f(mathbb{R}_{+})subsetmathbb{R}_{+}$. (use the fact that $xgeq 0$ iff x is a square)
- Deduce that $f$ is strictly increasing.
- For all $xinmathbb{R}setminusmathbb{Q}$, if $xneq f(x)$, choose a rational number between $x$ and $f(x)$.
answered Nov 22 at 15:49
曾靖國
3868
add a comment |
up vote
1
down vote
By applying $z = 0$, we have $f(0) = 0$.
$f(x) = 0$ for all $xinmathbb{R}$ is clearly a solution.
Assume that there exists $tinmathbb{R}$ such that $f(t)neq 0$. Since $f(y+zf(t)) = f(y) + tf(z)$, $f$ is surjective.
Let $m, ninmathbb{R}$ such that $f(m) = f(n)$. $f(y) + mf(t) = f(y+tf(m)) = f(y+tf(n)) = f(y) + nf(t)$, so $m = n$. Hence $f$ is bijective.
By applying $y = 0$, we have $f(zf(x)) = xf(z)$. By changing the roles of $x$ and $z$, we have $f(f(zf(x))) = f(xf(z)) = zf(x)$. So for all $linmathbb{R}$, by applying $z = frac{l}{f(t)}, x = t$, we obtain $f(f(l)) = l$.
By applying $z = 1$ in $f(zf(t)) = tf(z)$, we have $f(1) = 1$.
So $f$ satisfies $f(y + z) = f(y)+ f(z)$.
By applying $y = 0, z = f(m)$, we obtain $f(f(x)f(m)) = f(x)f(m)$ for all $x, minmathbb{R}$. Since $f$ is bijective, $f(xy) = f(x)f(y)$ for all $x, yinmathbb{R}$.
Therefore $f$ is an automorphsim of field $mathbb{R}$ ! And the identity is the only automorphism of field $mathbb{R}$.
It can be demonstrated by the following steps :
- Show that $f(x) = x$ for all $xinmathbb{Q}$.
- Show that $f(mathbb{R}_{+})subsetmathbb{R}_{+}$. (use the fact that $xgeq 0$ iff x is a square)
- Deduce that $f$ is strictly increasing.
- For all $xinmathbb{R}setminusmathbb{Q}$, if $xneq f(x)$, choose a rational number between $x$ and $f(x)$.
answered Nov 22 at 15:49
曾靖國
3868
add a comment |
up vote
1
down vote
up vote
1
down vote
By applying $z = 0$, we have $f(0) = 0$.
$f(x) = 0$ for all $xinmathbb{R}$ is clearly a solution.
Assume that there exists $tinmathbb{R}$ such that $f(t)neq 0$. Since $f(y+zf(t)) = f(y) + tf(z)$, $f$ is surjective.
Let $m, ninmathbb{R}$ such that $f(m) = f(n)$. $f(y) + mf(t) = f(y+tf(m)) = f(y+tf(n)) = f(y) + nf(t)$, so $m = n$. Hence $f$ is bijective.
By applying $y = 0$, we have $f(zf(x)) = xf(z)$. By changing the roles of $x$ and $z$, we have $f(f(zf(x))) = f(xf(z)) = zf(x)$. So for all $linmathbb{R}$, by applying $z = frac{l}{f(t)}, x = t$, we obtain $f(f(l)) = l$.
By applying $z = 1$ in $f(zf(t)) = tf(z)$, we have $f(1) = 1$.
So $f$ satisfies $f(y + z) = f(y)+ f(z)$.
By applying $y = 0, z = f(m)$, we obtain $f(f(x)f(m)) = f(x)f(m)$ for all $x, minmathbb{R}$. Since $f$ is bijective, $f(xy) = f(x)f(y)$ for all $x, yinmathbb{R}$.
Therefore $f$ is an automorphsim of field $mathbb{R}$ ! And the identity is the only automorphism of field $mathbb{R}$.
It can be demonstrated by the following steps :
- Show that $f(x) = x$ for all $xinmathbb{Q}$.
- Show that $f(mathbb{R}_{+})subsetmathbb{R}_{+}$. (use the fact that $xgeq 0$ iff x is a square)
- Deduce that $f$ is strictly increasing.
- For all $xinmathbb{R}setminusmathbb{Q}$, if $xneq f(x)$, choose a rational number between $x$ and $f(x)$.
answered Nov 22 at 15:49
曾靖國
3868
By applying $z = 0$, we have $f(0) = 0$.
$f(x) = 0$ for all $xinmathbb{R}$ is clearly a solution.
Assume that there exists $tinmathbb{R}$ such that $f(t)neq 0$. Since $f(y+zf(t)) = f(y) + tf(z)$, $f$ is surjective.
Let $m, ninmathbb{R}$ such that $f(m) = f(n)$. $f(y) + mf(t) = f(y+tf(m)) = f(y+tf(n)) = f(y) + nf(t)$, so $m = n$. Hence $f$ is bijective.
By applying $y = 0$, we have $f(zf(x)) = xf(z)$. By changing the roles of $x$ and $z$, we have $f(f(zf(x))) = f(xf(z)) = zf(x)$. So for all $linmathbb{R}$, by applying $z = frac{l}{f(t)}, x = t$, we obtain $f(f(l)) = l$.
By applying $z = 1$ in $f(zf(t)) = tf(z)$, we have $f(1) = 1$.
So $f$ satisfies $f(y + z) = f(y)+ f(z)$.
By applying $y = 0, z = f(m)$, we obtain $f(f(x)f(m)) = f(x)f(m)$ for all $x, minmathbb{R}$. Since $f$ is bijective, $f(xy) = f(x)f(y)$ for all $x, yinmathbb{R}$.
Therefore $f$ is an automorphsim of field $mathbb{R}$ ! And the identity is the only automorphism of field $mathbb{R}$.
It can be demonstrated by the following steps :
- Show that $f(x) = x$ for all $xinmathbb{Q}$.
- Show that $f(mathbb{R}_{+})subsetmathbb{R}_{+}$. (use the fact that $xgeq 0$ iff x is a square)
- Deduce that $f$ is strictly increasing.
- For all $xinmathbb{R}setminusmathbb{Q}$, if $xneq f(x)$, choose a rational number between $x$ and $f(x)$.
answered Nov 22 at 15:49
曾靖國
3868
edited Nov 22 at 16:06
answered Nov 22 at 15:49
曾靖國
3868
answered Nov 22 at 15:49
曾靖國
3868
answered Nov 22 at 15:49
曾靖國
3868
3868
add a comment |
add a comment |
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1
Does the function need to have all three properties at once? Or are you looking for all functions that satisfy a, and then all that satisfy b?
– 5xum
Nov 22 at 13:49
I'd narrow down the number of functions by noting that these would hold for $x=y=z$
– Rhys Hughes
Nov 22 at 14:00
For (a) can you show that $f(1) = 1$?. Hint: If $f(1) = p$ can you find $f(p^n)$ and $f(p^{-n})$ in terms of p? If you assume that f is continuous at 0 you can find $f(1)$.
– Paul
Nov 22 at 14:29