$f(x)=af(ax)+bf(bx)$Finding all continuous functions such that $f^2(x + y) − f^2(x − y) = 4f(x)f(y)$Attractive fixed-point?Does a non-trivial solution exist for $f'(x)=f(f(x))$?An example of the Sequential Quadratic Programming (SQP)Existence solution pendulum equation using the contraction principleUnderstanding of solution for a functional equation.continuous (on 3, 4 and 5) f is constant, if $f(x+2)+f(4x)=f(2x+1)+f(2x+2),forall xinmathbbR$Function with derivative-like property: $f(ab) = af(b) + bf(a)$$f:[0,1]rightarrow [0,1]$ and $fcirc f$ are not identically zero but $f(f(f(x)))=0$ for all $xin [0,1]$. Does $f$ exist?Continuous function and function relationship
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$f(x)=af(ax)+bf(bx)$
Finding all continuous functions such that $f^2(x + y) − f^2(x − y) = 4f(x)f(y)$Attractive fixed-point?Does a non-trivial solution exist for $f'(x)=f(f(x))$?An example of the Sequential Quadratic Programming (SQP)Existence solution pendulum equation using the contraction principleUnderstanding of solution for a functional equation.continuous (on 3, 4 and 5) f is constant, if $f(x+2)+f(4x)=f(2x+1)+f(2x+2),forall xinmathbbR$Function with derivative-like property: $f(ab) = af(b) + bf(a)$$f:[0,1]rightarrow [0,1]$ and $fcirc f$ are not identically zero but $f(f(f(x)))=0$ for all $xin [0,1]$. Does $f$ exist?Continuous function and function relationship
$begingroup$
$f(x)=af(ax)+bf(bx)$ . We know that $a+b=1$ and we have to prove that f is constant.
Let $x_0$ be the maxim point. $f(x_0) le (a+b)f(x_0)$. I don't know what can I do now.
functional-equations
asked Mar 10 at 20:49
GaboruGaboru
4407
$endgroup$
add a comment |
$begingroup$
$f(x)=af(ax)+bf(bx)$ . We know that $a+b=1$ and we have to prove that f is constant.
Let $x_0$ be the maxim point. $f(x_0) le (a+b)f(x_0)$. I don't know what can I do now.
functional-equations
asked Mar 10 at 20:49
GaboruGaboru
4407
$endgroup$
2
$begingroup$
"The" maximun point ? I think there misses some asumptions, otherwise "the" maximum point does not always exist.
$endgroup$
– TheSilverDoe
Mar 10 at 20:54
1
$begingroup$
Who is mister f? What are his properties? Is he continuous, differentiable?
$endgroup$
– HAMIDINE SOUMARE
Mar 10 at 20:55
add a comment |
$begingroup$
$f(x)=af(ax)+bf(bx)$ . We know that $a+b=1$ and we have to prove that f is constant.
Let $x_0$ be the maxim point. $f(x_0) le (a+b)f(x_0)$. I don't know what can I do now.
functional-equations
asked Mar 10 at 20:49
GaboruGaboru
4407
$endgroup$
$f(x)=af(ax)+bf(bx)$ . We know that $a+b=1$ and we have to prove that f is constant.
Let $x_0$ be the maxim point. $f(x_0) le (a+b)f(x_0)$. I don't know what can I do now.
functional-equations
functional-equations
asked Mar 10 at 20:49
GaboruGaboru
4407
asked Mar 10 at 20:49
GaboruGaboru
4407
asked Mar 10 at 20:49
GaboruGaboru
4407
asked Mar 10 at 20:49
GaboruGaboru
4407
asked Mar 10 at 20:49
GaboruGaboru
4407
4407
2
$begingroup$
"The" maximun point ? I think there misses some asumptions, otherwise "the" maximum point does not always exist.
$endgroup$
– TheSilverDoe
Mar 10 at 20:54
1
$begingroup$
Who is mister f? What are his properties? Is he continuous, differentiable?
$endgroup$
– HAMIDINE SOUMARE
Mar 10 at 20:55
add a comment |
2
$begingroup$
"The" maximun point ? I think there misses some asumptions, otherwise "the" maximum point does not always exist.
$endgroup$
– TheSilverDoe
Mar 10 at 20:54
1
$begingroup$
Who is mister f? What are his properties? Is he continuous, differentiable?
$endgroup$
– HAMIDINE SOUMARE
Mar 10 at 20:55
2
2
$begingroup$
"The" maximun point ? I think there misses some asumptions, otherwise "the" maximum point does not always exist.
$endgroup$
– TheSilverDoe
Mar 10 at 20:54
$begingroup$
"The" maximun point ? I think there misses some asumptions, otherwise "the" maximum point does not always exist.
$endgroup$
– TheSilverDoe
Mar 10 at 20:54
1
1
$begingroup$
Who is mister f? What are his properties? Is he continuous, differentiable?
$endgroup$
– HAMIDINE SOUMARE
Mar 10 at 20:55
$begingroup$
Who is mister f? What are his properties? Is he continuous, differentiable?
$endgroup$
– HAMIDINE SOUMARE
Mar 10 at 20:55
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hint: If $f$ is continuous, let $a = b = frac 12$. What do you get from that? Iterate and use continuity.
answered Mar 10 at 21:02
EnnarEnnar
14.8k32445
$endgroup$
$begingroup$
Thank you very much!
$endgroup$
– Gaboru
Mar 10 at 21:05
$begingroup$
It's easy to do for $a=b=1/2$ but I don't know how to do for different.
$endgroup$
– Gaboru
19 hours ago
$begingroup$
@Gaboru, I interpreted the exercise as: "If for all $a$, $b$ such that $a+b = 1$ we have that $f(x) = af(ax)+bf(bx)$, for all $x$, then $f$ is constant".
$endgroup$
– Ennar
18 hours ago
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Hint: If $f$ is continuous, let $a = b = frac 12$. What do you get from that? Iterate and use continuity.
answered Mar 10 at 21:02
EnnarEnnar
14.8k32445
$endgroup$
$begingroup$
Thank you very much!
$endgroup$
– Gaboru
Mar 10 at 21:05
$begingroup$
It's easy to do for $a=b=1/2$ but I don't know how to do for different.
$endgroup$
– Gaboru
19 hours ago
$begingroup$
@Gaboru, I interpreted the exercise as: "If for all $a$, $b$ such that $a+b = 1$ we have that $f(x) = af(ax)+bf(bx)$, for all $x$, then $f$ is constant".
$endgroup$
– Ennar
18 hours ago
add a comment |
$begingroup$
Hint: If $f$ is continuous, let $a = b = frac 12$. What do you get from that? Iterate and use continuity.
answered Mar 10 at 21:02
EnnarEnnar
14.8k32445
$endgroup$
$begingroup$
Thank you very much!
$endgroup$
– Gaboru
Mar 10 at 21:05
$begingroup$
It's easy to do for $a=b=1/2$ but I don't know how to do for different.
$endgroup$
– Gaboru
19 hours ago
$begingroup$
@Gaboru, I interpreted the exercise as: "If for all $a$, $b$ such that $a+b = 1$ we have that $f(x) = af(ax)+bf(bx)$, for all $x$, then $f$ is constant".
$endgroup$
– Ennar
18 hours ago
add a comment |
$begingroup$
Hint: If $f$ is continuous, let $a = b = frac 12$. What do you get from that? Iterate and use continuity.
answered Mar 10 at 21:02
EnnarEnnar
14.8k32445
$endgroup$
Hint: If $f$ is continuous, let $a = b = frac 12$. What do you get from that? Iterate and use continuity.
answered Mar 10 at 21:02
EnnarEnnar
14.8k32445
answered Mar 10 at 21:02
EnnarEnnar
14.8k32445
answered Mar 10 at 21:02
EnnarEnnar
14.8k32445
answered Mar 10 at 21:02
EnnarEnnar
14.8k32445
14.8k32445
$begingroup$
Thank you very much!
$endgroup$
– Gaboru
Mar 10 at 21:05
$begingroup$
It's easy to do for $a=b=1/2$ but I don't know how to do for different.
$endgroup$
– Gaboru
19 hours ago
$begingroup$
@Gaboru, I interpreted the exercise as: "If for all $a$, $b$ such that $a+b = 1$ we have that $f(x) = af(ax)+bf(bx)$, for all $x$, then $f$ is constant".
$endgroup$
– Ennar
18 hours ago
add a comment |
$begingroup$
Thank you very much!
$endgroup$
– Gaboru
Mar 10 at 21:05
$begingroup$
It's easy to do for $a=b=1/2$ but I don't know how to do for different.
$endgroup$
– Gaboru
19 hours ago
$begingroup$
@Gaboru, I interpreted the exercise as: "If for all $a$, $b$ such that $a+b = 1$ we have that $f(x) = af(ax)+bf(bx)$, for all $x$, then $f$ is constant".
$endgroup$
– Ennar
18 hours ago
$begingroup$
Thank you very much!
$endgroup$
– Gaboru
Mar 10 at 21:05
$begingroup$
Thank you very much!
$endgroup$
– Gaboru
Mar 10 at 21:05
$begingroup$
It's easy to do for $a=b=1/2$ but I don't know how to do for different.
$endgroup$
– Gaboru
19 hours ago
$begingroup$
It's easy to do for $a=b=1/2$ but I don't know how to do for different.
$endgroup$
– Gaboru
19 hours ago
$begingroup$
@Gaboru, I interpreted the exercise as: "If for all $a$, $b$ such that $a+b = 1$ we have that $f(x) = af(ax)+bf(bx)$, for all $x$, then $f$ is constant".
$endgroup$
– Ennar
18 hours ago
$begingroup$
@Gaboru, I interpreted the exercise as: "If for all $a$, $b$ such that $a+b = 1$ we have that $f(x) = af(ax)+bf(bx)$, for all $x$, then $f$ is constant".
$endgroup$
– Ennar
18 hours ago
add a comment |
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$begingroup$
"The" maximun point ? I think there misses some asumptions, otherwise "the" maximum point does not always exist.
$endgroup$
– TheSilverDoe
Mar 10 at 20:54
1
$begingroup$
Who is mister f? What are his properties? Is he continuous, differentiable?
$endgroup$
– HAMIDINE SOUMARE
Mar 10 at 20:55