Mixing
A function satisfying a given condition
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Is there a continuous real valued function $f$ satisfying $f(x+1)(f(x)+1)=1$ for all $x$ in the domain of $f$(possibly $mathbb{R})$? Clearly the image of $f$ does not include $0$ and $-1$. By the continuity of $f$ and the intermediate value theorem, $f$ has to be always positive or always negative since otherwise there would be a zero of $f$. Does such a continuous $f$ exist? How about discontinuous functions?
Please give a hint to proceed. Thank you.
real-analysis functions continuity functional-equations
asked Jan 10 at 21:56
FermatFermat
4,4001926
$endgroup$
add a comment |
$begingroup$
Is there a continuous real valued function $f$ satisfying $f(x+1)(f(x)+1)=1$ for all $x$ in the domain of $f$(possibly $mathbb{R})$? Clearly the image of $f$ does not include $0$ and $-1$. By the continuity of $f$ and the intermediate value theorem, $f$ has to be always positive or always negative since otherwise there would be a zero of $f$. Does such a continuous $f$ exist? How about discontinuous functions?
Please give a hint to proceed. Thank you.
real-analysis functions continuity functional-equations
asked Jan 10 at 21:56
FermatFermat
4,4001926
$endgroup$
add a comment |
$begingroup$
Is there a continuous real valued function $f$ satisfying $f(x+1)(f(x)+1)=1$ for all $x$ in the domain of $f$(possibly $mathbb{R})$? Clearly the image of $f$ does not include $0$ and $-1$. By the continuity of $f$ and the intermediate value theorem, $f$ has to be always positive or always negative since otherwise there would be a zero of $f$. Does such a continuous $f$ exist? How about discontinuous functions?
Please give a hint to proceed. Thank you.
real-analysis functions continuity functional-equations
asked Jan 10 at 21:56
FermatFermat
4,4001926
$endgroup$
Is there a continuous real valued function $f$ satisfying $f(x+1)(f(x)+1)=1$ for all $x$ in the domain of $f$(possibly $mathbb{R})$? Clearly the image of $f$ does not include $0$ and $-1$. By the continuity of $f$ and the intermediate value theorem, $f$ has to be always positive or always negative since otherwise there would be a zero of $f$. Does such a continuous $f$ exist? How about discontinuous functions?
Please give a hint to proceed. Thank you.
real-analysis functions continuity functional-equations
real-analysis functions continuity functional-equations
asked Jan 10 at 21:56
FermatFermat
4,4001926
asked Jan 10 at 21:56
FermatFermat
4,4001926
edited Jan 10 at 22:39
Fermat
asked Jan 10 at 21:56
FermatFermat
4,4001926
asked Jan 10 at 21:56
FermatFermat
4,4001926
asked Jan 10 at 21:56
FermatFermat
4,4001926
4,4001926
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Suppose $f:mathbb{R}tomathbb{R}$ is a continuous solution. Let us consider the sequence $f(n)$, $ninmathbb{Z}$. If we write $$f(n)=frac{a_n}{a_{n+1}},tag{*}$$ by the given recursion formula, we have
$$
f(n+1) = frac{a_{n+1}}{a_n+a_{n+1}}.
$$ This implies the sequence $a_n$ is a Fibonacci sequence, i.e. $a_{n+2} = a_{n+1}+a_n$. It is well-known that the general solution of $a_n$ is given by
$$
a_n = Aalpha^n + Bbeta^n
$$ for some $A,Binmathbb{R}$ where $alpha = frac{1+sqrt{5}}{2}$ and $beta=frac{1-sqrt{5}}{2}$. Plugging this into the expression in $(*)$, we get
$$
f(n) = frac{Aalpha^n + Bbeta^n}{Aalpha^{n+1} + Bbeta^{n+1}}.
$$ Assume that $A=0$. Since not both $A,B$ are $0$, we have $f(n) = beta^{-1}$. In the same way, $f(n)=alpha^{-1}$ holds if $B=0$. Now, assume that $A$ and $B$ are not zero. By taking limit as $ntopminfty$, we have
$$
lim_{ntoinfty}f(n) = alpha^{-1} = -beta>frac{1}{2},
$$ and
$$
lim_{nto-infty}f(n) =beta^{-1} = -alpha<-frac{3}{2}.
$$ But this implies by IVP that there exists $cinmathbb{R}$ such that $f(c)=0$. Since this is impossible as $f(x)ne 0$, there are only $2$ possibilities: Either $f(n) equiv alpha^{-1}$ or $f(n)equiv beta^{-1}$. Finally, notice that the above argument applies in the same way to $f(x+n)$ for arbitrary $xin mathbb{R}$. This implies $f(mathbb{R})subset {alpha^{-1}, beta^{-1}}$. By the continuity of $f$, this says that $f$ must be a constant function. Thus the only continuous solutions to the equation are
$$
f(x) equiv alpha^{-1} = frac{-1+sqrt{5}}{2}
$$ or
$$
f(x) equiv beta^{-1} = frac{-1-sqrt{5}}{2}.
$$
Note: Note that for all $ninmathbb{Z}$, $$f(n)=frac{a_n}{a_{n+1}}notinmathbb{Q} Leftrightarrow f(n+1)=frac{a_{n+1}}{a_n+a_{n+1}}notinmathbb{Q}.$$ Suppose that $f(0)= frac{a_0}{a_{1}}$ is given by an irrational number. If $a_{n+2}=0$ for some $nge 0$, then $f(n)=-1$ holds and this contradicts $f(0)$ is irrational. In the same way, if $a_n=0$ for some $n<0$, then $f(n)=0$ holds and this contradicts $f(0)$ is irrational. So, the irrational initial data $f(0)$ guarantees that $f(n)$ is well-defined for all $ninmathbb{Z}$.
Let $a_0 = 1$ and $a_1=pi$. If we solve it for $A,B$, we get
$$
A=frac{beta-pi}{beta-alpha},quad B=frac{pi-alpha}{beta-alpha}.
$$ If we let $f(x) = frac{1}{pi}$ for $xin [0,1)$, this implies
$$
f(x+n) = frac{(beta-pi)alpha^n + (pi-alpha)beta^n}{(beta-pi)alpha^{n+1} + (pi-alpha)beta^{n+1}}.
$$ is well-defined for $ninmathbb{Z}$ and satisfies the given functional equation. This shows that
$$
f(x)=frac{(beta-pi)alpha^{lfloor xrfloor} + (pi-alpha)beta^{lfloor xrfloor}}{(beta-pi)alpha^{{lfloor xrfloor}+1} + (pi-alpha)beta^{{lfloor xrfloor}+1}}
$$ is an example of a discontinuous solution $f:mathbb{R}tomathbb{R}$.
answered Jan 10 at 22:39
SongSong
11.9k628
$endgroup$
$begingroup$
Thanks @Song for your answer. So if $f$ is a continuous function, it must be a constant one. Is it possible to construct a discontinuous function satisfying the conditions?
$endgroup$
– Fermat
Jan 10 at 22:50
$begingroup$
@Fermat I'm sorry that I have to go to bed now .. (it's a late night here) But I'll think about those issues later.
$endgroup$
– Song
Jan 10 at 22:55
$begingroup$
You are welcome. Thank you very much for your help.
$endgroup$
– Fermat
Jan 10 at 22:56
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Suppose $f:mathbb{R}tomathbb{R}$ is a continuous solution. Let us consider the sequence $f(n)$, $ninmathbb{Z}$. If we write $$f(n)=frac{a_n}{a_{n+1}},tag{*}$$ by the given recursion formula, we have
$$
f(n+1) = frac{a_{n+1}}{a_n+a_{n+1}}.
$$ This implies the sequence $a_n$ is a Fibonacci sequence, i.e. $a_{n+2} = a_{n+1}+a_n$. It is well-known that the general solution of $a_n$ is given by
$$
a_n = Aalpha^n + Bbeta^n
$$ for some $A,Binmathbb{R}$ where $alpha = frac{1+sqrt{5}}{2}$ and $beta=frac{1-sqrt{5}}{2}$. Plugging this into the expression in $(*)$, we get
$$
f(n) = frac{Aalpha^n + Bbeta^n}{Aalpha^{n+1} + Bbeta^{n+1}}.
$$ Assume that $A=0$. Since not both $A,B$ are $0$, we have $f(n) = beta^{-1}$. In the same way, $f(n)=alpha^{-1}$ holds if $B=0$. Now, assume that $A$ and $B$ are not zero. By taking limit as $ntopminfty$, we have
$$
lim_{ntoinfty}f(n) = alpha^{-1} = -beta>frac{1}{2},
$$ and
$$
lim_{nto-infty}f(n) =beta^{-1} = -alpha<-frac{3}{2}.
$$ But this implies by IVP that there exists $cinmathbb{R}$ such that $f(c)=0$. Since this is impossible as $f(x)ne 0$, there are only $2$ possibilities: Either $f(n) equiv alpha^{-1}$ or $f(n)equiv beta^{-1}$. Finally, notice that the above argument applies in the same way to $f(x+n)$ for arbitrary $xin mathbb{R}$. This implies $f(mathbb{R})subset {alpha^{-1}, beta^{-1}}$. By the continuity of $f$, this says that $f$ must be a constant function. Thus the only continuous solutions to the equation are
$$
f(x) equiv alpha^{-1} = frac{-1+sqrt{5}}{2}
$$ or
$$
f(x) equiv beta^{-1} = frac{-1-sqrt{5}}{2}.
$$
Note: Note that for all $ninmathbb{Z}$, $$f(n)=frac{a_n}{a_{n+1}}notinmathbb{Q} Leftrightarrow f(n+1)=frac{a_{n+1}}{a_n+a_{n+1}}notinmathbb{Q}.$$ Suppose that $f(0)= frac{a_0}{a_{1}}$ is given by an irrational number. If $a_{n+2}=0$ for some $nge 0$, then $f(n)=-1$ holds and this contradicts $f(0)$ is irrational. In the same way, if $a_n=0$ for some $n<0$, then $f(n)=0$ holds and this contradicts $f(0)$ is irrational. So, the irrational initial data $f(0)$ guarantees that $f(n)$ is well-defined for all $ninmathbb{Z}$.
Let $a_0 = 1$ and $a_1=pi$. If we solve it for $A,B$, we get
$$
A=frac{beta-pi}{beta-alpha},quad B=frac{pi-alpha}{beta-alpha}.
$$ If we let $f(x) = frac{1}{pi}$ for $xin [0,1)$, this implies
$$
f(x+n) = frac{(beta-pi)alpha^n + (pi-alpha)beta^n}{(beta-pi)alpha^{n+1} + (pi-alpha)beta^{n+1}}.
$$ is well-defined for $ninmathbb{Z}$ and satisfies the given functional equation. This shows that
$$
f(x)=frac{(beta-pi)alpha^{lfloor xrfloor} + (pi-alpha)beta^{lfloor xrfloor}}{(beta-pi)alpha^{{lfloor xrfloor}+1} + (pi-alpha)beta^{{lfloor xrfloor}+1}}
$$ is an example of a discontinuous solution $f:mathbb{R}tomathbb{R}$.
answered Jan 10 at 22:39
SongSong
11.9k628
$endgroup$
$begingroup$
Thanks @Song for your answer. So if $f$ is a continuous function, it must be a constant one. Is it possible to construct a discontinuous function satisfying the conditions?
$endgroup$
– Fermat
Jan 10 at 22:50
$begingroup$
@Fermat I'm sorry that I have to go to bed now .. (it's a late night here) But I'll think about those issues later.
$endgroup$
– Song
Jan 10 at 22:55
$begingroup$
You are welcome. Thank you very much for your help.
$endgroup$
– Fermat
Jan 10 at 22:56
add a comment |
$begingroup$
Suppose $f:mathbb{R}tomathbb{R}$ is a continuous solution. Let us consider the sequence $f(n)$, $ninmathbb{Z}$. If we write $$f(n)=frac{a_n}{a_{n+1}},tag{*}$$ by the given recursion formula, we have
$$
f(n+1) = frac{a_{n+1}}{a_n+a_{n+1}}.
$$ This implies the sequence $a_n$ is a Fibonacci sequence, i.e. $a_{n+2} = a_{n+1}+a_n$. It is well-known that the general solution of $a_n$ is given by
$$
a_n = Aalpha^n + Bbeta^n
$$ for some $A,Binmathbb{R}$ where $alpha = frac{1+sqrt{5}}{2}$ and $beta=frac{1-sqrt{5}}{2}$. Plugging this into the expression in $(*)$, we get
$$
f(n) = frac{Aalpha^n + Bbeta^n}{Aalpha^{n+1} + Bbeta^{n+1}}.
$$ Assume that $A=0$. Since not both $A,B$ are $0$, we have $f(n) = beta^{-1}$. In the same way, $f(n)=alpha^{-1}$ holds if $B=0$. Now, assume that $A$ and $B$ are not zero. By taking limit as $ntopminfty$, we have
$$
lim_{ntoinfty}f(n) = alpha^{-1} = -beta>frac{1}{2},
$$ and
$$
lim_{nto-infty}f(n) =beta^{-1} = -alpha<-frac{3}{2}.
$$ But this implies by IVP that there exists $cinmathbb{R}$ such that $f(c)=0$. Since this is impossible as $f(x)ne 0$, there are only $2$ possibilities: Either $f(n) equiv alpha^{-1}$ or $f(n)equiv beta^{-1}$. Finally, notice that the above argument applies in the same way to $f(x+n)$ for arbitrary $xin mathbb{R}$. This implies $f(mathbb{R})subset {alpha^{-1}, beta^{-1}}$. By the continuity of $f$, this says that $f$ must be a constant function. Thus the only continuous solutions to the equation are
$$
f(x) equiv alpha^{-1} = frac{-1+sqrt{5}}{2}
$$ or
$$
f(x) equiv beta^{-1} = frac{-1-sqrt{5}}{2}.
$$
Note: Note that for all $ninmathbb{Z}$, $$f(n)=frac{a_n}{a_{n+1}}notinmathbb{Q} Leftrightarrow f(n+1)=frac{a_{n+1}}{a_n+a_{n+1}}notinmathbb{Q}.$$ Suppose that $f(0)= frac{a_0}{a_{1}}$ is given by an irrational number. If $a_{n+2}=0$ for some $nge 0$, then $f(n)=-1$ holds and this contradicts $f(0)$ is irrational. In the same way, if $a_n=0$ for some $n<0$, then $f(n)=0$ holds and this contradicts $f(0)$ is irrational. So, the irrational initial data $f(0)$ guarantees that $f(n)$ is well-defined for all $ninmathbb{Z}$.
Let $a_0 = 1$ and $a_1=pi$. If we solve it for $A,B$, we get
$$
A=frac{beta-pi}{beta-alpha},quad B=frac{pi-alpha}{beta-alpha}.
$$ If we let $f(x) = frac{1}{pi}$ for $xin [0,1)$, this implies
$$
f(x+n) = frac{(beta-pi)alpha^n + (pi-alpha)beta^n}{(beta-pi)alpha^{n+1} + (pi-alpha)beta^{n+1}}.
$$ is well-defined for $ninmathbb{Z}$ and satisfies the given functional equation. This shows that
$$
f(x)=frac{(beta-pi)alpha^{lfloor xrfloor} + (pi-alpha)beta^{lfloor xrfloor}}{(beta-pi)alpha^{{lfloor xrfloor}+1} + (pi-alpha)beta^{{lfloor xrfloor}+1}}
$$ is an example of a discontinuous solution $f:mathbb{R}tomathbb{R}$.
answered Jan 10 at 22:39
SongSong
11.9k628
$endgroup$
$begingroup$
Thanks @Song for your answer. So if $f$ is a continuous function, it must be a constant one. Is it possible to construct a discontinuous function satisfying the conditions?
$endgroup$
– Fermat
Jan 10 at 22:50
$begingroup$
@Fermat I'm sorry that I have to go to bed now .. (it's a late night here) But I'll think about those issues later.
$endgroup$
– Song
Jan 10 at 22:55
$begingroup$
You are welcome. Thank you very much for your help.
$endgroup$
– Fermat
Jan 10 at 22:56
add a comment |
$begingroup$
Suppose $f:mathbb{R}tomathbb{R}$ is a continuous solution. Let us consider the sequence $f(n)$, $ninmathbb{Z}$. If we write $$f(n)=frac{a_n}{a_{n+1}},tag{*}$$ by the given recursion formula, we have
$$
f(n+1) = frac{a_{n+1}}{a_n+a_{n+1}}.
$$ This implies the sequence $a_n$ is a Fibonacci sequence, i.e. $a_{n+2} = a_{n+1}+a_n$. It is well-known that the general solution of $a_n$ is given by
$$
a_n = Aalpha^n + Bbeta^n
$$ for some $A,Binmathbb{R}$ where $alpha = frac{1+sqrt{5}}{2}$ and $beta=frac{1-sqrt{5}}{2}$. Plugging this into the expression in $(*)$, we get
$$
f(n) = frac{Aalpha^n + Bbeta^n}{Aalpha^{n+1} + Bbeta^{n+1}}.
$$ Assume that $A=0$. Since not both $A,B$ are $0$, we have $f(n) = beta^{-1}$. In the same way, $f(n)=alpha^{-1}$ holds if $B=0$. Now, assume that $A$ and $B$ are not zero. By taking limit as $ntopminfty$, we have
$$
lim_{ntoinfty}f(n) = alpha^{-1} = -beta>frac{1}{2},
$$ and
$$
lim_{nto-infty}f(n) =beta^{-1} = -alpha<-frac{3}{2}.
$$ But this implies by IVP that there exists $cinmathbb{R}$ such that $f(c)=0$. Since this is impossible as $f(x)ne 0$, there are only $2$ possibilities: Either $f(n) equiv alpha^{-1}$ or $f(n)equiv beta^{-1}$. Finally, notice that the above argument applies in the same way to $f(x+n)$ for arbitrary $xin mathbb{R}$. This implies $f(mathbb{R})subset {alpha^{-1}, beta^{-1}}$. By the continuity of $f$, this says that $f$ must be a constant function. Thus the only continuous solutions to the equation are
$$
f(x) equiv alpha^{-1} = frac{-1+sqrt{5}}{2}
$$ or
$$
f(x) equiv beta^{-1} = frac{-1-sqrt{5}}{2}.
$$
Note: Note that for all $ninmathbb{Z}$, $$f(n)=frac{a_n}{a_{n+1}}notinmathbb{Q} Leftrightarrow f(n+1)=frac{a_{n+1}}{a_n+a_{n+1}}notinmathbb{Q}.$$ Suppose that $f(0)= frac{a_0}{a_{1}}$ is given by an irrational number. If $a_{n+2}=0$ for some $nge 0$, then $f(n)=-1$ holds and this contradicts $f(0)$ is irrational. In the same way, if $a_n=0$ for some $n<0$, then $f(n)=0$ holds and this contradicts $f(0)$ is irrational. So, the irrational initial data $f(0)$ guarantees that $f(n)$ is well-defined for all $ninmathbb{Z}$.
Let $a_0 = 1$ and $a_1=pi$. If we solve it for $A,B$, we get
$$
A=frac{beta-pi}{beta-alpha},quad B=frac{pi-alpha}{beta-alpha}.
$$ If we let $f(x) = frac{1}{pi}$ for $xin [0,1)$, this implies
$$
f(x+n) = frac{(beta-pi)alpha^n + (pi-alpha)beta^n}{(beta-pi)alpha^{n+1} + (pi-alpha)beta^{n+1}}.
$$ is well-defined for $ninmathbb{Z}$ and satisfies the given functional equation. This shows that
$$
f(x)=frac{(beta-pi)alpha^{lfloor xrfloor} + (pi-alpha)beta^{lfloor xrfloor}}{(beta-pi)alpha^{{lfloor xrfloor}+1} + (pi-alpha)beta^{{lfloor xrfloor}+1}}
$$ is an example of a discontinuous solution $f:mathbb{R}tomathbb{R}$.
answered Jan 10 at 22:39
SongSong
11.9k628
$endgroup$
Suppose $f:mathbb{R}tomathbb{R}$ is a continuous solution. Let us consider the sequence $f(n)$, $ninmathbb{Z}$. If we write $$f(n)=frac{a_n}{a_{n+1}},tag{*}$$ by the given recursion formula, we have
$$
f(n+1) = frac{a_{n+1}}{a_n+a_{n+1}}.
$$ This implies the sequence $a_n$ is a Fibonacci sequence, i.e. $a_{n+2} = a_{n+1}+a_n$. It is well-known that the general solution of $a_n$ is given by
$$
a_n = Aalpha^n + Bbeta^n
$$ for some $A,Binmathbb{R}$ where $alpha = frac{1+sqrt{5}}{2}$ and $beta=frac{1-sqrt{5}}{2}$. Plugging this into the expression in $(*)$, we get
$$
f(n) = frac{Aalpha^n + Bbeta^n}{Aalpha^{n+1} + Bbeta^{n+1}}.
$$ Assume that $A=0$. Since not both $A,B$ are $0$, we have $f(n) = beta^{-1}$. In the same way, $f(n)=alpha^{-1}$ holds if $B=0$. Now, assume that $A$ and $B$ are not zero. By taking limit as $ntopminfty$, we have
$$
lim_{ntoinfty}f(n) = alpha^{-1} = -beta>frac{1}{2},
$$ and
$$
lim_{nto-infty}f(n) =beta^{-1} = -alpha<-frac{3}{2}.
$$ But this implies by IVP that there exists $cinmathbb{R}$ such that $f(c)=0$. Since this is impossible as $f(x)ne 0$, there are only $2$ possibilities: Either $f(n) equiv alpha^{-1}$ or $f(n)equiv beta^{-1}$. Finally, notice that the above argument applies in the same way to $f(x+n)$ for arbitrary $xin mathbb{R}$. This implies $f(mathbb{R})subset {alpha^{-1}, beta^{-1}}$. By the continuity of $f$, this says that $f$ must be a constant function. Thus the only continuous solutions to the equation are
$$
f(x) equiv alpha^{-1} = frac{-1+sqrt{5}}{2}
$$ or
$$
f(x) equiv beta^{-1} = frac{-1-sqrt{5}}{2}.
$$
Note: Note that for all $ninmathbb{Z}$, $$f(n)=frac{a_n}{a_{n+1}}notinmathbb{Q} Leftrightarrow f(n+1)=frac{a_{n+1}}{a_n+a_{n+1}}notinmathbb{Q}.$$ Suppose that $f(0)= frac{a_0}{a_{1}}$ is given by an irrational number. If $a_{n+2}=0$ for some $nge 0$, then $f(n)=-1$ holds and this contradicts $f(0)$ is irrational. In the same way, if $a_n=0$ for some $n<0$, then $f(n)=0$ holds and this contradicts $f(0)$ is irrational. So, the irrational initial data $f(0)$ guarantees that $f(n)$ is well-defined for all $ninmathbb{Z}$.
Let $a_0 = 1$ and $a_1=pi$. If we solve it for $A,B$, we get
$$
A=frac{beta-pi}{beta-alpha},quad B=frac{pi-alpha}{beta-alpha}.
$$ If we let $f(x) = frac{1}{pi}$ for $xin [0,1)$, this implies
$$
f(x+n) = frac{(beta-pi)alpha^n + (pi-alpha)beta^n}{(beta-pi)alpha^{n+1} + (pi-alpha)beta^{n+1}}.
$$ is well-defined for $ninmathbb{Z}$ and satisfies the given functional equation. This shows that
$$
f(x)=frac{(beta-pi)alpha^{lfloor xrfloor} + (pi-alpha)beta^{lfloor xrfloor}}{(beta-pi)alpha^{{lfloor xrfloor}+1} + (pi-alpha)beta^{{lfloor xrfloor}+1}}
$$ is an example of a discontinuous solution $f:mathbb{R}tomathbb{R}$.
answered Jan 10 at 22:39
SongSong
11.9k628
edited Jan 11 at 10:54
answered Jan 10 at 22:39
SongSong
11.9k628
answered Jan 10 at 22:39
SongSong
11.9k628
answered Jan 10 at 22:39
SongSong
11.9k628
11.9k628
$begingroup$
Thanks @Song for your answer. So if $f$ is a continuous function, it must be a constant one. Is it possible to construct a discontinuous function satisfying the conditions?
$endgroup$
– Fermat
Jan 10 at 22:50
$begingroup$
@Fermat I'm sorry that I have to go to bed now .. (it's a late night here) But I'll think about those issues later.
$endgroup$
– Song
Jan 10 at 22:55
$begingroup$
You are welcome. Thank you very much for your help.
$endgroup$
– Fermat
Jan 10 at 22:56
add a comment |
$begingroup$
Thanks @Song for your answer. So if $f$ is a continuous function, it must be a constant one. Is it possible to construct a discontinuous function satisfying the conditions?
$endgroup$
– Fermat
Jan 10 at 22:50
$begingroup$
@Fermat I'm sorry that I have to go to bed now .. (it's a late night here) But I'll think about those issues later.
$endgroup$
– Song
Jan 10 at 22:55
$begingroup$
You are welcome. Thank you very much for your help.
$endgroup$
– Fermat
Jan 10 at 22:56
$begingroup$
Thanks @Song for your answer. So if $f$ is a continuous function, it must be a constant one. Is it possible to construct a discontinuous function satisfying the conditions?
$endgroup$
– Fermat
Jan 10 at 22:50
$begingroup$
Thanks @Song for your answer. So if $f$ is a continuous function, it must be a constant one. Is it possible to construct a discontinuous function satisfying the conditions?
$endgroup$
– Fermat
Jan 10 at 22:50
$begingroup$
@Fermat I'm sorry that I have to go to bed now .. (it's a late night here) But I'll think about those issues later.
$endgroup$
– Song
Jan 10 at 22:55
$begingroup$
@Fermat I'm sorry that I have to go to bed now .. (it's a late night here) But I'll think about those issues later.
$endgroup$
– Song
Jan 10 at 22:55
$begingroup$
You are welcome. Thank you very much for your help.
$endgroup$
– Fermat
Jan 10 at 22:56
$begingroup$
You are welcome. Thank you very much for your help.
$endgroup$
– Fermat
Jan 10 at 22:56
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