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If $g'$ is a function of $g$, then $g$ is monotonic
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Is this true or false?
Let $g:[0,1] to Bbb R$ be a function continuous on $[0,1]$ and differentiable on $(0,1)$, such that $g'$ is a function of $g$, i.e. for every $x, y in (0,1)$, if $g(x) = g(y)$ then $g'(x) = g'(y)$. Then $g$ is monotonic.
I posted my proof as a self-answer, and wonder if there are any easier proofs.
real-analysis
asked Aug 2 at 9:45
Kenny Lau
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Is this true or false?
Let $g:[0,1] to Bbb R$ be a function continuous on $[0,1]$ and differentiable on $(0,1)$, such that $g'$ is a function of $g$, i.e. for every $x, y in (0,1)$, if $g(x) = g(y)$ then $g'(x) = g'(y)$. Then $g$ is monotonic.
I posted my proof as a self-answer, and wonder if there are any easier proofs.
real-analysis
asked Aug 2 at 9:45
Kenny Lau
17.7k2156
add a comment |Â
up vote
10
down vote
favorite
up vote
10
down vote
favorite
Is this true or false?
Let $g:[0,1] to Bbb R$ be a function continuous on $[0,1]$ and differentiable on $(0,1)$, such that $g'$ is a function of $g$, i.e. for every $x, y in (0,1)$, if $g(x) = g(y)$ then $g'(x) = g'(y)$. Then $g$ is monotonic.
I posted my proof as a self-answer, and wonder if there are any easier proofs.
real-analysis
asked Aug 2 at 9:45
Kenny Lau
17.7k2156
Is this true or false?
Let $g:[0,1] to Bbb R$ be a function continuous on $[0,1]$ and differentiable on $(0,1)$, such that $g'$ is a function of $g$, i.e. for every $x, y in (0,1)$, if $g(x) = g(y)$ then $g'(x) = g'(y)$. Then $g$ is monotonic.
I posted my proof as a self-answer, and wonder if there are any easier proofs.
real-analysis
asked Aug 2 at 9:45
Kenny Lau
17.7k2156
asked Aug 2 at 9:45
Kenny Lau
17.7k2156
asked Aug 2 at 9:45
Kenny Lau
17.7k2156
asked Aug 2 at 9:45
Kenny Lau
17.7k2156
17.7k2156
add a comment |Â
add a comment |Â
2 Answers
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If $g$ is not monotone, then neither is the restriction $defepsepsilong|_smash(0,1)$. Non-monotonicity on $(0,1)$ implies the existence of $0<a<b<1$ with (w.l.o.g.)
$$gamma :=g(a)=g(b)implies g'(a)=g'(b)=:gamma'>0$$
(I can go into details here if needed, but it is not too hard to see using the intermediate value theorem and the mean value theorem, there are just a lot of annoying details pay attention to). So there is some $defepsepsiloneps>0$ with
$$g(a+x)>gammaquadtextandquad g(b-x)<gamma,qquadtextfor all $xin[0,eps]$.$$
Define $M:=xin [a+eps,b-eps]mid g(x)< gamma$. Note that $M$ is non-empty, hence we can define $a^*:=inf Min(a,b)$. By continuity, we have $g(a^*)=gammaimplies g'(a^*)=gamma '>0$, implying the existence of $ delta>0$ with
$$g(a^*+x) > gamma,qquadtextfor all $xin[0,delta]$$$
in contradiction to the choice $a^*:=inf M$.
$qquadqquadqquadqquad$
answered Aug 2 at 16:42
M. Winter
17.5k62664
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Lemma: for any $d in (0,1)$, if $g'(d) ne 0$ and $g(d) ne g(0)$ and $g(d) ne g(1)$, then $ x in [0,1] mid g(x) = g(d) $ is closed, hence compact; discrete, hence finite; and finally is a singleton.
Proof: WLOG $g'(d) > 0$ (the proof of $g'(d) < 0$ is similar). It is clear that the set is closed, hence compact. Now, if $g(x) = g(d)$, then $g'(x) = g'(d) > 0$, so taking $varepsilon = g'(x)/2$, we obtain $delta$ such that $left| dfrac g(x+h) - g(x) h - g'(x) right| < dfrac g'(x) 2$ whenever $h in (-delta, delta)$, so this is a neighbourhood of $x$ that contains only $x$ from the set, so the set is discrete. Since the set is compact and discrete, it follows that it is finite. Now that it is finite, if it has two distinct points, then there are two distinct points $x_1$ and $x_2$ from the set that are next to each other. But again using epsilon-delta on the derivative of the two points we would find two points between $x_1$ and $x_2$ such that the function is above $g(d)$ at one point and below $g(d)$ at another, and by IVT on the function it would have to pass through $g(d)$ again, contradicting the fact that $x_1$ and $x_2$ are the closest two points from the set.
If the function has positive derivative $g'(x)$ at a point $x$, then there is an open neighbourhood of the point where the function is trapped between two lines with slope $frac12 g'(x)$ and $frac32 g'(x)$:
WLOG assume $g(0) le g(1)$, and we shall prove that $g$ is increasing.
Step 1: $g(1)$ is a global maximum.
Proof: Otherwise, there is $x_0 in (0,1)$ with $g(x_0) > g(1)$ being the global maximum. To establish a contradiction, we find $x_R in (x_0, 1)$ such that $g(1) < g(x_R) < g(x_0)$ and $g'(x_R) < 0$. This is a contradiction, as $g(0) < g(x_R) < g(x_0)$, so by IVT there should be $x_L in (0,x_0)$ such that $g(x_L) = g(x_R)$, but the lemma says that the set $ x in [0,1] mid g(x) = g(x_R) $ is a singleton.
We now find the required $x_R$. First, let $x_1$ be the minimum value after $x_0$ such that $g(x_1) = frac12[g(x_0) + g(1)]$, which exists because $ x in [x_0,1] mid g(x) = frac12[g(x_0) + g(1)] $ is closed (hence compact) and non-empty (by IVT). Trivially we have $x_0 < x_1$, and that $g(x) > frac12[g(x_0) + g(1)]$ whenever $x in (x_0, x_1)$. We now obtain $x_R$ through MVT at the points $x_0$ and $x_1$.
Step 2: $g(0)$ is a global minimum.
Proof: Similar to Step 1.
Step 3: $g$ is increasing.
Otherwise, there would be $c<d$ such that $g(c) > g(d)$. We let $e$ be the smallest value after $c$ such that $g(e) = g(d)$, and then it is clear that $c < e$ and that $g(x) in (g(d), g(c))$ whenever $x in (c,e)$. We use MVT on $c$ and $e$ to obtain $x_R$ such that $g(d) < g(x_R) < g(c)$ and $g'(x_R) < 0$ as before, and derive a contradiction as before, noting that there must be $x_L in (0,c)$ such that $g(x_L) = g(x_R)$.
answered Aug 2 at 9:45
Kenny Lau
17.7k2156
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2 Answers
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
If $g$ is not monotone, then neither is the restriction $defepsepsilong|_smash(0,1)$. Non-monotonicity on $(0,1)$ implies the existence of $0<a<b<1$ with (w.l.o.g.)
$$gamma :=g(a)=g(b)implies g'(a)=g'(b)=:gamma'>0$$
(I can go into details here if needed, but it is not too hard to see using the intermediate value theorem and the mean value theorem, there are just a lot of annoying details pay attention to). So there is some $defepsepsiloneps>0$ with
$$g(a+x)>gammaquadtextandquad g(b-x)<gamma,qquadtextfor all $xin[0,eps]$.$$
Define $M:=xin [a+eps,b-eps]mid g(x)< gamma$. Note that $M$ is non-empty, hence we can define $a^*:=inf Min(a,b)$. By continuity, we have $g(a^*)=gammaimplies g'(a^*)=gamma '>0$, implying the existence of $ delta>0$ with
$$g(a^*+x) > gamma,qquadtextfor all $xin[0,delta]$$$
in contradiction to the choice $a^*:=inf M$.
$qquadqquadqquadqquad$
answered Aug 2 at 16:42
M. Winter
17.5k62664
add a comment |Â
up vote
3
down vote
If $g$ is not monotone, then neither is the restriction $defepsepsilong|_smash(0,1)$. Non-monotonicity on $(0,1)$ implies the existence of $0<a<b<1$ with (w.l.o.g.)
$$gamma :=g(a)=g(b)implies g'(a)=g'(b)=:gamma'>0$$
(I can go into details here if needed, but it is not too hard to see using the intermediate value theorem and the mean value theorem, there are just a lot of annoying details pay attention to). So there is some $defepsepsiloneps>0$ with
$$g(a+x)>gammaquadtextandquad g(b-x)<gamma,qquadtextfor all $xin[0,eps]$.$$
Define $M:=xin [a+eps,b-eps]mid g(x)< gamma$. Note that $M$ is non-empty, hence we can define $a^*:=inf Min(a,b)$. By continuity, we have $g(a^*)=gammaimplies g'(a^*)=gamma '>0$, implying the existence of $ delta>0$ with
$$g(a^*+x) > gamma,qquadtextfor all $xin[0,delta]$$$
in contradiction to the choice $a^*:=inf M$.
$qquadqquadqquadqquad$
answered Aug 2 at 16:42
M. Winter
17.5k62664
add a comment |Â
up vote
3
down vote
up vote
3
down vote
If $g$ is not monotone, then neither is the restriction $defepsepsilong|_smash(0,1)$. Non-monotonicity on $(0,1)$ implies the existence of $0<a<b<1$ with (w.l.o.g.)
$$gamma :=g(a)=g(b)implies g'(a)=g'(b)=:gamma'>0$$
(I can go into details here if needed, but it is not too hard to see using the intermediate value theorem and the mean value theorem, there are just a lot of annoying details pay attention to). So there is some $defepsepsiloneps>0$ with
$$g(a+x)>gammaquadtextandquad g(b-x)<gamma,qquadtextfor all $xin[0,eps]$.$$
Define $M:=xin [a+eps,b-eps]mid g(x)< gamma$. Note that $M$ is non-empty, hence we can define $a^*:=inf Min(a,b)$. By continuity, we have $g(a^*)=gammaimplies g'(a^*)=gamma '>0$, implying the existence of $ delta>0$ with
$$g(a^*+x) > gamma,qquadtextfor all $xin[0,delta]$$$
in contradiction to the choice $a^*:=inf M$.
$qquadqquadqquadqquad$
answered Aug 2 at 16:42
M. Winter
17.5k62664
If $g$ is not monotone, then neither is the restriction $defepsepsilong|_smash(0,1)$. Non-monotonicity on $(0,1)$ implies the existence of $0<a<b<1$ with (w.l.o.g.)
$$gamma :=g(a)=g(b)implies g'(a)=g'(b)=:gamma'>0$$
(I can go into details here if needed, but it is not too hard to see using the intermediate value theorem and the mean value theorem, there are just a lot of annoying details pay attention to). So there is some $defepsepsiloneps>0$ with
$$g(a+x)>gammaquadtextandquad g(b-x)<gamma,qquadtextfor all $xin[0,eps]$.$$
Define $M:=xin [a+eps,b-eps]mid g(x)< gamma$. Note that $M$ is non-empty, hence we can define $a^*:=inf Min(a,b)$. By continuity, we have $g(a^*)=gammaimplies g'(a^*)=gamma '>0$, implying the existence of $ delta>0$ with
$$g(a^*+x) > gamma,qquadtextfor all $xin[0,delta]$$$
in contradiction to the choice $a^*:=inf M$.
$qquadqquadqquadqquad$
answered Aug 2 at 16:42
M. Winter
17.5k62664
edited Aug 2 at 20:06
answered Aug 2 at 16:42
M. Winter
17.5k62664
answered Aug 2 at 16:42
M. Winter
17.5k62664
answered Aug 2 at 16:42
M. Winter
17.5k62664
17.5k62664
add a comment |Â
add a comment |Â
up vote
0
down vote
Lemma: for any $d in (0,1)$, if $g'(d) ne 0$ and $g(d) ne g(0)$ and $g(d) ne g(1)$, then $ x in [0,1] mid g(x) = g(d) $ is closed, hence compact; discrete, hence finite; and finally is a singleton.
Proof: WLOG $g'(d) > 0$ (the proof of $g'(d) < 0$ is similar). It is clear that the set is closed, hence compact. Now, if $g(x) = g(d)$, then $g'(x) = g'(d) > 0$, so taking $varepsilon = g'(x)/2$, we obtain $delta$ such that $left| dfrac g(x+h) - g(x) h - g'(x) right| < dfrac g'(x) 2$ whenever $h in (-delta, delta)$, so this is a neighbourhood of $x$ that contains only $x$ from the set, so the set is discrete. Since the set is compact and discrete, it follows that it is finite. Now that it is finite, if it has two distinct points, then there are two distinct points $x_1$ and $x_2$ from the set that are next to each other. But again using epsilon-delta on the derivative of the two points we would find two points between $x_1$ and $x_2$ such that the function is above $g(d)$ at one point and below $g(d)$ at another, and by IVT on the function it would have to pass through $g(d)$ again, contradicting the fact that $x_1$ and $x_2$ are the closest two points from the set.
If the function has positive derivative $g'(x)$ at a point $x$, then there is an open neighbourhood of the point where the function is trapped between two lines with slope $frac12 g'(x)$ and $frac32 g'(x)$:
WLOG assume $g(0) le g(1)$, and we shall prove that $g$ is increasing.
Step 1: $g(1)$ is a global maximum.
Proof: Otherwise, there is $x_0 in (0,1)$ with $g(x_0) > g(1)$ being the global maximum. To establish a contradiction, we find $x_R in (x_0, 1)$ such that $g(1) < g(x_R) < g(x_0)$ and $g'(x_R) < 0$. This is a contradiction, as $g(0) < g(x_R) < g(x_0)$, so by IVT there should be $x_L in (0,x_0)$ such that $g(x_L) = g(x_R)$, but the lemma says that the set $ x in [0,1] mid g(x) = g(x_R) $ is a singleton.
We now find the required $x_R$. First, let $x_1$ be the minimum value after $x_0$ such that $g(x_1) = frac12[g(x_0) + g(1)]$, which exists because $ x in [x_0,1] mid g(x) = frac12[g(x_0) + g(1)] $ is closed (hence compact) and non-empty (by IVT). Trivially we have $x_0 < x_1$, and that $g(x) > frac12[g(x_0) + g(1)]$ whenever $x in (x_0, x_1)$. We now obtain $x_R$ through MVT at the points $x_0$ and $x_1$.
Step 2: $g(0)$ is a global minimum.
Proof: Similar to Step 1.
Step 3: $g$ is increasing.
Otherwise, there would be $c<d$ such that $g(c) > g(d)$. We let $e$ be the smallest value after $c$ such that $g(e) = g(d)$, and then it is clear that $c < e$ and that $g(x) in (g(d), g(c))$ whenever $x in (c,e)$. We use MVT on $c$ and $e$ to obtain $x_R$ such that $g(d) < g(x_R) < g(c)$ and $g'(x_R) < 0$ as before, and derive a contradiction as before, noting that there must be $x_L in (0,c)$ such that $g(x_L) = g(x_R)$.
answered Aug 2 at 9:45
Kenny Lau
17.7k2156
add a comment |Â
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Lemma: for any $d in (0,1)$, if $g'(d) ne 0$ and $g(d) ne g(0)$ and $g(d) ne g(1)$, then $ x in [0,1] mid g(x) = g(d) $ is closed, hence compact; discrete, hence finite; and finally is a singleton.
Proof: WLOG $g'(d) > 0$ (the proof of $g'(d) < 0$ is similar). It is clear that the set is closed, hence compact. Now, if $g(x) = g(d)$, then $g'(x) = g'(d) > 0$, so taking $varepsilon = g'(x)/2$, we obtain $delta$ such that $left| dfrac g(x+h) - g(x) h - g'(x) right| < dfrac g'(x) 2$ whenever $h in (-delta, delta)$, so this is a neighbourhood of $x$ that contains only $x$ from the set, so the set is discrete. Since the set is compact and discrete, it follows that it is finite. Now that it is finite, if it has two distinct points, then there are two distinct points $x_1$ and $x_2$ from the set that are next to each other. But again using epsilon-delta on the derivative of the two points we would find two points between $x_1$ and $x_2$ such that the function is above $g(d)$ at one point and below $g(d)$ at another, and by IVT on the function it would have to pass through $g(d)$ again, contradicting the fact that $x_1$ and $x_2$ are the closest two points from the set.
If the function has positive derivative $g'(x)$ at a point $x$, then there is an open neighbourhood of the point where the function is trapped between two lines with slope $frac12 g'(x)$ and $frac32 g'(x)$:
WLOG assume $g(0) le g(1)$, and we shall prove that $g$ is increasing.
Step 1: $g(1)$ is a global maximum.
Proof: Otherwise, there is $x_0 in (0,1)$ with $g(x_0) > g(1)$ being the global maximum. To establish a contradiction, we find $x_R in (x_0, 1)$ such that $g(1) < g(x_R) < g(x_0)$ and $g'(x_R) < 0$. This is a contradiction, as $g(0) < g(x_R) < g(x_0)$, so by IVT there should be $x_L in (0,x_0)$ such that $g(x_L) = g(x_R)$, but the lemma says that the set $ x in [0,1] mid g(x) = g(x_R) $ is a singleton.
We now find the required $x_R$. First, let $x_1$ be the minimum value after $x_0$ such that $g(x_1) = frac12[g(x_0) + g(1)]$, which exists because $ x in [x_0,1] mid g(x) = frac12[g(x_0) + g(1)] $ is closed (hence compact) and non-empty (by IVT). Trivially we have $x_0 < x_1$, and that $g(x) > frac12[g(x_0) + g(1)]$ whenever $x in (x_0, x_1)$. We now obtain $x_R$ through MVT at the points $x_0$ and $x_1$.
Step 2: $g(0)$ is a global minimum.
Proof: Similar to Step 1.
Step 3: $g$ is increasing.
Otherwise, there would be $c<d$ such that $g(c) > g(d)$. We let $e$ be the smallest value after $c$ such that $g(e) = g(d)$, and then it is clear that $c < e$ and that $g(x) in (g(d), g(c))$ whenever $x in (c,e)$. We use MVT on $c$ and $e$ to obtain $x_R$ such that $g(d) < g(x_R) < g(c)$ and $g'(x_R) < 0$ as before, and derive a contradiction as before, noting that there must be $x_L in (0,c)$ such that $g(x_L) = g(x_R)$.
answered Aug 2 at 9:45
Kenny Lau
17.7k2156
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Lemma: for any $d in (0,1)$, if $g'(d) ne 0$ and $g(d) ne g(0)$ and $g(d) ne g(1)$, then $ x in [0,1] mid g(x) = g(d) $ is closed, hence compact; discrete, hence finite; and finally is a singleton.
Proof: WLOG $g'(d) > 0$ (the proof of $g'(d) < 0$ is similar). It is clear that the set is closed, hence compact. Now, if $g(x) = g(d)$, then $g'(x) = g'(d) > 0$, so taking $varepsilon = g'(x)/2$, we obtain $delta$ such that $left| dfrac g(x+h) - g(x) h - g'(x) right| < dfrac g'(x) 2$ whenever $h in (-delta, delta)$, so this is a neighbourhood of $x$ that contains only $x$ from the set, so the set is discrete. Since the set is compact and discrete, it follows that it is finite. Now that it is finite, if it has two distinct points, then there are two distinct points $x_1$ and $x_2$ from the set that are next to each other. But again using epsilon-delta on the derivative of the two points we would find two points between $x_1$ and $x_2$ such that the function is above $g(d)$ at one point and below $g(d)$ at another, and by IVT on the function it would have to pass through $g(d)$ again, contradicting the fact that $x_1$ and $x_2$ are the closest two points from the set.
If the function has positive derivative $g'(x)$ at a point $x$, then there is an open neighbourhood of the point where the function is trapped between two lines with slope $frac12 g'(x)$ and $frac32 g'(x)$:
WLOG assume $g(0) le g(1)$, and we shall prove that $g$ is increasing.
Step 1: $g(1)$ is a global maximum.
Proof: Otherwise, there is $x_0 in (0,1)$ with $g(x_0) > g(1)$ being the global maximum. To establish a contradiction, we find $x_R in (x_0, 1)$ such that $g(1) < g(x_R) < g(x_0)$ and $g'(x_R) < 0$. This is a contradiction, as $g(0) < g(x_R) < g(x_0)$, so by IVT there should be $x_L in (0,x_0)$ such that $g(x_L) = g(x_R)$, but the lemma says that the set $ x in [0,1] mid g(x) = g(x_R) $ is a singleton.
We now find the required $x_R$. First, let $x_1$ be the minimum value after $x_0$ such that $g(x_1) = frac12[g(x_0) + g(1)]$, which exists because $ x in [x_0,1] mid g(x) = frac12[g(x_0) + g(1)] $ is closed (hence compact) and non-empty (by IVT). Trivially we have $x_0 < x_1$, and that $g(x) > frac12[g(x_0) + g(1)]$ whenever $x in (x_0, x_1)$. We now obtain $x_R$ through MVT at the points $x_0$ and $x_1$.
Step 2: $g(0)$ is a global minimum.
Proof: Similar to Step 1.
Step 3: $g$ is increasing.
Otherwise, there would be $c<d$ such that $g(c) > g(d)$. We let $e$ be the smallest value after $c$ such that $g(e) = g(d)$, and then it is clear that $c < e$ and that $g(x) in (g(d), g(c))$ whenever $x in (c,e)$. We use MVT on $c$ and $e$ to obtain $x_R$ such that $g(d) < g(x_R) < g(c)$ and $g'(x_R) < 0$ as before, and derive a contradiction as before, noting that there must be $x_L in (0,c)$ such that $g(x_L) = g(x_R)$.
answered Aug 2 at 9:45
Kenny Lau
17.7k2156
Lemma: for any $d in (0,1)$, if $g'(d) ne 0$ and $g(d) ne g(0)$ and $g(d) ne g(1)$, then $ x in [0,1] mid g(x) = g(d) $ is closed, hence compact; discrete, hence finite; and finally is a singleton.
Proof: WLOG $g'(d) > 0$ (the proof of $g'(d) < 0$ is similar). It is clear that the set is closed, hence compact. Now, if $g(x) = g(d)$, then $g'(x) = g'(d) > 0$, so taking $varepsilon = g'(x)/2$, we obtain $delta$ such that $left| dfrac g(x+h) - g(x) h - g'(x) right| < dfrac g'(x) 2$ whenever $h in (-delta, delta)$, so this is a neighbourhood of $x$ that contains only $x$ from the set, so the set is discrete. Since the set is compact and discrete, it follows that it is finite. Now that it is finite, if it has two distinct points, then there are two distinct points $x_1$ and $x_2$ from the set that are next to each other. But again using epsilon-delta on the derivative of the two points we would find two points between $x_1$ and $x_2$ such that the function is above $g(d)$ at one point and below $g(d)$ at another, and by IVT on the function it would have to pass through $g(d)$ again, contradicting the fact that $x_1$ and $x_2$ are the closest two points from the set.
If the function has positive derivative $g'(x)$ at a point $x$, then there is an open neighbourhood of the point where the function is trapped between two lines with slope $frac12 g'(x)$ and $frac32 g'(x)$:
WLOG assume $g(0) le g(1)$, and we shall prove that $g$ is increasing.
Step 1: $g(1)$ is a global maximum.
Proof: Otherwise, there is $x_0 in (0,1)$ with $g(x_0) > g(1)$ being the global maximum. To establish a contradiction, we find $x_R in (x_0, 1)$ such that $g(1) < g(x_R) < g(x_0)$ and $g'(x_R) < 0$. This is a contradiction, as $g(0) < g(x_R) < g(x_0)$, so by IVT there should be $x_L in (0,x_0)$ such that $g(x_L) = g(x_R)$, but the lemma says that the set $ x in [0,1] mid g(x) = g(x_R) $ is a singleton.
We now find the required $x_R$. First, let $x_1$ be the minimum value after $x_0$ such that $g(x_1) = frac12[g(x_0) + g(1)]$, which exists because $ x in [x_0,1] mid g(x) = frac12[g(x_0) + g(1)] $ is closed (hence compact) and non-empty (by IVT). Trivially we have $x_0 < x_1$, and that $g(x) > frac12[g(x_0) + g(1)]$ whenever $x in (x_0, x_1)$. We now obtain $x_R$ through MVT at the points $x_0$ and $x_1$.
Step 2: $g(0)$ is a global minimum.
Proof: Similar to Step 1.
Step 3: $g$ is increasing.
Otherwise, there would be $c<d$ such that $g(c) > g(d)$. We let $e$ be the smallest value after $c$ such that $g(e) = g(d)$, and then it is clear that $c < e$ and that $g(x) in (g(d), g(c))$ whenever $x in (c,e)$. We use MVT on $c$ and $e$ to obtain $x_R$ such that $g(d) < g(x_R) < g(c)$ and $g'(x_R) < 0$ as before, and derive a contradiction as before, noting that there must be $x_L in (0,c)$ such that $g(x_L) = g(x_R)$.
answered Aug 2 at 9:45
Kenny Lau
17.7k2156
answered Aug 2 at 9:45
Kenny Lau
17.7k2156
answered Aug 2 at 9:45
Kenny Lau
17.7k2156
answered Aug 2 at 9:45
Kenny Lau
17.7k2156
17.7k2156
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