Equivalent definitions of convexity
$begingroup$
We defined the convexity of an $f: mathbb{R} to mathbb{R}$ function like this:
$f$ is convex on the interval $I$ if $forall a,bin
I$, $a<b$, $forall x in I$
$$f(x)leqslant f(a)+frac{f(b)-f(a)}{b-a}(x-a)$$
And then we had the following theorem:
The following statements are equivalent:
(i) $f$ is convex on $I$
(ii) $forall a in I$, $m_a(x)=frac{f(x)-f(a)}{x-a}$ is monotone increasing on $Isetminus{a}$
I tried to prove (i)$implies$(ii) like this: Let $a,x,y in I$. If $a<x<y$, then
$$f(x) leqslant f(a) + frac{f(y)-f(a)}{y-a}(x-a)$$
$$frac{f(x)-f(a)}{x-a} leqslant frac{f(y)-f(a)}{y-a}$$
$$m_a(x) leqslant m_a(y)$$
Which is fine, but when $x<a<y$, we have that $x-a leqslant 0$, so
$$frac{f(x)-f(a)}{x-a} geqslant frac{f(y)-f(a)}{y-a}$$
$$m_a(x) geqslant m_a(y)$$
And it's not good. Did I miss something or is the theorem false as it is stated?
real-analysis convex-analysis
asked Dec 20 '18 at 13:07
BotondBotond
5,8532832
$endgroup$
|
show 2 more comments
$begingroup$
We defined the convexity of an $f: mathbb{R} to mathbb{R}$ function like this:
$f$ is convex on the interval $I$ if $forall a,bin
I$, $a<b$, $forall x in I$
$$f(x)leqslant f(a)+frac{f(b)-f(a)}{b-a}(x-a)$$
And then we had the following theorem:
The following statements are equivalent:
(i) $f$ is convex on $I$
(ii) $forall a in I$, $m_a(x)=frac{f(x)-f(a)}{x-a}$ is monotone increasing on $Isetminus{a}$
I tried to prove (i)$implies$(ii) like this: Let $a,x,y in I$. If $a<x<y$, then
$$f(x) leqslant f(a) + frac{f(y)-f(a)}{y-a}(x-a)$$
$$frac{f(x)-f(a)}{x-a} leqslant frac{f(y)-f(a)}{y-a}$$
$$m_a(x) leqslant m_a(y)$$
Which is fine, but when $x<a<y$, we have that $x-a leqslant 0$, so
$$frac{f(x)-f(a)}{x-a} geqslant frac{f(y)-f(a)}{y-a}$$
$$m_a(x) geqslant m_a(y)$$
And it's not good. Did I miss something or is the theorem false as it is stated?
real-analysis convex-analysis
asked Dec 20 '18 at 13:07
BotondBotond
5,8532832
$endgroup$
1
$begingroup$
To me, the original definition implies the additional constraint that x also be an element of the interval I (although it might be more precise to make the constraint explicit in the definition statement). Your counter example violates that constraint.
$endgroup$
– John Polcari
Dec 20 '18 at 13:22
$begingroup$
@JohnPolcari You are right, I did miss that out. But I don't see why did my counterexample violate that constraint?
$endgroup$
– Botond
Dec 20 '18 at 13:26
1
$begingroup$
Because there is nothing in your counter example that prevents x from being outside I (think about the case where a is the left edge of I). Actually, as I think about it more, I suspect that the original definition should require x to actually be between a and b.
$endgroup$
– John Polcari
Dec 20 '18 at 13:33
$begingroup$
@JohnPolcari Yes, you are right. It was not mentioned in the formal definition, just before it, that's why I didn't see it. Thank you!
$endgroup$
– Botond
Dec 20 '18 at 13:40
$begingroup$
Actually, it looks like the definition can be further simplified so that a and b are specifically defined as the edges of I rather than some arbitrary pair of points in I.
$endgroup$
– John Polcari
Dec 20 '18 at 14:00
|
show 2 more comments
$begingroup$
We defined the convexity of an $f: mathbb{R} to mathbb{R}$ function like this:
$f$ is convex on the interval $I$ if $forall a,bin
I$, $a<b$, $forall x in I$
$$f(x)leqslant f(a)+frac{f(b)-f(a)}{b-a}(x-a)$$
And then we had the following theorem:
The following statements are equivalent:
(i) $f$ is convex on $I$
(ii) $forall a in I$, $m_a(x)=frac{f(x)-f(a)}{x-a}$ is monotone increasing on $Isetminus{a}$
I tried to prove (i)$implies$(ii) like this: Let $a,x,y in I$. If $a<x<y$, then
$$f(x) leqslant f(a) + frac{f(y)-f(a)}{y-a}(x-a)$$
$$frac{f(x)-f(a)}{x-a} leqslant frac{f(y)-f(a)}{y-a}$$
$$m_a(x) leqslant m_a(y)$$
Which is fine, but when $x<a<y$, we have that $x-a leqslant 0$, so
$$frac{f(x)-f(a)}{x-a} geqslant frac{f(y)-f(a)}{y-a}$$
$$m_a(x) geqslant m_a(y)$$
And it's not good. Did I miss something or is the theorem false as it is stated?
real-analysis convex-analysis
asked Dec 20 '18 at 13:07
BotondBotond
5,8532832
$endgroup$
We defined the convexity of an $f: mathbb{R} to mathbb{R}$ function like this:
$f$ is convex on the interval $I$ if $forall a,bin
I$, $a<b$, $forall x in I$
$$f(x)leqslant f(a)+frac{f(b)-f(a)}{b-a}(x-a)$$
And then we had the following theorem:
The following statements are equivalent:
(i) $f$ is convex on $I$
(ii) $forall a in I$, $m_a(x)=frac{f(x)-f(a)}{x-a}$ is monotone increasing on $Isetminus{a}$
I tried to prove (i)$implies$(ii) like this: Let $a,x,y in I$. If $a<x<y$, then
$$f(x) leqslant f(a) + frac{f(y)-f(a)}{y-a}(x-a)$$
$$frac{f(x)-f(a)}{x-a} leqslant frac{f(y)-f(a)}{y-a}$$
$$m_a(x) leqslant m_a(y)$$
Which is fine, but when $x<a<y$, we have that $x-a leqslant 0$, so
$$frac{f(x)-f(a)}{x-a} geqslant frac{f(y)-f(a)}{y-a}$$
$$m_a(x) geqslant m_a(y)$$
And it's not good. Did I miss something or is the theorem false as it is stated?
real-analysis convex-analysis
real-analysis convex-analysis
asked Dec 20 '18 at 13:07
BotondBotond
5,8532832
asked Dec 20 '18 at 13:07
BotondBotond
5,8532832
edited Dec 20 '18 at 13:25
Botond
asked Dec 20 '18 at 13:07
BotondBotond
5,8532832
asked Dec 20 '18 at 13:07
BotondBotond
5,8532832
asked Dec 20 '18 at 13:07
BotondBotond
5,8532832
5,8532832
1
$begingroup$
To me, the original definition implies the additional constraint that x also be an element of the interval I (although it might be more precise to make the constraint explicit in the definition statement). Your counter example violates that constraint.
$endgroup$
– John Polcari
Dec 20 '18 at 13:22
$begingroup$
@JohnPolcari You are right, I did miss that out. But I don't see why did my counterexample violate that constraint?
$endgroup$
– Botond
Dec 20 '18 at 13:26
1
$begingroup$
Because there is nothing in your counter example that prevents x from being outside I (think about the case where a is the left edge of I). Actually, as I think about it more, I suspect that the original definition should require x to actually be between a and b.
$endgroup$
– John Polcari
Dec 20 '18 at 13:33
$begingroup$
@JohnPolcari Yes, you are right. It was not mentioned in the formal definition, just before it, that's why I didn't see it. Thank you!
$endgroup$
– Botond
Dec 20 '18 at 13:40
$begingroup$
Actually, it looks like the definition can be further simplified so that a and b are specifically defined as the edges of I rather than some arbitrary pair of points in I.
$endgroup$
– John Polcari
Dec 20 '18 at 14:00
|
show 2 more comments
1
$begingroup$
To me, the original definition implies the additional constraint that x also be an element of the interval I (although it might be more precise to make the constraint explicit in the definition statement). Your counter example violates that constraint.
$endgroup$
– John Polcari
Dec 20 '18 at 13:22
$begingroup$
@JohnPolcari You are right, I did miss that out. But I don't see why did my counterexample violate that constraint?
$endgroup$
– Botond
Dec 20 '18 at 13:26
1
$begingroup$
Because there is nothing in your counter example that prevents x from being outside I (think about the case where a is the left edge of I). Actually, as I think about it more, I suspect that the original definition should require x to actually be between a and b.
$endgroup$
– John Polcari
Dec 20 '18 at 13:33
$begingroup$
@JohnPolcari Yes, you are right. It was not mentioned in the formal definition, just before it, that's why I didn't see it. Thank you!
$endgroup$
– Botond
Dec 20 '18 at 13:40
$begingroup$
Actually, it looks like the definition can be further simplified so that a and b are specifically defined as the edges of I rather than some arbitrary pair of points in I.
$endgroup$
– John Polcari
Dec 20 '18 at 14:00
1
1
$begingroup$
To me, the original definition implies the additional constraint that x also be an element of the interval I (although it might be more precise to make the constraint explicit in the definition statement). Your counter example violates that constraint.
$endgroup$
– John Polcari
Dec 20 '18 at 13:22
$begingroup$
To me, the original definition implies the additional constraint that x also be an element of the interval I (although it might be more precise to make the constraint explicit in the definition statement). Your counter example violates that constraint.
$endgroup$
– John Polcari
Dec 20 '18 at 13:22
$begingroup$
@JohnPolcari You are right, I did miss that out. But I don't see why did my counterexample violate that constraint?
$endgroup$
– Botond
Dec 20 '18 at 13:26
$begingroup$
@JohnPolcari You are right, I did miss that out. But I don't see why did my counterexample violate that constraint?
$endgroup$
– Botond
Dec 20 '18 at 13:26
1
1
$begingroup$
Because there is nothing in your counter example that prevents x from being outside I (think about the case where a is the left edge of I). Actually, as I think about it more, I suspect that the original definition should require x to actually be between a and b.
$endgroup$
– John Polcari
Dec 20 '18 at 13:33
$begingroup$
Because there is nothing in your counter example that prevents x from being outside I (think about the case where a is the left edge of I). Actually, as I think about it more, I suspect that the original definition should require x to actually be between a and b.
$endgroup$
– John Polcari
Dec 20 '18 at 13:33
$begingroup$
@JohnPolcari Yes, you are right. It was not mentioned in the formal definition, just before it, that's why I didn't see it. Thank you!
$endgroup$
– Botond
Dec 20 '18 at 13:40
$begingroup$
@JohnPolcari Yes, you are right. It was not mentioned in the formal definition, just before it, that's why I didn't see it. Thank you!
$endgroup$
– Botond
Dec 20 '18 at 13:40
$begingroup$
Actually, it looks like the definition can be further simplified so that a and b are specifically defined as the edges of I rather than some arbitrary pair of points in I.
$endgroup$
– John Polcari
Dec 20 '18 at 14:00
$begingroup$
Actually, it looks like the definition can be further simplified so that a and b are specifically defined as the edges of I rather than some arbitrary pair of points in I.
$endgroup$
– John Polcari
Dec 20 '18 at 14:00
|
show 2 more comments
2 Answers
2
active
oldest
votes
$begingroup$
If $x<a<y$, one should use
$$
f(a) leqslant f(x) + frac{f(y)-f(x)}{y-x}(a-x)tag{*}
$$
instead of
$$
f(x) leqslant f(a) + frac{f(y)-f(a)}{y-a}(x-a),
$$ which holds for $a<x<y$. Then one gets
$$
m_a(x)leq m_x(y).
$$
One can also show that
$$
m_x(y) leq m_a(y)
$$ which is equivalent to $(*)$.
edited Dec 20 '18 at 13:57
Botond
5,8532832
answered Dec 20 '18 at 13:19
SongSong
13.8k633
$endgroup$
$begingroup$
But why isn't the $f(x) leqslant dots$ working?
$endgroup$
– Botond
Dec 20 '18 at 13:27
1
$begingroup$
I think the definition is wrong. It should require that $ xin [a,b]$, rather than $xin I$.
$endgroup$
– Song
Dec 20 '18 at 13:31
$begingroup$
You are right. It was stated in the very first "definition": The graph of $f$ is under the line passing through $(a,f(a))$ and $(b,f(b))$ on $[a,b]$, but somehow It was missing from the formal definition. Thank you!
$endgroup$
– Botond
Dec 20 '18 at 13:37
add a comment |
$begingroup$
You want to look up the secant lemma, as that is what is likely being a alluded too in the second part of their definition. The problem is, they wrote it incorrectly. For example, by their current definition, $x$ is convex on $I=[a,b]$ since $frac{d^2}{(dx)^2}(x)geq 0$, however, letting $f(x)=x$ in the second “equivalent” definition, we find it simplifies to $1$ which is not monotonic.
The issue comes from the actual secant lemma being much more general, and on you mixing strict vs non strict inequalities in both parts.
answered Dec 20 '18 at 13:27
R.JacksonR.Jackson
1688
$endgroup$
$begingroup$
I think $f(x)=x$ is both convex and concave. Also, $m_a(x)=frac{x-a}{x-a}=1$, which is increasing, because $1=m_a(x)geq m_a(y)=1$. So it looks ok for me.
$endgroup$
– Botond
Dec 20 '18 at 13:31
$begingroup$
It’s not monotonically increasing, however, since its constant.
$endgroup$
– R.Jackson
Dec 20 '18 at 13:33
$begingroup$
I think it is monotonically increasing and decreasing at the same time, but neither strictly increasing nor strictly decreasing.
$endgroup$
– Botond
Dec 20 '18 at 13:39
$begingroup$
Right, nvm, I had the definition of monotonic wrong it just means non-decreasing.
$endgroup$
– R.Jackson
Dec 20 '18 at 13:45
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If $x<a<y$, one should use
$$
f(a) leqslant f(x) + frac{f(y)-f(x)}{y-x}(a-x)tag{*}
$$
instead of
$$
f(x) leqslant f(a) + frac{f(y)-f(a)}{y-a}(x-a),
$$ which holds for $a<x<y$. Then one gets
$$
m_a(x)leq m_x(y).
$$
One can also show that
$$
m_x(y) leq m_a(y)
$$ which is equivalent to $(*)$.
edited Dec 20 '18 at 13:57
Botond
5,8532832
answered Dec 20 '18 at 13:19
SongSong
13.8k633
$endgroup$
$begingroup$
But why isn't the $f(x) leqslant dots$ working?
$endgroup$
– Botond
Dec 20 '18 at 13:27
1
$begingroup$
I think the definition is wrong. It should require that $ xin [a,b]$, rather than $xin I$.
$endgroup$
– Song
Dec 20 '18 at 13:31
$begingroup$
You are right. It was stated in the very first "definition": The graph of $f$ is under the line passing through $(a,f(a))$ and $(b,f(b))$ on $[a,b]$, but somehow It was missing from the formal definition. Thank you!
$endgroup$
– Botond
Dec 20 '18 at 13:37
add a comment |
$begingroup$
If $x<a<y$, one should use
$$
f(a) leqslant f(x) + frac{f(y)-f(x)}{y-x}(a-x)tag{*}
$$
instead of
$$
f(x) leqslant f(a) + frac{f(y)-f(a)}{y-a}(x-a),
$$ which holds for $a<x<y$. Then one gets
$$
m_a(x)leq m_x(y).
$$
One can also show that
$$
m_x(y) leq m_a(y)
$$ which is equivalent to $(*)$.
edited Dec 20 '18 at 13:57
Botond
5,8532832
answered Dec 20 '18 at 13:19
SongSong
13.8k633
$endgroup$
$begingroup$
But why isn't the $f(x) leqslant dots$ working?
$endgroup$
– Botond
Dec 20 '18 at 13:27
1
$begingroup$
I think the definition is wrong. It should require that $ xin [a,b]$, rather than $xin I$.
$endgroup$
– Song
Dec 20 '18 at 13:31
$begingroup$
You are right. It was stated in the very first "definition": The graph of $f$ is under the line passing through $(a,f(a))$ and $(b,f(b))$ on $[a,b]$, but somehow It was missing from the formal definition. Thank you!
$endgroup$
– Botond
Dec 20 '18 at 13:37
add a comment |
$begingroup$
If $x<a<y$, one should use
$$
f(a) leqslant f(x) + frac{f(y)-f(x)}{y-x}(a-x)tag{*}
$$
instead of
$$
f(x) leqslant f(a) + frac{f(y)-f(a)}{y-a}(x-a),
$$ which holds for $a<x<y$. Then one gets
$$
m_a(x)leq m_x(y).
$$
One can also show that
$$
m_x(y) leq m_a(y)
$$ which is equivalent to $(*)$.
edited Dec 20 '18 at 13:57
Botond
5,8532832
answered Dec 20 '18 at 13:19
SongSong
13.8k633
$endgroup$
If $x<a<y$, one should use
$$
f(a) leqslant f(x) + frac{f(y)-f(x)}{y-x}(a-x)tag{*}
$$
instead of
$$
f(x) leqslant f(a) + frac{f(y)-f(a)}{y-a}(x-a),
$$ which holds for $a<x<y$. Then one gets
$$
m_a(x)leq m_x(y).
$$
One can also show that
$$
m_x(y) leq m_a(y)
$$ which is equivalent to $(*)$.
edited Dec 20 '18 at 13:57
Botond
5,8532832
answered Dec 20 '18 at 13:19
SongSong
13.8k633
edited Dec 20 '18 at 13:57
Botond
5,8532832
edited Dec 20 '18 at 13:57
Botond
5,8532832
edited Dec 20 '18 at 13:57
Botond
5,8532832
5,8532832
answered Dec 20 '18 at 13:19
SongSong
13.8k633
answered Dec 20 '18 at 13:19
SongSong
13.8k633
answered Dec 20 '18 at 13:19
SongSong
13.8k633
13.8k633
$begingroup$
But why isn't the $f(x) leqslant dots$ working?
$endgroup$
– Botond
Dec 20 '18 at 13:27
1
$begingroup$
I think the definition is wrong. It should require that $ xin [a,b]$, rather than $xin I$.
$endgroup$
– Song
Dec 20 '18 at 13:31
$begingroup$
You are right. It was stated in the very first "definition": The graph of $f$ is under the line passing through $(a,f(a))$ and $(b,f(b))$ on $[a,b]$, but somehow It was missing from the formal definition. Thank you!
$endgroup$
– Botond
Dec 20 '18 at 13:37
add a comment |
$begingroup$
But why isn't the $f(x) leqslant dots$ working?
$endgroup$
– Botond
Dec 20 '18 at 13:27
1
$begingroup$
I think the definition is wrong. It should require that $ xin [a,b]$, rather than $xin I$.
$endgroup$
– Song
Dec 20 '18 at 13:31
$begingroup$
You are right. It was stated in the very first "definition": The graph of $f$ is under the line passing through $(a,f(a))$ and $(b,f(b))$ on $[a,b]$, but somehow It was missing from the formal definition. Thank you!
$endgroup$
– Botond
Dec 20 '18 at 13:37
$begingroup$
But why isn't the $f(x) leqslant dots$ working?
$endgroup$
– Botond
Dec 20 '18 at 13:27
$begingroup$
But why isn't the $f(x) leqslant dots$ working?
$endgroup$
– Botond
Dec 20 '18 at 13:27
1
1
$begingroup$
I think the definition is wrong. It should require that $ xin [a,b]$, rather than $xin I$.
$endgroup$
– Song
Dec 20 '18 at 13:31
$begingroup$
I think the definition is wrong. It should require that $ xin [a,b]$, rather than $xin I$.
$endgroup$
– Song
Dec 20 '18 at 13:31
$begingroup$
You are right. It was stated in the very first "definition": The graph of $f$ is under the line passing through $(a,f(a))$ and $(b,f(b))$ on $[a,b]$, but somehow It was missing from the formal definition. Thank you!
$endgroup$
– Botond
Dec 20 '18 at 13:37
$begingroup$
You are right. It was stated in the very first "definition": The graph of $f$ is under the line passing through $(a,f(a))$ and $(b,f(b))$ on $[a,b]$, but somehow It was missing from the formal definition. Thank you!
$endgroup$
– Botond
Dec 20 '18 at 13:37
add a comment |
$begingroup$
You want to look up the secant lemma, as that is what is likely being a alluded too in the second part of their definition. The problem is, they wrote it incorrectly. For example, by their current definition, $x$ is convex on $I=[a,b]$ since $frac{d^2}{(dx)^2}(x)geq 0$, however, letting $f(x)=x$ in the second “equivalent” definition, we find it simplifies to $1$ which is not monotonic.
The issue comes from the actual secant lemma being much more general, and on you mixing strict vs non strict inequalities in both parts.
answered Dec 20 '18 at 13:27
R.JacksonR.Jackson
1688
$endgroup$
$begingroup$
I think $f(x)=x$ is both convex and concave. Also, $m_a(x)=frac{x-a}{x-a}=1$, which is increasing, because $1=m_a(x)geq m_a(y)=1$. So it looks ok for me.
$endgroup$
– Botond
Dec 20 '18 at 13:31
$begingroup$
It’s not monotonically increasing, however, since its constant.
$endgroup$
– R.Jackson
Dec 20 '18 at 13:33
$begingroup$
I think it is monotonically increasing and decreasing at the same time, but neither strictly increasing nor strictly decreasing.
$endgroup$
– Botond
Dec 20 '18 at 13:39
$begingroup$
Right, nvm, I had the definition of monotonic wrong it just means non-decreasing.
$endgroup$
– R.Jackson
Dec 20 '18 at 13:45
add a comment |
$begingroup$
You want to look up the secant lemma, as that is what is likely being a alluded too in the second part of their definition. The problem is, they wrote it incorrectly. For example, by their current definition, $x$ is convex on $I=[a,b]$ since $frac{d^2}{(dx)^2}(x)geq 0$, however, letting $f(x)=x$ in the second “equivalent” definition, we find it simplifies to $1$ which is not monotonic.
The issue comes from the actual secant lemma being much more general, and on you mixing strict vs non strict inequalities in both parts.
answered Dec 20 '18 at 13:27
R.JacksonR.Jackson
1688
$endgroup$
$begingroup$
I think $f(x)=x$ is both convex and concave. Also, $m_a(x)=frac{x-a}{x-a}=1$, which is increasing, because $1=m_a(x)geq m_a(y)=1$. So it looks ok for me.
$endgroup$
– Botond
Dec 20 '18 at 13:31
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It’s not monotonically increasing, however, since its constant.
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– R.Jackson
Dec 20 '18 at 13:33
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I think it is monotonically increasing and decreasing at the same time, but neither strictly increasing nor strictly decreasing.
$endgroup$
– Botond
Dec 20 '18 at 13:39
$begingroup$
Right, nvm, I had the definition of monotonic wrong it just means non-decreasing.
$endgroup$
– R.Jackson
Dec 20 '18 at 13:45
add a comment |
$begingroup$
You want to look up the secant lemma, as that is what is likely being a alluded too in the second part of their definition. The problem is, they wrote it incorrectly. For example, by their current definition, $x$ is convex on $I=[a,b]$ since $frac{d^2}{(dx)^2}(x)geq 0$, however, letting $f(x)=x$ in the second “equivalent” definition, we find it simplifies to $1$ which is not monotonic.
The issue comes from the actual secant lemma being much more general, and on you mixing strict vs non strict inequalities in both parts.
answered Dec 20 '18 at 13:27
R.JacksonR.Jackson
1688
$endgroup$
You want to look up the secant lemma, as that is what is likely being a alluded too in the second part of their definition. The problem is, they wrote it incorrectly. For example, by their current definition, $x$ is convex on $I=[a,b]$ since $frac{d^2}{(dx)^2}(x)geq 0$, however, letting $f(x)=x$ in the second “equivalent” definition, we find it simplifies to $1$ which is not monotonic.
The issue comes from the actual secant lemma being much more general, and on you mixing strict vs non strict inequalities in both parts.
answered Dec 20 '18 at 13:27
R.JacksonR.Jackson
1688
answered Dec 20 '18 at 13:27
R.JacksonR.Jackson
1688
answered Dec 20 '18 at 13:27
R.JacksonR.Jackson
1688
answered Dec 20 '18 at 13:27
R.JacksonR.Jackson
1688
1688
$begingroup$
I think $f(x)=x$ is both convex and concave. Also, $m_a(x)=frac{x-a}{x-a}=1$, which is increasing, because $1=m_a(x)geq m_a(y)=1$. So it looks ok for me.
$endgroup$
– Botond
Dec 20 '18 at 13:31
$begingroup$
It’s not monotonically increasing, however, since its constant.
$endgroup$
– R.Jackson
Dec 20 '18 at 13:33
$begingroup$
I think it is monotonically increasing and decreasing at the same time, but neither strictly increasing nor strictly decreasing.
$endgroup$
– Botond
Dec 20 '18 at 13:39
$begingroup$
Right, nvm, I had the definition of monotonic wrong it just means non-decreasing.
$endgroup$
– R.Jackson
Dec 20 '18 at 13:45
add a comment |
$begingroup$
I think $f(x)=x$ is both convex and concave. Also, $m_a(x)=frac{x-a}{x-a}=1$, which is increasing, because $1=m_a(x)geq m_a(y)=1$. So it looks ok for me.
$endgroup$
– Botond
Dec 20 '18 at 13:31
$begingroup$
It’s not monotonically increasing, however, since its constant.
$endgroup$
– R.Jackson
Dec 20 '18 at 13:33
$begingroup$
I think it is monotonically increasing and decreasing at the same time, but neither strictly increasing nor strictly decreasing.
$endgroup$
– Botond
Dec 20 '18 at 13:39
$begingroup$
Right, nvm, I had the definition of monotonic wrong it just means non-decreasing.
$endgroup$
– R.Jackson
Dec 20 '18 at 13:45
$begingroup$
I think $f(x)=x$ is both convex and concave. Also, $m_a(x)=frac{x-a}{x-a}=1$, which is increasing, because $1=m_a(x)geq m_a(y)=1$. So it looks ok for me.
$endgroup$
– Botond
Dec 20 '18 at 13:31
$begingroup$
I think $f(x)=x$ is both convex and concave. Also, $m_a(x)=frac{x-a}{x-a}=1$, which is increasing, because $1=m_a(x)geq m_a(y)=1$. So it looks ok for me.
$endgroup$
– Botond
Dec 20 '18 at 13:31
$begingroup$
It’s not monotonically increasing, however, since its constant.
$endgroup$
– R.Jackson
Dec 20 '18 at 13:33
$begingroup$
It’s not monotonically increasing, however, since its constant.
$endgroup$
– R.Jackson
Dec 20 '18 at 13:33
$begingroup$
I think it is monotonically increasing and decreasing at the same time, but neither strictly increasing nor strictly decreasing.
$endgroup$
– Botond
Dec 20 '18 at 13:39
$begingroup$
I think it is monotonically increasing and decreasing at the same time, but neither strictly increasing nor strictly decreasing.
$endgroup$
– Botond
Dec 20 '18 at 13:39
$begingroup$
Right, nvm, I had the definition of monotonic wrong it just means non-decreasing.
$endgroup$
– R.Jackson
Dec 20 '18 at 13:45
$begingroup$
Right, nvm, I had the definition of monotonic wrong it just means non-decreasing.
$endgroup$
– R.Jackson
Dec 20 '18 at 13:45
add a comment |
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1
$begingroup$
To me, the original definition implies the additional constraint that x also be an element of the interval I (although it might be more precise to make the constraint explicit in the definition statement). Your counter example violates that constraint.
$endgroup$
– John Polcari
Dec 20 '18 at 13:22
$begingroup$
@JohnPolcari You are right, I did miss that out. But I don't see why did my counterexample violate that constraint?
$endgroup$
– Botond
Dec 20 '18 at 13:26
1
$begingroup$
Because there is nothing in your counter example that prevents x from being outside I (think about the case where a is the left edge of I). Actually, as I think about it more, I suspect that the original definition should require x to actually be between a and b.
$endgroup$
– John Polcari
Dec 20 '18 at 13:33
$begingroup$
@JohnPolcari Yes, you are right. It was not mentioned in the formal definition, just before it, that's why I didn't see it. Thank you!
$endgroup$
– Botond
Dec 20 '18 at 13:40
$begingroup$
Actually, it looks like the definition can be further simplified so that a and b are specifically defined as the edges of I rather than some arbitrary pair of points in I.
$endgroup$
– John Polcari
Dec 20 '18 at 14:00