Equivalent definitions of convexity












0












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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
















0












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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














0












0








0





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 20 '18 at 13:25







Botond

















asked Dec 20 '18 at 13:07









BotondBotond

6,54531034




6,54531034








  • 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




    $begingroup$
    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










2 Answers
2






active

oldest

votes


















1












$begingroup$

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 $(*)$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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





















0












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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












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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

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 $(*)$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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


















1












$begingroup$

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 $(*)$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
















1












1








1





$begingroup$

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 $(*)$.






share|cite|improve this answer











$endgroup$



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 $(*)$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 20 '18 at 13:57









Botond

6,54531034




6,54531034










answered Dec 20 '18 at 13:19









SongSong

18.6k21651




18.6k21651












  • $begingroup$
    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






  • 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













0












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















0












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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














0












0








0





$begingroup$

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.






share|cite|improve this answer









$endgroup$



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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 20 '18 at 13:27









R.JacksonR.Jackson

1738




1738












  • $begingroup$
    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










  • $begingroup$
    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


















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