How to prove this property using convexity?
$begingroup$
Suppose that $f:[a,b]tomathbb{R}$ be a twice-differentiable function, and that there exists $cin(a,b)$ such that $f'(c)=frac{f(b)-f(a)}{b-a}$.
Show that if $f''(x)>0$ for all $xin[a,b]$ and $f''$ is strictly increasing on $[a,b]$, then $c>frac{a+b}{2}$.
Using Taylor's theorem, i solved the problem.
But, i'd like to prove that using convexity.
Give some comments or hints. Thank you!
real-analysis derivatives convexity-inequality
asked Jan 11 at 5:10
PrimaveraPrimavera
32619
$endgroup$
add a comment |
$begingroup$
Suppose that $f:[a,b]tomathbb{R}$ be a twice-differentiable function, and that there exists $cin(a,b)$ such that $f'(c)=frac{f(b)-f(a)}{b-a}$.
Show that if $f''(x)>0$ for all $xin[a,b]$ and $f''$ is strictly increasing on $[a,b]$, then $c>frac{a+b}{2}$.
Using Taylor's theorem, i solved the problem.
But, i'd like to prove that using convexity.
Give some comments or hints. Thank you!
real-analysis derivatives convexity-inequality
asked Jan 11 at 5:10
PrimaveraPrimavera
32619
$endgroup$
$begingroup$
There is a stronger condition than convexity: $f''$ is increasing. It is unlikely that convexity will yield this result.
$endgroup$
– Kavi Rama Murthy
Jan 11 at 5:53
$begingroup$
@KaviRamaMurthy I see. It seemed to be related to convexity. Thanks.
$endgroup$
– Primavera
Jan 11 at 7:10
add a comment |
$begingroup$
Suppose that $f:[a,b]tomathbb{R}$ be a twice-differentiable function, and that there exists $cin(a,b)$ such that $f'(c)=frac{f(b)-f(a)}{b-a}$.
Show that if $f''(x)>0$ for all $xin[a,b]$ and $f''$ is strictly increasing on $[a,b]$, then $c>frac{a+b}{2}$.
Using Taylor's theorem, i solved the problem.
But, i'd like to prove that using convexity.
Give some comments or hints. Thank you!
real-analysis derivatives convexity-inequality
asked Jan 11 at 5:10
PrimaveraPrimavera
32619
$endgroup$
Suppose that $f:[a,b]tomathbb{R}$ be a twice-differentiable function, and that there exists $cin(a,b)$ such that $f'(c)=frac{f(b)-f(a)}{b-a}$.
Show that if $f''(x)>0$ for all $xin[a,b]$ and $f''$ is strictly increasing on $[a,b]$, then $c>frac{a+b}{2}$.
Using Taylor's theorem, i solved the problem.
But, i'd like to prove that using convexity.
Give some comments or hints. Thank you!
real-analysis derivatives convexity-inequality
real-analysis derivatives convexity-inequality
asked Jan 11 at 5:10
PrimaveraPrimavera
32619
asked Jan 11 at 5:10
PrimaveraPrimavera
32619
asked Jan 11 at 5:10
PrimaveraPrimavera
32619
asked Jan 11 at 5:10
PrimaveraPrimavera
32619
asked Jan 11 at 5:10
PrimaveraPrimavera
32619
32619
$begingroup$
There is a stronger condition than convexity: $f''$ is increasing. It is unlikely that convexity will yield this result.
$endgroup$
– Kavi Rama Murthy
Jan 11 at 5:53
$begingroup$
@KaviRamaMurthy I see. It seemed to be related to convexity. Thanks.
$endgroup$
– Primavera
Jan 11 at 7:10
add a comment |
$begingroup$
There is a stronger condition than convexity: $f''$ is increasing. It is unlikely that convexity will yield this result.
$endgroup$
– Kavi Rama Murthy
Jan 11 at 5:53
$begingroup$
@KaviRamaMurthy I see. It seemed to be related to convexity. Thanks.
$endgroup$
– Primavera
Jan 11 at 7:10
$begingroup$
There is a stronger condition than convexity: $f''$ is increasing. It is unlikely that convexity will yield this result.
$endgroup$
– Kavi Rama Murthy
Jan 11 at 5:53
$begingroup$
There is a stronger condition than convexity: $f''$ is increasing. It is unlikely that convexity will yield this result.
$endgroup$
– Kavi Rama Murthy
Jan 11 at 5:53
$begingroup$
@KaviRamaMurthy I see. It seemed to be related to convexity. Thanks.
$endgroup$
– Primavera
Jan 11 at 7:10
$begingroup$
@KaviRamaMurthy I see. It seemed to be related to convexity. Thanks.
$endgroup$
– Primavera
Jan 11 at 7:10
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
As indicated in Kavi Rama Murthy’s comment, convexity is not enough. Consider $g(x)=f(a+b-x)$. Then $g$ satisfies the same conditions as $f$ except that $g’’$ is decreasing, while
$$
frac{g(b)-g(a)}{b-a}=g’(a+b-c),
$$
with $a+b-c<(a+b)/2$.
answered Jan 11 at 6:49
Julián AguirreJulián Aguirre
69.5k24297
$endgroup$
$begingroup$
Sorry, I don't understand. How to deduce the last inequality?
$endgroup$
– Primavera
Jan 11 at 7:53
$begingroup$
Well, $c>(a+b)/2$.
$endgroup$
– Julián Aguirre
Jan 11 at 13:44
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
As indicated in Kavi Rama Murthy’s comment, convexity is not enough. Consider $g(x)=f(a+b-x)$. Then $g$ satisfies the same conditions as $f$ except that $g’’$ is decreasing, while
$$
frac{g(b)-g(a)}{b-a}=g’(a+b-c),
$$
with $a+b-c<(a+b)/2$.
answered Jan 11 at 6:49
Julián AguirreJulián Aguirre
69.5k24297
$endgroup$
$begingroup$
Sorry, I don't understand. How to deduce the last inequality?
$endgroup$
– Primavera
Jan 11 at 7:53
$begingroup$
Well, $c>(a+b)/2$.
$endgroup$
– Julián Aguirre
Jan 11 at 13:44
add a comment |
$begingroup$
As indicated in Kavi Rama Murthy’s comment, convexity is not enough. Consider $g(x)=f(a+b-x)$. Then $g$ satisfies the same conditions as $f$ except that $g’’$ is decreasing, while
$$
frac{g(b)-g(a)}{b-a}=g’(a+b-c),
$$
with $a+b-c<(a+b)/2$.
answered Jan 11 at 6:49
Julián AguirreJulián Aguirre
69.5k24297
$endgroup$
$begingroup$
Sorry, I don't understand. How to deduce the last inequality?
$endgroup$
– Primavera
Jan 11 at 7:53
$begingroup$
Well, $c>(a+b)/2$.
$endgroup$
– Julián Aguirre
Jan 11 at 13:44
add a comment |
$begingroup$
As indicated in Kavi Rama Murthy’s comment, convexity is not enough. Consider $g(x)=f(a+b-x)$. Then $g$ satisfies the same conditions as $f$ except that $g’’$ is decreasing, while
$$
frac{g(b)-g(a)}{b-a}=g’(a+b-c),
$$
with $a+b-c<(a+b)/2$.
answered Jan 11 at 6:49
Julián AguirreJulián Aguirre
69.5k24297
$endgroup$
As indicated in Kavi Rama Murthy’s comment, convexity is not enough. Consider $g(x)=f(a+b-x)$. Then $g$ satisfies the same conditions as $f$ except that $g’’$ is decreasing, while
$$
frac{g(b)-g(a)}{b-a}=g’(a+b-c),
$$
with $a+b-c<(a+b)/2$.
answered Jan 11 at 6:49
Julián AguirreJulián Aguirre
69.5k24297
answered Jan 11 at 6:49
Julián AguirreJulián Aguirre
69.5k24297
answered Jan 11 at 6:49
Julián AguirreJulián Aguirre
69.5k24297
answered Jan 11 at 6:49
Julián AguirreJulián Aguirre
69.5k24297
69.5k24297
$begingroup$
Sorry, I don't understand. How to deduce the last inequality?
$endgroup$
– Primavera
Jan 11 at 7:53
$begingroup$
Well, $c>(a+b)/2$.
$endgroup$
– Julián Aguirre
Jan 11 at 13:44
add a comment |
$begingroup$
Sorry, I don't understand. How to deduce the last inequality?
$endgroup$
– Primavera
Jan 11 at 7:53
$begingroup$
Well, $c>(a+b)/2$.
$endgroup$
– Julián Aguirre
Jan 11 at 13:44
$begingroup$
Sorry, I don't understand. How to deduce the last inequality?
$endgroup$
– Primavera
Jan 11 at 7:53
$begingroup$
Sorry, I don't understand. How to deduce the last inequality?
$endgroup$
– Primavera
Jan 11 at 7:53
$begingroup$
Well, $c>(a+b)/2$.
$endgroup$
– Julián Aguirre
Jan 11 at 13:44
$begingroup$
Well, $c>(a+b)/2$.
$endgroup$
– Julián Aguirre
Jan 11 at 13:44
add a comment |
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$begingroup$
There is a stronger condition than convexity: $f''$ is increasing. It is unlikely that convexity will yield this result.
$endgroup$
– Kavi Rama Murthy
Jan 11 at 5:53
$begingroup$
@KaviRamaMurthy I see. It seemed to be related to convexity. Thanks.
$endgroup$
– Primavera
Jan 11 at 7:10