How to prove the property of convex function in higher dimension
Suppose $f:mathbb{R}^nmapsto mathbb{R}$ is convex, could anyone tell me how to prove the following fact?
(1) If $fin C^1$, then for any $u,v$
$$
f(v)geqslant langlenabla f(u),v-urangle
$$
(2) If $fin C^2$, then for any $u,v$
$$
langlenabla^2f(u)v,vranglegeqslant0
$$
analysis convex-analysis
edited May 14 '13 at 14:09
Alex Ravsky
39.3k32181
asked May 14 '13 at 13:11
hxhxhx88
2,2111232
add a comment |
Suppose $f:mathbb{R}^nmapsto mathbb{R}$ is convex, could anyone tell me how to prove the following fact?
(1) If $fin C^1$, then for any $u,v$
$$
f(v)geqslant langlenabla f(u),v-urangle
$$
(2) If $fin C^2$, then for any $u,v$
$$
langlenabla^2f(u)v,vranglegeqslant0
$$
analysis convex-analysis
edited May 14 '13 at 14:09
Alex Ravsky
39.3k32181
asked May 14 '13 at 13:11
hxhxhx88
2,2111232
add a comment |
Suppose $f:mathbb{R}^nmapsto mathbb{R}$ is convex, could anyone tell me how to prove the following fact?
(1) If $fin C^1$, then for any $u,v$
$$
f(v)geqslant langlenabla f(u),v-urangle
$$
(2) If $fin C^2$, then for any $u,v$
$$
langlenabla^2f(u)v,vranglegeqslant0
$$
analysis convex-analysis
edited May 14 '13 at 14:09
Alex Ravsky
39.3k32181
asked May 14 '13 at 13:11
hxhxhx88
2,2111232
Suppose $f:mathbb{R}^nmapsto mathbb{R}$ is convex, could anyone tell me how to prove the following fact?
(1) If $fin C^1$, then for any $u,v$
$$
f(v)geqslant langlenabla f(u),v-urangle
$$
(2) If $fin C^2$, then for any $u,v$
$$
langlenabla^2f(u)v,vranglegeqslant0
$$
analysis convex-analysis
analysis convex-analysis
edited May 14 '13 at 14:09
Alex Ravsky
39.3k32181
asked May 14 '13 at 13:11
hxhxhx88
2,2111232
edited May 14 '13 at 14:09
Alex Ravsky
39.3k32181
asked May 14 '13 at 13:11
hxhxhx88
2,2111232
edited May 14 '13 at 14:09
Alex Ravsky
39.3k32181
edited May 14 '13 at 14:09
Alex Ravsky
39.3k32181
edited May 14 '13 at 14:09
Alex Ravsky
39.3k32181
39.3k32181
asked May 14 '13 at 13:11
hxhxhx88
2,2111232
asked May 14 '13 at 13:11
hxhxhx88
2,2111232
asked May 14 '13 at 13:11
hxhxhx88
2,2111232
2,2111232
add a comment |
add a comment |
1 Answer
1
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oldest
votes
Useful fact: the restriction of a convex function to a line is a convex function.
Formally, for any point $a$ and any nonzero vector $w$ the function $varphi(t) = f(a+tw)$ is convex on $mathbb R$.
- If $f$ is $C^1$, then so is $varphi$. We have $varphi(t)-varphi(0)=varphi'(xi),t>varphi'(0),t$ by the Mean Value Theorem and because $varphi'$ is increasing. In terms of $f$ this translates into the inequality you want.
- If $f$ is $C^2$, then so is $varphi$. Consequently, $varphi''(0)ge 0$ which translates into the inequality you want.
answered May 14 '13 at 23:45
75064
3,0001922
add a comment |
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1 Answer
1
active
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votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Useful fact: the restriction of a convex function to a line is a convex function.
Formally, for any point $a$ and any nonzero vector $w$ the function $varphi(t) = f(a+tw)$ is convex on $mathbb R$.
- If $f$ is $C^1$, then so is $varphi$. We have $varphi(t)-varphi(0)=varphi'(xi),t>varphi'(0),t$ by the Mean Value Theorem and because $varphi'$ is increasing. In terms of $f$ this translates into the inequality you want.
- If $f$ is $C^2$, then so is $varphi$. Consequently, $varphi''(0)ge 0$ which translates into the inequality you want.
answered May 14 '13 at 23:45
75064
3,0001922
add a comment |
Useful fact: the restriction of a convex function to a line is a convex function.
Formally, for any point $a$ and any nonzero vector $w$ the function $varphi(t) = f(a+tw)$ is convex on $mathbb R$.
- If $f$ is $C^1$, then so is $varphi$. We have $varphi(t)-varphi(0)=varphi'(xi),t>varphi'(0),t$ by the Mean Value Theorem and because $varphi'$ is increasing. In terms of $f$ this translates into the inequality you want.
- If $f$ is $C^2$, then so is $varphi$. Consequently, $varphi''(0)ge 0$ which translates into the inequality you want.
answered May 14 '13 at 23:45
75064
3,0001922
add a comment |
Useful fact: the restriction of a convex function to a line is a convex function.
Formally, for any point $a$ and any nonzero vector $w$ the function $varphi(t) = f(a+tw)$ is convex on $mathbb R$.
- If $f$ is $C^1$, then so is $varphi$. We have $varphi(t)-varphi(0)=varphi'(xi),t>varphi'(0),t$ by the Mean Value Theorem and because $varphi'$ is increasing. In terms of $f$ this translates into the inequality you want.
- If $f$ is $C^2$, then so is $varphi$. Consequently, $varphi''(0)ge 0$ which translates into the inequality you want.
answered May 14 '13 at 23:45
75064
3,0001922
Useful fact: the restriction of a convex function to a line is a convex function.
Formally, for any point $a$ and any nonzero vector $w$ the function $varphi(t) = f(a+tw)$ is convex on $mathbb R$.
- If $f$ is $C^1$, then so is $varphi$. We have $varphi(t)-varphi(0)=varphi'(xi),t>varphi'(0),t$ by the Mean Value Theorem and because $varphi'$ is increasing. In terms of $f$ this translates into the inequality you want.
- If $f$ is $C^2$, then so is $varphi$. Consequently, $varphi''(0)ge 0$ which translates into the inequality you want.
answered May 14 '13 at 23:45
75064
3,0001922
answered May 14 '13 at 23:45
75064
3,0001922
answered May 14 '13 at 23:45
75064
3,0001922
answered May 14 '13 at 23:45
75064
3,0001922
3,0001922
add a comment |
add a comment |
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