Generalization of Jensen's inequality to multivariate functions
Is there a generalization of Jensen's inequality for convex multivariate functions? By convex, let's say $f$ is a multivariate function defined on the convex set $A$, and for all $x,y in A$ and $lambda in [0,1]$,
$$f(lambda x + (1-lambda)y) leq lambda f(x) + (1-lambda)f(y).$$
Then, letting $x_1,ldots,x_n$ denote points in $A$, the result would be something to the effect of saying that for any $n$ points in $A$,
$$frac{sum_{i=1}^n f(x_i)}{n} geq f left(frac{sum_{i=1}^n{x_i}}{n} right). $$
I do see a few articles that may be related:
- Perlman, Michael D. "Jensen's inequality for a convex vector-valued function on an infinite-dimensional space." Journal of Multivariate Analysis 4.1 (1974): 52-65.
- Merkle, Milan. "Jensen's inequality for multivariate medians." Journal of Mathematical Analysis and Applications 370.1 (2010): 258-269.
- Aras-Gazic, G., et al. "GENERALIZATION OF JENSEN’S INEQUALITY BY HERMITE POLYNOMIALS AND RELATED RESULTS." Mathematical reports 17.2 (2015): 201-223.
- Agnew, Robert A. "Multivariate version of a Jensen-type inequality." J. Inequal. in Pure and Appl. Math 6.4 (2005).
I do not think the first is particularly related if I'm interested in finite dimensional spaces, and my function is not vector-valued in any case. The second may be more related, but it seems to be generalizing in slightly different directions. The third is beyond my comprehension and the fourth, again, seems to be working on a slightly different generalization.
Are there no less technical generalizations of Jensen's to multivariate functions out there?
real-analysis inequality
asked Jul 4 '16 at 8:33
Shane
923718
|
show 9 more comments
Is there a generalization of Jensen's inequality for convex multivariate functions? By convex, let's say $f$ is a multivariate function defined on the convex set $A$, and for all $x,y in A$ and $lambda in [0,1]$,
$$f(lambda x + (1-lambda)y) leq lambda f(x) + (1-lambda)f(y).$$
Then, letting $x_1,ldots,x_n$ denote points in $A$, the result would be something to the effect of saying that for any $n$ points in $A$,
$$frac{sum_{i=1}^n f(x_i)}{n} geq f left(frac{sum_{i=1}^n{x_i}}{n} right). $$
I do see a few articles that may be related:
- Perlman, Michael D. "Jensen's inequality for a convex vector-valued function on an infinite-dimensional space." Journal of Multivariate Analysis 4.1 (1974): 52-65.
- Merkle, Milan. "Jensen's inequality for multivariate medians." Journal of Mathematical Analysis and Applications 370.1 (2010): 258-269.
- Aras-Gazic, G., et al. "GENERALIZATION OF JENSEN’S INEQUALITY BY HERMITE POLYNOMIALS AND RELATED RESULTS." Mathematical reports 17.2 (2015): 201-223.
- Agnew, Robert A. "Multivariate version of a Jensen-type inequality." J. Inequal. in Pure and Appl. Math 6.4 (2005).
I do not think the first is particularly related if I'm interested in finite dimensional spaces, and my function is not vector-valued in any case. The second may be more related, but it seems to be generalizing in slightly different directions. The third is beyond my comprehension and the fourth, again, seems to be working on a slightly different generalization.
Are there no less technical generalizations of Jensen's to multivariate functions out there?
real-analysis inequality
asked Jul 4 '16 at 8:33
Shane
923718
@Did I'm not sure I follow. I'm thinking $f:mathbb{R}^n rightarrow mathbb{R}$. $u$ and $v$ (or $x$ and $y$) are vectors, then, but where am I using any notion of their ordering? Or could you perhaps point me someplace to better understand your first sentence?
– Shane
Jul 4 '16 at 8:53
1
From that proof: "$ldots varphi$ be a convex function on the real numbers. Since $varphi$ is convex, at each real number $x ldots$". Isn't this univariate?
– Shane
Jul 4 '16 at 9:37
2
Just yell if the mystery does not dissolve by itself...
– Did
Jul 4 '16 at 10:16
2
Rereading your question and the comments, I must admit being a little lost. Is your goal to show that, if $f$ is a multivariate function defined on some convex set $A$, and if, for all $x$ and $y$ in $A$ and all $lambda$ in $[0,1]$, $f(lambda x + (1-lambda)y) leq lambda f(x) + (1-lambda)f(y)$, then, for all $n$ and all points $x_1$, $ldots$, $x_n$ in $A$, one has $frac1nsumlimits_{i=1}^n f(x_i)geqslant f left(frac1nsumlimits_{i=1}^n{x_i} right)$? Because this one has self-contained short proofs, for example, by induction on $n$.
– Did
Jul 10 '16 at 15:24
1
To deduce the $n+1$ case of the inequality from the $n$ case, use $lambda=frac1{n+1}$, $x=x_{n+1}$, $y=frac1nsumlimits_{k=1}^nx_k$, then $z=frac1{n+1}sumlimits_{k=1}^{n+1}x_k=lambda x+(1-lambda)y$ hence $f(z)leqslantlambda f(x)+(1-lambda)f(y)$, now by the recurrence hypothesis, $f(y)leqslantfrac1nsumlimits_{k=1}^nf(x_k)$, hence $f(z)leqslantlambda f(x_{n+1})+(1-lambda)frac1nsumlimits_{k=1}^nf(x_k)$, qed.
– Did
Jul 11 '16 at 19:44
|
show 9 more comments
Is there a generalization of Jensen's inequality for convex multivariate functions? By convex, let's say $f$ is a multivariate function defined on the convex set $A$, and for all $x,y in A$ and $lambda in [0,1]$,
$$f(lambda x + (1-lambda)y) leq lambda f(x) + (1-lambda)f(y).$$
Then, letting $x_1,ldots,x_n$ denote points in $A$, the result would be something to the effect of saying that for any $n$ points in $A$,
$$frac{sum_{i=1}^n f(x_i)}{n} geq f left(frac{sum_{i=1}^n{x_i}}{n} right). $$
I do see a few articles that may be related:
- Perlman, Michael D. "Jensen's inequality for a convex vector-valued function on an infinite-dimensional space." Journal of Multivariate Analysis 4.1 (1974): 52-65.
- Merkle, Milan. "Jensen's inequality for multivariate medians." Journal of Mathematical Analysis and Applications 370.1 (2010): 258-269.
- Aras-Gazic, G., et al. "GENERALIZATION OF JENSEN’S INEQUALITY BY HERMITE POLYNOMIALS AND RELATED RESULTS." Mathematical reports 17.2 (2015): 201-223.
- Agnew, Robert A. "Multivariate version of a Jensen-type inequality." J. Inequal. in Pure and Appl. Math 6.4 (2005).
I do not think the first is particularly related if I'm interested in finite dimensional spaces, and my function is not vector-valued in any case. The second may be more related, but it seems to be generalizing in slightly different directions. The third is beyond my comprehension and the fourth, again, seems to be working on a slightly different generalization.
Are there no less technical generalizations of Jensen's to multivariate functions out there?
real-analysis inequality
asked Jul 4 '16 at 8:33
Shane
923718
Is there a generalization of Jensen's inequality for convex multivariate functions? By convex, let's say $f$ is a multivariate function defined on the convex set $A$, and for all $x,y in A$ and $lambda in [0,1]$,
$$f(lambda x + (1-lambda)y) leq lambda f(x) + (1-lambda)f(y).$$
Then, letting $x_1,ldots,x_n$ denote points in $A$, the result would be something to the effect of saying that for any $n$ points in $A$,
$$frac{sum_{i=1}^n f(x_i)}{n} geq f left(frac{sum_{i=1}^n{x_i}}{n} right). $$
I do see a few articles that may be related:
- Perlman, Michael D. "Jensen's inequality for a convex vector-valued function on an infinite-dimensional space." Journal of Multivariate Analysis 4.1 (1974): 52-65.
- Merkle, Milan. "Jensen's inequality for multivariate medians." Journal of Mathematical Analysis and Applications 370.1 (2010): 258-269.
- Aras-Gazic, G., et al. "GENERALIZATION OF JENSEN’S INEQUALITY BY HERMITE POLYNOMIALS AND RELATED RESULTS." Mathematical reports 17.2 (2015): 201-223.
- Agnew, Robert A. "Multivariate version of a Jensen-type inequality." J. Inequal. in Pure and Appl. Math 6.4 (2005).
I do not think the first is particularly related if I'm interested in finite dimensional spaces, and my function is not vector-valued in any case. The second may be more related, but it seems to be generalizing in slightly different directions. The third is beyond my comprehension and the fourth, again, seems to be working on a slightly different generalization.
Are there no less technical generalizations of Jensen's to multivariate functions out there?
real-analysis inequality
real-analysis inequality
asked Jul 4 '16 at 8:33
Shane
923718
asked Jul 4 '16 at 8:33
Shane
923718
asked Jul 4 '16 at 8:33
Shane
923718
asked Jul 4 '16 at 8:33
Shane
923718
asked Jul 4 '16 at 8:33
Shane
923718
923718
@Did I'm not sure I follow. I'm thinking $f:mathbb{R}^n rightarrow mathbb{R}$. $u$ and $v$ (or $x$ and $y$) are vectors, then, but where am I using any notion of their ordering? Or could you perhaps point me someplace to better understand your first sentence?
– Shane
Jul 4 '16 at 8:53
1
From that proof: "$ldots varphi$ be a convex function on the real numbers. Since $varphi$ is convex, at each real number $x ldots$". Isn't this univariate?
– Shane
Jul 4 '16 at 9:37
2
Just yell if the mystery does not dissolve by itself...
– Did
Jul 4 '16 at 10:16
2
Rereading your question and the comments, I must admit being a little lost. Is your goal to show that, if $f$ is a multivariate function defined on some convex set $A$, and if, for all $x$ and $y$ in $A$ and all $lambda$ in $[0,1]$, $f(lambda x + (1-lambda)y) leq lambda f(x) + (1-lambda)f(y)$, then, for all $n$ and all points $x_1$, $ldots$, $x_n$ in $A$, one has $frac1nsumlimits_{i=1}^n f(x_i)geqslant f left(frac1nsumlimits_{i=1}^n{x_i} right)$? Because this one has self-contained short proofs, for example, by induction on $n$.
– Did
Jul 10 '16 at 15:24
1
To deduce the $n+1$ case of the inequality from the $n$ case, use $lambda=frac1{n+1}$, $x=x_{n+1}$, $y=frac1nsumlimits_{k=1}^nx_k$, then $z=frac1{n+1}sumlimits_{k=1}^{n+1}x_k=lambda x+(1-lambda)y$ hence $f(z)leqslantlambda f(x)+(1-lambda)f(y)$, now by the recurrence hypothesis, $f(y)leqslantfrac1nsumlimits_{k=1}^nf(x_k)$, hence $f(z)leqslantlambda f(x_{n+1})+(1-lambda)frac1nsumlimits_{k=1}^nf(x_k)$, qed.
– Did
Jul 11 '16 at 19:44
|
show 9 more comments
@Did I'm not sure I follow. I'm thinking $f:mathbb{R}^n rightarrow mathbb{R}$. $u$ and $v$ (or $x$ and $y$) are vectors, then, but where am I using any notion of their ordering? Or could you perhaps point me someplace to better understand your first sentence?
– Shane
Jul 4 '16 at 8:53
1
From that proof: "$ldots varphi$ be a convex function on the real numbers. Since $varphi$ is convex, at each real number $x ldots$". Isn't this univariate?
– Shane
Jul 4 '16 at 9:37
2
Just yell if the mystery does not dissolve by itself...
– Did
Jul 4 '16 at 10:16
2
Rereading your question and the comments, I must admit being a little lost. Is your goal to show that, if $f$ is a multivariate function defined on some convex set $A$, and if, for all $x$ and $y$ in $A$ and all $lambda$ in $[0,1]$, $f(lambda x + (1-lambda)y) leq lambda f(x) + (1-lambda)f(y)$, then, for all $n$ and all points $x_1$, $ldots$, $x_n$ in $A$, one has $frac1nsumlimits_{i=1}^n f(x_i)geqslant f left(frac1nsumlimits_{i=1}^n{x_i} right)$? Because this one has self-contained short proofs, for example, by induction on $n$.
– Did
Jul 10 '16 at 15:24
1
To deduce the $n+1$ case of the inequality from the $n$ case, use $lambda=frac1{n+1}$, $x=x_{n+1}$, $y=frac1nsumlimits_{k=1}^nx_k$, then $z=frac1{n+1}sumlimits_{k=1}^{n+1}x_k=lambda x+(1-lambda)y$ hence $f(z)leqslantlambda f(x)+(1-lambda)f(y)$, now by the recurrence hypothesis, $f(y)leqslantfrac1nsumlimits_{k=1}^nf(x_k)$, hence $f(z)leqslantlambda f(x_{n+1})+(1-lambda)frac1nsumlimits_{k=1}^nf(x_k)$, qed.
– Did
Jul 11 '16 at 19:44
@Did I'm not sure I follow. I'm thinking $f:mathbb{R}^n rightarrow mathbb{R}$. $u$ and $v$ (or $x$ and $y$) are vectors, then, but where am I using any notion of their ordering? Or could you perhaps point me someplace to better understand your first sentence?
– Shane
Jul 4 '16 at 8:53
@Did I'm not sure I follow. I'm thinking $f:mathbb{R}^n rightarrow mathbb{R}$. $u$ and $v$ (or $x$ and $y$) are vectors, then, but where am I using any notion of their ordering? Or could you perhaps point me someplace to better understand your first sentence?
– Shane
Jul 4 '16 at 8:53
1
1
From that proof: "$ldots varphi$ be a convex function on the real numbers. Since $varphi$ is convex, at each real number $x ldots$". Isn't this univariate?
– Shane
Jul 4 '16 at 9:37
From that proof: "$ldots varphi$ be a convex function on the real numbers. Since $varphi$ is convex, at each real number $x ldots$". Isn't this univariate?
– Shane
Jul 4 '16 at 9:37
2
2
Just yell if the mystery does not dissolve by itself...
– Did
Jul 4 '16 at 10:16
Just yell if the mystery does not dissolve by itself...
– Did
Jul 4 '16 at 10:16
2
2
Rereading your question and the comments, I must admit being a little lost. Is your goal to show that, if $f$ is a multivariate function defined on some convex set $A$, and if, for all $x$ and $y$ in $A$ and all $lambda$ in $[0,1]$, $f(lambda x + (1-lambda)y) leq lambda f(x) + (1-lambda)f(y)$, then, for all $n$ and all points $x_1$, $ldots$, $x_n$ in $A$, one has $frac1nsumlimits_{i=1}^n f(x_i)geqslant f left(frac1nsumlimits_{i=1}^n{x_i} right)$? Because this one has self-contained short proofs, for example, by induction on $n$.
– Did
Jul 10 '16 at 15:24
Rereading your question and the comments, I must admit being a little lost. Is your goal to show that, if $f$ is a multivariate function defined on some convex set $A$, and if, for all $x$ and $y$ in $A$ and all $lambda$ in $[0,1]$, $f(lambda x + (1-lambda)y) leq lambda f(x) + (1-lambda)f(y)$, then, for all $n$ and all points $x_1$, $ldots$, $x_n$ in $A$, one has $frac1nsumlimits_{i=1}^n f(x_i)geqslant f left(frac1nsumlimits_{i=1}^n{x_i} right)$? Because this one has self-contained short proofs, for example, by induction on $n$.
– Did
Jul 10 '16 at 15:24
1
1
To deduce the $n+1$ case of the inequality from the $n$ case, use $lambda=frac1{n+1}$, $x=x_{n+1}$, $y=frac1nsumlimits_{k=1}^nx_k$, then $z=frac1{n+1}sumlimits_{k=1}^{n+1}x_k=lambda x+(1-lambda)y$ hence $f(z)leqslantlambda f(x)+(1-lambda)f(y)$, now by the recurrence hypothesis, $f(y)leqslantfrac1nsumlimits_{k=1}^nf(x_k)$, hence $f(z)leqslantlambda f(x_{n+1})+(1-lambda)frac1nsumlimits_{k=1}^nf(x_k)$, qed.
– Did
Jul 11 '16 at 19:44
To deduce the $n+1$ case of the inequality from the $n$ case, use $lambda=frac1{n+1}$, $x=x_{n+1}$, $y=frac1nsumlimits_{k=1}^nx_k$, then $z=frac1{n+1}sumlimits_{k=1}^{n+1}x_k=lambda x+(1-lambda)y$ hence $f(z)leqslantlambda f(x)+(1-lambda)f(y)$, now by the recurrence hypothesis, $f(y)leqslantfrac1nsumlimits_{k=1}^nf(x_k)$, hence $f(z)leqslantlambda f(x_{n+1})+(1-lambda)frac1nsumlimits_{k=1}^nf(x_k)$, qed.
– Did
Jul 11 '16 at 19:44
|
show 9 more comments
1 Answer
1
active
oldest
votes
Jensen's inequality always holds when you're dealing with a convex function whose domain is finite-dimensional. By restricting the domain of $f$ to $S := text{span} {x_1, ldots, x_n}$, we put the question into that setting and know that $f(mathbb{E} X) leq mathbb{E} f(X)$ for any random vector $X$ taking values in $S$. In particular, it holds when the distribution of $X$ is uniform on ${x_1, ldots, x_n}$.
answered Jul 21 '17 at 1:27
student45
11110
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1848487%2fgeneralization-of-jensens-inequality-to-multivariate-functions%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Jensen's inequality always holds when you're dealing with a convex function whose domain is finite-dimensional. By restricting the domain of $f$ to $S := text{span} {x_1, ldots, x_n}$, we put the question into that setting and know that $f(mathbb{E} X) leq mathbb{E} f(X)$ for any random vector $X$ taking values in $S$. In particular, it holds when the distribution of $X$ is uniform on ${x_1, ldots, x_n}$.
answered Jul 21 '17 at 1:27
student45
11110
add a comment |
Jensen's inequality always holds when you're dealing with a convex function whose domain is finite-dimensional. By restricting the domain of $f$ to $S := text{span} {x_1, ldots, x_n}$, we put the question into that setting and know that $f(mathbb{E} X) leq mathbb{E} f(X)$ for any random vector $X$ taking values in $S$. In particular, it holds when the distribution of $X$ is uniform on ${x_1, ldots, x_n}$.
answered Jul 21 '17 at 1:27
student45
11110
add a comment |
Jensen's inequality always holds when you're dealing with a convex function whose domain is finite-dimensional. By restricting the domain of $f$ to $S := text{span} {x_1, ldots, x_n}$, we put the question into that setting and know that $f(mathbb{E} X) leq mathbb{E} f(X)$ for any random vector $X$ taking values in $S$. In particular, it holds when the distribution of $X$ is uniform on ${x_1, ldots, x_n}$.
answered Jul 21 '17 at 1:27
student45
11110
Jensen's inequality always holds when you're dealing with a convex function whose domain is finite-dimensional. By restricting the domain of $f$ to $S := text{span} {x_1, ldots, x_n}$, we put the question into that setting and know that $f(mathbb{E} X) leq mathbb{E} f(X)$ for any random vector $X$ taking values in $S$. In particular, it holds when the distribution of $X$ is uniform on ${x_1, ldots, x_n}$.
answered Jul 21 '17 at 1:27
student45
11110
answered Jul 21 '17 at 1:27
student45
11110
answered Jul 21 '17 at 1:27
student45
11110
answered Jul 21 '17 at 1:27
student45
11110
11110
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1848487%2fgeneralization-of-jensens-inequality-to-multivariate-functions%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
@Did I'm not sure I follow. I'm thinking $f:mathbb{R}^n rightarrow mathbb{R}$. $u$ and $v$ (or $x$ and $y$) are vectors, then, but where am I using any notion of their ordering? Or could you perhaps point me someplace to better understand your first sentence?
– Shane
Jul 4 '16 at 8:53
1
From that proof: "$ldots varphi$ be a convex function on the real numbers. Since $varphi$ is convex, at each real number $x ldots$". Isn't this univariate?
– Shane
Jul 4 '16 at 9:37
2
Just yell if the mystery does not dissolve by itself...
– Did
Jul 4 '16 at 10:16
2
Rereading your question and the comments, I must admit being a little lost. Is your goal to show that, if $f$ is a multivariate function defined on some convex set $A$, and if, for all $x$ and $y$ in $A$ and all $lambda$ in $[0,1]$, $f(lambda x + (1-lambda)y) leq lambda f(x) + (1-lambda)f(y)$, then, for all $n$ and all points $x_1$, $ldots$, $x_n$ in $A$, one has $frac1nsumlimits_{i=1}^n f(x_i)geqslant f left(frac1nsumlimits_{i=1}^n{x_i} right)$? Because this one has self-contained short proofs, for example, by induction on $n$.
– Did
Jul 10 '16 at 15:24
1
To deduce the $n+1$ case of the inequality from the $n$ case, use $lambda=frac1{n+1}$, $x=x_{n+1}$, $y=frac1nsumlimits_{k=1}^nx_k$, then $z=frac1{n+1}sumlimits_{k=1}^{n+1}x_k=lambda x+(1-lambda)y$ hence $f(z)leqslantlambda f(x)+(1-lambda)f(y)$, now by the recurrence hypothesis, $f(y)leqslantfrac1nsumlimits_{k=1}^nf(x_k)$, hence $f(z)leqslantlambda f(x_{n+1})+(1-lambda)frac1nsumlimits_{k=1}^nf(x_k)$, qed.
– Did
Jul 11 '16 at 19:44