How to show negative entropy function $f(x)=sum_{i=1}^nx_ilog(x_i)$ is strongly convex?
$begingroup$
Let $x in mathbb{R}^n$ belongs to $S$ where
$$
S= {x in mathbb{R}^n mid x succ 0, |x|_{infty} leq M}
$$
where $succ$ is the generalized inequality which means all elements of $x$ are positive and $log$ is natural logarithm. Use the following theorem to show that $f(x)=sum_{i=1}^nx_ilog(x_i)$ is $frac{1}{M}$-strongly convex over $S$.
Theorem: f is $alpha$-strongly convex if and only if $nabla^2f(x) succeq alpha I$ for all $x$.
Definition:$f$ is $alpha$-strongly convex if there exist a constant $alpha$ such that
$$ f(y) geq f(x)+left<f '(x),y-xright>+frac{alpha}{2}|y-x|^2$$
or
$$ left<f'(y)-f '(x),y-xright> geq alpha|y-x|^2$$
for all $x,y$.
linear-algebra convex-analysis hessian-matrix
asked Dec 28 '18 at 18:57
SaeedSaeed
1,036310
$endgroup$
add a comment |
$begingroup$
Let $x in mathbb{R}^n$ belongs to $S$ where
$$
S= {x in mathbb{R}^n mid x succ 0, |x|_{infty} leq M}
$$
where $succ$ is the generalized inequality which means all elements of $x$ are positive and $log$ is natural logarithm. Use the following theorem to show that $f(x)=sum_{i=1}^nx_ilog(x_i)$ is $frac{1}{M}$-strongly convex over $S$.
Theorem: f is $alpha$-strongly convex if and only if $nabla^2f(x) succeq alpha I$ for all $x$.
Definition:$f$ is $alpha$-strongly convex if there exist a constant $alpha$ such that
$$ f(y) geq f(x)+left<f '(x),y-xright>+frac{alpha}{2}|y-x|^2$$
or
$$ left<f'(y)-f '(x),y-xright> geq alpha|y-x|^2$$
for all $x,y$.
linear-algebra convex-analysis hessian-matrix
asked Dec 28 '18 at 18:57
SaeedSaeed
1,036310
$endgroup$
add a comment |
$begingroup$
Let $x in mathbb{R}^n$ belongs to $S$ where
$$
S= {x in mathbb{R}^n mid x succ 0, |x|_{infty} leq M}
$$
where $succ$ is the generalized inequality which means all elements of $x$ are positive and $log$ is natural logarithm. Use the following theorem to show that $f(x)=sum_{i=1}^nx_ilog(x_i)$ is $frac{1}{M}$-strongly convex over $S$.
Theorem: f is $alpha$-strongly convex if and only if $nabla^2f(x) succeq alpha I$ for all $x$.
Definition:$f$ is $alpha$-strongly convex if there exist a constant $alpha$ such that
$$ f(y) geq f(x)+left<f '(x),y-xright>+frac{alpha}{2}|y-x|^2$$
or
$$ left<f'(y)-f '(x),y-xright> geq alpha|y-x|^2$$
for all $x,y$.
linear-algebra convex-analysis hessian-matrix
asked Dec 28 '18 at 18:57
SaeedSaeed
1,036310
$endgroup$
Let $x in mathbb{R}^n$ belongs to $S$ where
$$
S= {x in mathbb{R}^n mid x succ 0, |x|_{infty} leq M}
$$
where $succ$ is the generalized inequality which means all elements of $x$ are positive and $log$ is natural logarithm. Use the following theorem to show that $f(x)=sum_{i=1}^nx_ilog(x_i)$ is $frac{1}{M}$-strongly convex over $S$.
Theorem: f is $alpha$-strongly convex if and only if $nabla^2f(x) succeq alpha I$ for all $x$.
Definition:$f$ is $alpha$-strongly convex if there exist a constant $alpha$ such that
$$ f(y) geq f(x)+left<f '(x),y-xright>+frac{alpha}{2}|y-x|^2$$
or
$$ left<f'(y)-f '(x),y-xright> geq alpha|y-x|^2$$
for all $x,y$.
linear-algebra convex-analysis hessian-matrix
linear-algebra convex-analysis hessian-matrix
asked Dec 28 '18 at 18:57
SaeedSaeed
1,036310
asked Dec 28 '18 at 18:57
SaeedSaeed
1,036310
edited Jan 17 at 19:57
Saeed
asked Dec 28 '18 at 18:57
SaeedSaeed
1,036310
asked Dec 28 '18 at 18:57
SaeedSaeed
1,036310
asked Dec 28 '18 at 18:57
SaeedSaeed
1,036310
1,036310
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
First of all, It seems to me that in the definition of $alpha$-strongly convex function should have a coefficient of $alpha$ but not $frac{alpha}{2}$.
Now here goes the proof
$$
begin{aligned}
frac{partial^2 f}{partial x_i partial_j} & = &frac{partial}{partial x_j}[log(x_i) + 1] \
& = & left {
begin{aligned}
1/x_i quad text{if } i =j \
0 quad text{if } i neq j
end{aligned}
right .
end{aligned}
$$
Since $0<x_ileq M$, then $1/x_i geq 1/M$
Thus $nabla^2 f geq frac{1}{M} I$, which is the theorem for $1/M$-strongly convex function.
answered Jan 17 at 18:13
MoonKnightMoonKnight
1,429611
$endgroup$
$begingroup$
I added two equivalent definition of $alpha$-strongly convex functions. In the last two lines $1/x_i geq 1/M$, $nabla^2 f geq 1/M$ and $1/M$ strongly convex.
$endgroup$
– Saeed
Jan 17 at 20:05
$begingroup$
you are right. It should be $1/M$ strongly convex.
$endgroup$
– MoonKnight
Jan 17 at 21:53
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
$begingroup$
First of all, It seems to me that in the definition of $alpha$-strongly convex function should have a coefficient of $alpha$ but not $frac{alpha}{2}$.
Now here goes the proof
$$
begin{aligned}
frac{partial^2 f}{partial x_i partial_j} & = &frac{partial}{partial x_j}[log(x_i) + 1] \
& = & left {
begin{aligned}
1/x_i quad text{if } i =j \
0 quad text{if } i neq j
end{aligned}
right .
end{aligned}
$$
Since $0<x_ileq M$, then $1/x_i geq 1/M$
Thus $nabla^2 f geq frac{1}{M} I$, which is the theorem for $1/M$-strongly convex function.
answered Jan 17 at 18:13
MoonKnightMoonKnight
1,429611
$endgroup$
$begingroup$
I added two equivalent definition of $alpha$-strongly convex functions. In the last two lines $1/x_i geq 1/M$, $nabla^2 f geq 1/M$ and $1/M$ strongly convex.
$endgroup$
– Saeed
Jan 17 at 20:05
$begingroup$
you are right. It should be $1/M$ strongly convex.
$endgroup$
– MoonKnight
Jan 17 at 21:53
add a comment |
$begingroup$
First of all, It seems to me that in the definition of $alpha$-strongly convex function should have a coefficient of $alpha$ but not $frac{alpha}{2}$.
Now here goes the proof
$$
begin{aligned}
frac{partial^2 f}{partial x_i partial_j} & = &frac{partial}{partial x_j}[log(x_i) + 1] \
& = & left {
begin{aligned}
1/x_i quad text{if } i =j \
0 quad text{if } i neq j
end{aligned}
right .
end{aligned}
$$
Since $0<x_ileq M$, then $1/x_i geq 1/M$
Thus $nabla^2 f geq frac{1}{M} I$, which is the theorem for $1/M$-strongly convex function.
answered Jan 17 at 18:13
MoonKnightMoonKnight
1,429611
$endgroup$
$begingroup$
I added two equivalent definition of $alpha$-strongly convex functions. In the last two lines $1/x_i geq 1/M$, $nabla^2 f geq 1/M$ and $1/M$ strongly convex.
$endgroup$
– Saeed
Jan 17 at 20:05
$begingroup$
you are right. It should be $1/M$ strongly convex.
$endgroup$
– MoonKnight
Jan 17 at 21:53
add a comment |
$begingroup$
First of all, It seems to me that in the definition of $alpha$-strongly convex function should have a coefficient of $alpha$ but not $frac{alpha}{2}$.
Now here goes the proof
$$
begin{aligned}
frac{partial^2 f}{partial x_i partial_j} & = &frac{partial}{partial x_j}[log(x_i) + 1] \
& = & left {
begin{aligned}
1/x_i quad text{if } i =j \
0 quad text{if } i neq j
end{aligned}
right .
end{aligned}
$$
Since $0<x_ileq M$, then $1/x_i geq 1/M$
Thus $nabla^2 f geq frac{1}{M} I$, which is the theorem for $1/M$-strongly convex function.
answered Jan 17 at 18:13
MoonKnightMoonKnight
1,429611
$endgroup$
First of all, It seems to me that in the definition of $alpha$-strongly convex function should have a coefficient of $alpha$ but not $frac{alpha}{2}$.
Now here goes the proof
$$
begin{aligned}
frac{partial^2 f}{partial x_i partial_j} & = &frac{partial}{partial x_j}[log(x_i) + 1] \
& = & left {
begin{aligned}
1/x_i quad text{if } i =j \
0 quad text{if } i neq j
end{aligned}
right .
end{aligned}
$$
Since $0<x_ileq M$, then $1/x_i geq 1/M$
Thus $nabla^2 f geq frac{1}{M} I$, which is the theorem for $1/M$-strongly convex function.
answered Jan 17 at 18:13
MoonKnightMoonKnight
1,429611
edited Jan 17 at 21:53
answered Jan 17 at 18:13
MoonKnightMoonKnight
1,429611
answered Jan 17 at 18:13
MoonKnightMoonKnight
1,429611
answered Jan 17 at 18:13
MoonKnightMoonKnight
1,429611
1,429611
$begingroup$
I added two equivalent definition of $alpha$-strongly convex functions. In the last two lines $1/x_i geq 1/M$, $nabla^2 f geq 1/M$ and $1/M$ strongly convex.
$endgroup$
– Saeed
Jan 17 at 20:05
$begingroup$
you are right. It should be $1/M$ strongly convex.
$endgroup$
– MoonKnight
Jan 17 at 21:53
add a comment |
$begingroup$
I added two equivalent definition of $alpha$-strongly convex functions. In the last two lines $1/x_i geq 1/M$, $nabla^2 f geq 1/M$ and $1/M$ strongly convex.
$endgroup$
– Saeed
Jan 17 at 20:05
$begingroup$
you are right. It should be $1/M$ strongly convex.
$endgroup$
– MoonKnight
Jan 17 at 21:53
$begingroup$
I added two equivalent definition of $alpha$-strongly convex functions. In the last two lines $1/x_i geq 1/M$, $nabla^2 f geq 1/M$ and $1/M$ strongly convex.
$endgroup$
– Saeed
Jan 17 at 20:05
$begingroup$
I added two equivalent definition of $alpha$-strongly convex functions. In the last two lines $1/x_i geq 1/M$, $nabla^2 f geq 1/M$ and $1/M$ strongly convex.
$endgroup$
– Saeed
Jan 17 at 20:05
$begingroup$
you are right. It should be $1/M$ strongly convex.
$endgroup$
– MoonKnight
Jan 17 at 21:53
$begingroup$
you are right. It should be $1/M$ strongly convex.
$endgroup$
– MoonKnight
Jan 17 at 21:53
add a comment |
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