Mixing
Proving lipschitz continuity based on bounded gradient
$begingroup$
I would like to prove that if $S$ is a non-empty convex subset, and $f:S xrightarrow{} R $ is a convex and differentiable function, that the following holds true:
$ |f(x)-f(y)| le L|x-y |_{2} Longleftrightarrow | nabla f(x) |_{2} le L $
I already have $rightarrow$, but I am unfinished with $leftarrow$. The following is what I currently have:
From convexity:
$f(x) - f(y) le nabla f(x)^T(x-y) $
According to Cauchy-Schwartz, the following also holds true:
$ |nabla f(x)^T(x-y)| le |nabla f(x)|_{2} |x-y|_{2}$
It is given that $| nabla f(x) |_{2} le L$, so the following is true:
$ |nabla f(x)^T(x-y)| le L |x-y|_{2} $
Hence, because $forall a: |a| ge a$, we have:
$ nabla f(x)^T(x-y) le |nabla f(x)^T(x-y)| le L |x-y|_{2}$
So far I have proven the following:
$ f(x)-f(y) le L|x-y |_{2} $
But I am missing the absolute value, I have made some attempts but they all involve making fallacies with inequality signs... Does anyone have an idea?
analysis convex-analysis
asked Jan 22 at 11:40
Gilles GGilles G
112
$endgroup$
add a comment |
$begingroup$
I would like to prove that if $S$ is a non-empty convex subset, and $f:S xrightarrow{} R $ is a convex and differentiable function, that the following holds true:
$ |f(x)-f(y)| le L|x-y |_{2} Longleftrightarrow | nabla f(x) |_{2} le L $
I already have $rightarrow$, but I am unfinished with $leftarrow$. The following is what I currently have:
From convexity:
$f(x) - f(y) le nabla f(x)^T(x-y) $
According to Cauchy-Schwartz, the following also holds true:
$ |nabla f(x)^T(x-y)| le |nabla f(x)|_{2} |x-y|_{2}$
It is given that $| nabla f(x) |_{2} le L$, so the following is true:
$ |nabla f(x)^T(x-y)| le L |x-y|_{2} $
Hence, because $forall a: |a| ge a$, we have:
$ nabla f(x)^T(x-y) le |nabla f(x)^T(x-y)| le L |x-y|_{2}$
So far I have proven the following:
$ f(x)-f(y) le L|x-y |_{2} $
But I am missing the absolute value, I have made some attempts but they all involve making fallacies with inequality signs... Does anyone have an idea?
analysis convex-analysis
asked Jan 22 at 11:40
Gilles GGilles G
112
$endgroup$
$begingroup$
The easiest way is always the fundamental theorem of calculus; $$f(x+h)-f(x)=int_0^1 frac{d}{dt}[f(x+th)], dt.$$
$endgroup$
– Giuseppe Negro
Jan 22 at 11:50
add a comment |
$begingroup$
I would like to prove that if $S$ is a non-empty convex subset, and $f:S xrightarrow{} R $ is a convex and differentiable function, that the following holds true:
$ |f(x)-f(y)| le L|x-y |_{2} Longleftrightarrow | nabla f(x) |_{2} le L $
I already have $rightarrow$, but I am unfinished with $leftarrow$. The following is what I currently have:
From convexity:
$f(x) - f(y) le nabla f(x)^T(x-y) $
According to Cauchy-Schwartz, the following also holds true:
$ |nabla f(x)^T(x-y)| le |nabla f(x)|_{2} |x-y|_{2}$
It is given that $| nabla f(x) |_{2} le L$, so the following is true:
$ |nabla f(x)^T(x-y)| le L |x-y|_{2} $
Hence, because $forall a: |a| ge a$, we have:
$ nabla f(x)^T(x-y) le |nabla f(x)^T(x-y)| le L |x-y|_{2}$
So far I have proven the following:
$ f(x)-f(y) le L|x-y |_{2} $
But I am missing the absolute value, I have made some attempts but they all involve making fallacies with inequality signs... Does anyone have an idea?
analysis convex-analysis
asked Jan 22 at 11:40
Gilles GGilles G
112
$endgroup$
I would like to prove that if $S$ is a non-empty convex subset, and $f:S xrightarrow{} R $ is a convex and differentiable function, that the following holds true:
$ |f(x)-f(y)| le L|x-y |_{2} Longleftrightarrow | nabla f(x) |_{2} le L $
I already have $rightarrow$, but I am unfinished with $leftarrow$. The following is what I currently have:
From convexity:
$f(x) - f(y) le nabla f(x)^T(x-y) $
According to Cauchy-Schwartz, the following also holds true:
$ |nabla f(x)^T(x-y)| le |nabla f(x)|_{2} |x-y|_{2}$
It is given that $| nabla f(x) |_{2} le L$, so the following is true:
$ |nabla f(x)^T(x-y)| le L |x-y|_{2} $
Hence, because $forall a: |a| ge a$, we have:
$ nabla f(x)^T(x-y) le |nabla f(x)^T(x-y)| le L |x-y|_{2}$
So far I have proven the following:
$ f(x)-f(y) le L|x-y |_{2} $
But I am missing the absolute value, I have made some attempts but they all involve making fallacies with inequality signs... Does anyone have an idea?
analysis convex-analysis
analysis convex-analysis
asked Jan 22 at 11:40
Gilles GGilles G
112
asked Jan 22 at 11:40
Gilles GGilles G
112
asked Jan 22 at 11:40
Gilles GGilles G
112
asked Jan 22 at 11:40
Gilles GGilles G
112
asked Jan 22 at 11:40
Gilles GGilles G
112
112
$begingroup$
The easiest way is always the fundamental theorem of calculus; $$f(x+h)-f(x)=int_0^1 frac{d}{dt}[f(x+th)], dt.$$
$endgroup$
– Giuseppe Negro
Jan 22 at 11:50
add a comment |
$begingroup$
The easiest way is always the fundamental theorem of calculus; $$f(x+h)-f(x)=int_0^1 frac{d}{dt}[f(x+th)], dt.$$
$endgroup$
– Giuseppe Negro
Jan 22 at 11:50
$begingroup$
The easiest way is always the fundamental theorem of calculus; $$f(x+h)-f(x)=int_0^1 frac{d}{dt}[f(x+th)], dt.$$
$endgroup$
– Giuseppe Negro
Jan 22 at 11:50
$begingroup$
The easiest way is always the fundamental theorem of calculus; $$f(x+h)-f(x)=int_0^1 frac{d}{dt}[f(x+th)], dt.$$
$endgroup$
– Giuseppe Negro
Jan 22 at 11:50
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hint
Convexity of $f$ is not necessary here. Since $S$ is convex, by mean value theorem, there is $c_{x,y}in (x,y)={x+t(y-x)mid tin (0,1)}$ s.t. $$f(x)-f(y)=nabla f(c_{x,y})cdot (y-x).$$
answered Jan 22 at 11:46
DylanDylan
3118
$endgroup$
add a comment |
$begingroup$
I can't really tell if these answers other people gave me are very useful or not.
However, I think i found the right answer: you have to work symetrically as above with $f(y) - f(x) le nabla f(y)^T(y-x)$.
You will then end up at $f(x)-f(y) ge -L|x-y |_{2}$, then you will have proven the theorem...
answered Jan 22 at 17:10
Gilles GGilles G
112
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
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oldest
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votes
$begingroup$
Hint
Convexity of $f$ is not necessary here. Since $S$ is convex, by mean value theorem, there is $c_{x,y}in (x,y)={x+t(y-x)mid tin (0,1)}$ s.t. $$f(x)-f(y)=nabla f(c_{x,y})cdot (y-x).$$
answered Jan 22 at 11:46
DylanDylan
3118
$endgroup$
add a comment |
$begingroup$
Hint
Convexity of $f$ is not necessary here. Since $S$ is convex, by mean value theorem, there is $c_{x,y}in (x,y)={x+t(y-x)mid tin (0,1)}$ s.t. $$f(x)-f(y)=nabla f(c_{x,y})cdot (y-x).$$
answered Jan 22 at 11:46
DylanDylan
3118
$endgroup$
add a comment |
$begingroup$
Hint
Convexity of $f$ is not necessary here. Since $S$ is convex, by mean value theorem, there is $c_{x,y}in (x,y)={x+t(y-x)mid tin (0,1)}$ s.t. $$f(x)-f(y)=nabla f(c_{x,y})cdot (y-x).$$
answered Jan 22 at 11:46
DylanDylan
3118
$endgroup$
Hint
Convexity of $f$ is not necessary here. Since $S$ is convex, by mean value theorem, there is $c_{x,y}in (x,y)={x+t(y-x)mid tin (0,1)}$ s.t. $$f(x)-f(y)=nabla f(c_{x,y})cdot (y-x).$$
answered Jan 22 at 11:46
DylanDylan
3118
answered Jan 22 at 11:46
DylanDylan
3118
answered Jan 22 at 11:46
DylanDylan
3118
answered Jan 22 at 11:46
DylanDylan
3118
3118
add a comment |
add a comment |
$begingroup$
I can't really tell if these answers other people gave me are very useful or not.
However, I think i found the right answer: you have to work symetrically as above with $f(y) - f(x) le nabla f(y)^T(y-x)$.
You will then end up at $f(x)-f(y) ge -L|x-y |_{2}$, then you will have proven the theorem...
answered Jan 22 at 17:10
Gilles GGilles G
112
$endgroup$
add a comment |
$begingroup$
I can't really tell if these answers other people gave me are very useful or not.
However, I think i found the right answer: you have to work symetrically as above with $f(y) - f(x) le nabla f(y)^T(y-x)$.
You will then end up at $f(x)-f(y) ge -L|x-y |_{2}$, then you will have proven the theorem...
answered Jan 22 at 17:10
Gilles GGilles G
112
$endgroup$
add a comment |
$begingroup$
I can't really tell if these answers other people gave me are very useful or not.
However, I think i found the right answer: you have to work symetrically as above with $f(y) - f(x) le nabla f(y)^T(y-x)$.
You will then end up at $f(x)-f(y) ge -L|x-y |_{2}$, then you will have proven the theorem...
answered Jan 22 at 17:10
Gilles GGilles G
112
$endgroup$
I can't really tell if these answers other people gave me are very useful or not.
However, I think i found the right answer: you have to work symetrically as above with $f(y) - f(x) le nabla f(y)^T(y-x)$.
You will then end up at $f(x)-f(y) ge -L|x-y |_{2}$, then you will have proven the theorem...
answered Jan 22 at 17:10
Gilles GGilles G
112
answered Jan 22 at 17:10
Gilles GGilles G
112
answered Jan 22 at 17:10
Gilles GGilles G
112
answered Jan 22 at 17:10
Gilles GGilles G
112
112
add a comment |
add a comment |
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$begingroup$
The easiest way is always the fundamental theorem of calculus; $$f(x+h)-f(x)=int_0^1 frac{d}{dt}[f(x+th)], dt.$$
$endgroup$
– Giuseppe Negro
Jan 22 at 11:50