Equivalent Definition of Strong Convexity
Let $f in mathcal{C}^2(mathbb{R}^n).$ Recall that we defined $f$ to be strongly convex if there exists $beta > 0$ such that $langle D^{2}f|_{x}y,yranglegebeta$ for every $x, y in mathbb{R}^{n}$ or equivalently if the function $g(x)=f(x)-frac{beta}{2}|x|^{2}$ is convex. Show that if there exists $gamma>0$ such that
$$f(tx+(1-t)y) leq tf(x)+(1-t)f(y)-gamma t(1-t)|x-y|^{2}tag{*}$$
for all $x,y in mathbb{R}^{n}, t in [0,1]$ then the function $g(x)=f(x)-gamma|x|^2$ is convex.
I tried showing that the function $g(x)=f(x)-gamma |x|^2$ is convex using the above assumption but did not strike any luck. Any hints on how to proceed for this problem are appreciated. Than you for you help.
real-analysis multivariable-calculus convex-analysis convex-optimization
edited Nov 26 at 15:35
A.Γ.
21.8k22455
asked Nov 26 at 9:20
Joe Man Analysis
33419
add a comment |
Let $f in mathcal{C}^2(mathbb{R}^n).$ Recall that we defined $f$ to be strongly convex if there exists $beta > 0$ such that $langle D^{2}f|_{x}y,yranglegebeta$ for every $x, y in mathbb{R}^{n}$ or equivalently if the function $g(x)=f(x)-frac{beta}{2}|x|^{2}$ is convex. Show that if there exists $gamma>0$ such that
$$f(tx+(1-t)y) leq tf(x)+(1-t)f(y)-gamma t(1-t)|x-y|^{2}tag{*}$$
for all $x,y in mathbb{R}^{n}, t in [0,1]$ then the function $g(x)=f(x)-gamma|x|^2$ is convex.
I tried showing that the function $g(x)=f(x)-gamma |x|^2$ is convex using the above assumption but did not strike any luck. Any hints on how to proceed for this problem are appreciated. Than you for you help.
real-analysis multivariable-calculus convex-analysis convex-optimization
edited Nov 26 at 15:35
A.Γ.
21.8k22455
asked Nov 26 at 9:20
Joe Man Analysis
33419
Your statement "Show that if ..." had no then. I edited the question adding what I think you were trying to show. Tell me if I am wrong.
– A.Γ.
Nov 26 at 15:38
add a comment |
Let $f in mathcal{C}^2(mathbb{R}^n).$ Recall that we defined $f$ to be strongly convex if there exists $beta > 0$ such that $langle D^{2}f|_{x}y,yranglegebeta$ for every $x, y in mathbb{R}^{n}$ or equivalently if the function $g(x)=f(x)-frac{beta}{2}|x|^{2}$ is convex. Show that if there exists $gamma>0$ such that
$$f(tx+(1-t)y) leq tf(x)+(1-t)f(y)-gamma t(1-t)|x-y|^{2}tag{*}$$
for all $x,y in mathbb{R}^{n}, t in [0,1]$ then the function $g(x)=f(x)-gamma|x|^2$ is convex.
I tried showing that the function $g(x)=f(x)-gamma |x|^2$ is convex using the above assumption but did not strike any luck. Any hints on how to proceed for this problem are appreciated. Than you for you help.
real-analysis multivariable-calculus convex-analysis convex-optimization
edited Nov 26 at 15:35
A.Γ.
21.8k22455
asked Nov 26 at 9:20
Joe Man Analysis
33419
Let $f in mathcal{C}^2(mathbb{R}^n).$ Recall that we defined $f$ to be strongly convex if there exists $beta > 0$ such that $langle D^{2}f|_{x}y,yranglegebeta$ for every $x, y in mathbb{R}^{n}$ or equivalently if the function $g(x)=f(x)-frac{beta}{2}|x|^{2}$ is convex. Show that if there exists $gamma>0$ such that
$$f(tx+(1-t)y) leq tf(x)+(1-t)f(y)-gamma t(1-t)|x-y|^{2}tag{*}$$
for all $x,y in mathbb{R}^{n}, t in [0,1]$ then the function $g(x)=f(x)-gamma|x|^2$ is convex.
I tried showing that the function $g(x)=f(x)-gamma |x|^2$ is convex using the above assumption but did not strike any luck. Any hints on how to proceed for this problem are appreciated. Than you for you help.
real-analysis multivariable-calculus convex-analysis convex-optimization
real-analysis multivariable-calculus convex-analysis convex-optimization
edited Nov 26 at 15:35
A.Γ.
21.8k22455
asked Nov 26 at 9:20
Joe Man Analysis
33419
edited Nov 26 at 15:35
A.Γ.
21.8k22455
asked Nov 26 at 9:20
Joe Man Analysis
33419
edited Nov 26 at 15:35
A.Γ.
21.8k22455
edited Nov 26 at 15:35
A.Γ.
21.8k22455
edited Nov 26 at 15:35
A.Γ.
21.8k22455
21.8k22455
asked Nov 26 at 9:20
Joe Man Analysis
33419
asked Nov 26 at 9:20
Joe Man Analysis
33419
asked Nov 26 at 9:20
Joe Man Analysis
33419
33419
Your statement "Show that if ..." had no then. I edited the question adding what I think you were trying to show. Tell me if I am wrong.
– A.Γ.
Nov 26 at 15:38
add a comment |
Your statement "Show that if ..." had no then. I edited the question adding what I think you were trying to show. Tell me if I am wrong.
– A.Γ.
Nov 26 at 15:38
Your statement "Show that if ..." had no then. I edited the question adding what I think you were trying to show. Tell me if I am wrong.
– A.Γ.
Nov 26 at 15:38
Your statement "Show that if ..." had no then. I edited the question adding what I think you were trying to show. Tell me if I am wrong.
– A.Γ.
Nov 26 at 15:38
add a comment |
1 Answer
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"Equivalently" is not really right here as the definition of strong convexity does not assume $f in mathcal{C}^2$. However, $f$ satisfies (*) iff $g$ is convex.
The proof: (*) is equivalent to
$$
g(tx+(1-t)y)le tg(x)+(1-t)g(y)+gammatext{Res}
$$
where
$$
text{Res}=t|x|^2+(1-t)|y|^2-|tx+(1-t)y|^2-t(1-t)|x-y|^2.
$$
It is straightforward calculation to show that $text{Res}=0$ for $ell^2$-norm. For example, the $|x|^2$ term is
$$
t|x|^2-t^2|x|^2-t(1-t)|x|^2=(t-t^2-t+t^2)|x|^2=0.
$$
I leave it for you to verify the rest.
answered Nov 26 at 15:26
A.Γ.
21.8k22455
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
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oldest
votes
"Equivalently" is not really right here as the definition of strong convexity does not assume $f in mathcal{C}^2$. However, $f$ satisfies (*) iff $g$ is convex.
The proof: (*) is equivalent to
$$
g(tx+(1-t)y)le tg(x)+(1-t)g(y)+gammatext{Res}
$$
where
$$
text{Res}=t|x|^2+(1-t)|y|^2-|tx+(1-t)y|^2-t(1-t)|x-y|^2.
$$
It is straightforward calculation to show that $text{Res}=0$ for $ell^2$-norm. For example, the $|x|^2$ term is
$$
t|x|^2-t^2|x|^2-t(1-t)|x|^2=(t-t^2-t+t^2)|x|^2=0.
$$
I leave it for you to verify the rest.
answered Nov 26 at 15:26
A.Γ.
21.8k22455
add a comment |
"Equivalently" is not really right here as the definition of strong convexity does not assume $f in mathcal{C}^2$. However, $f$ satisfies (*) iff $g$ is convex.
The proof: (*) is equivalent to
$$
g(tx+(1-t)y)le tg(x)+(1-t)g(y)+gammatext{Res}
$$
where
$$
text{Res}=t|x|^2+(1-t)|y|^2-|tx+(1-t)y|^2-t(1-t)|x-y|^2.
$$
It is straightforward calculation to show that $text{Res}=0$ for $ell^2$-norm. For example, the $|x|^2$ term is
$$
t|x|^2-t^2|x|^2-t(1-t)|x|^2=(t-t^2-t+t^2)|x|^2=0.
$$
I leave it for you to verify the rest.
answered Nov 26 at 15:26
A.Γ.
21.8k22455
add a comment |
"Equivalently" is not really right here as the definition of strong convexity does not assume $f in mathcal{C}^2$. However, $f$ satisfies (*) iff $g$ is convex.
The proof: (*) is equivalent to
$$
g(tx+(1-t)y)le tg(x)+(1-t)g(y)+gammatext{Res}
$$
where
$$
text{Res}=t|x|^2+(1-t)|y|^2-|tx+(1-t)y|^2-t(1-t)|x-y|^2.
$$
It is straightforward calculation to show that $text{Res}=0$ for $ell^2$-norm. For example, the $|x|^2$ term is
$$
t|x|^2-t^2|x|^2-t(1-t)|x|^2=(t-t^2-t+t^2)|x|^2=0.
$$
I leave it for you to verify the rest.
answered Nov 26 at 15:26
A.Γ.
21.8k22455
"Equivalently" is not really right here as the definition of strong convexity does not assume $f in mathcal{C}^2$. However, $f$ satisfies (*) iff $g$ is convex.
The proof: (*) is equivalent to
$$
g(tx+(1-t)y)le tg(x)+(1-t)g(y)+gammatext{Res}
$$
where
$$
text{Res}=t|x|^2+(1-t)|y|^2-|tx+(1-t)y|^2-t(1-t)|x-y|^2.
$$
It is straightforward calculation to show that $text{Res}=0$ for $ell^2$-norm. For example, the $|x|^2$ term is
$$
t|x|^2-t^2|x|^2-t(1-t)|x|^2=(t-t^2-t+t^2)|x|^2=0.
$$
I leave it for you to verify the rest.
answered Nov 26 at 15:26
A.Γ.
21.8k22455
answered Nov 26 at 15:26
A.Γ.
21.8k22455
answered Nov 26 at 15:26
A.Γ.
21.8k22455
answered Nov 26 at 15:26
A.Γ.
21.8k22455
21.8k22455
add a comment |
add a comment |
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Your statement "Show that if ..." had no then. I edited the question adding what I think you were trying to show. Tell me if I am wrong.
– A.Γ.
Nov 26 at 15:38