Convex or Concave Function
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the problem is: Is the function $phi(f)=frac{sqrt{f}}{2}cdotlog f$ for $fin L^2(mathbb{R}_{>0})$ convex or concave.
My idee or presumption is, that the function $phi$ is concave, because we have the inequalitiy[sqrt{alphacdot f+(1-alpha)cdot g}geq alphacdot sqrt{f}+(1-alpha)cdotsqrt{g}quadtext{for}~alphain(0,1).]
functional-analysis
asked Jan 22 at 14:27
FuncAna09FuncAna09
93
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add a comment |
$begingroup$
the problem is: Is the function $phi(f)=frac{sqrt{f}}{2}cdotlog f$ for $fin L^2(mathbb{R}_{>0})$ convex or concave.
My idee or presumption is, that the function $phi$ is concave, because we have the inequalitiy[sqrt{alphacdot f+(1-alpha)cdot g}geq alphacdot sqrt{f}+(1-alpha)cdotsqrt{g}quadtext{for}~alphain(0,1).]
functional-analysis
asked Jan 22 at 14:27
FuncAna09FuncAna09
93
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How do you define convexity of $phi$?
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– gerw
Jan 22 at 15:20
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We call a map convex, if [phi(alphacdot f+(1-alpha)cdot g)leq alphacdot phi(f)+(1-alpha)cdot phi(f)] for $alphain (0,1)$
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– FuncAna09
Jan 22 at 17:59
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But $phi( something )$ is a function. Is $ge$ to be understood pointwise?
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– gerw
Jan 22 at 18:00
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Sorry, the correct definition is $phi(f(x)).$ f is also a map.
$endgroup$
– FuncAna09
Jan 22 at 18:06
add a comment |
$begingroup$
the problem is: Is the function $phi(f)=frac{sqrt{f}}{2}cdotlog f$ for $fin L^2(mathbb{R}_{>0})$ convex or concave.
My idee or presumption is, that the function $phi$ is concave, because we have the inequalitiy[sqrt{alphacdot f+(1-alpha)cdot g}geq alphacdot sqrt{f}+(1-alpha)cdotsqrt{g}quadtext{for}~alphain(0,1).]
functional-analysis
asked Jan 22 at 14:27
FuncAna09FuncAna09
93
$endgroup$
the problem is: Is the function $phi(f)=frac{sqrt{f}}{2}cdotlog f$ for $fin L^2(mathbb{R}_{>0})$ convex or concave.
My idee or presumption is, that the function $phi$ is concave, because we have the inequalitiy[sqrt{alphacdot f+(1-alpha)cdot g}geq alphacdot sqrt{f}+(1-alpha)cdotsqrt{g}quadtext{for}~alphain(0,1).]
functional-analysis
functional-analysis
asked Jan 22 at 14:27
FuncAna09FuncAna09
93
asked Jan 22 at 14:27
FuncAna09FuncAna09
93
asked Jan 22 at 14:27
FuncAna09FuncAna09
93
asked Jan 22 at 14:27
FuncAna09FuncAna09
93
asked Jan 22 at 14:27
FuncAna09FuncAna09
93
93
$begingroup$
How do you define convexity of $phi$?
$endgroup$
– gerw
Jan 22 at 15:20
$begingroup$
We call a map convex, if [phi(alphacdot f+(1-alpha)cdot g)leq alphacdot phi(f)+(1-alpha)cdot phi(f)] for $alphain (0,1)$
$endgroup$
– FuncAna09
Jan 22 at 17:59
$begingroup$
But $phi( something )$ is a function. Is $ge$ to be understood pointwise?
$endgroup$
– gerw
Jan 22 at 18:00
$begingroup$
Sorry, the correct definition is $phi(f(x)).$ f is also a map.
$endgroup$
– FuncAna09
Jan 22 at 18:06
add a comment |
$begingroup$
How do you define convexity of $phi$?
$endgroup$
– gerw
Jan 22 at 15:20
$begingroup$
We call a map convex, if [phi(alphacdot f+(1-alpha)cdot g)leq alphacdot phi(f)+(1-alpha)cdot phi(f)] for $alphain (0,1)$
$endgroup$
– FuncAna09
Jan 22 at 17:59
$begingroup$
But $phi( something )$ is a function. Is $ge$ to be understood pointwise?
$endgroup$
– gerw
Jan 22 at 18:00
$begingroup$
Sorry, the correct definition is $phi(f(x)).$ f is also a map.
$endgroup$
– FuncAna09
Jan 22 at 18:06
$begingroup$
How do you define convexity of $phi$?
$endgroup$
– gerw
Jan 22 at 15:20
$begingroup$
How do you define convexity of $phi$?
$endgroup$
– gerw
Jan 22 at 15:20
$begingroup$
We call a map convex, if [phi(alphacdot f+(1-alpha)cdot g)leq alphacdot phi(f)+(1-alpha)cdot phi(f)] for $alphain (0,1)$
$endgroup$
– FuncAna09
Jan 22 at 17:59
$begingroup$
We call a map convex, if [phi(alphacdot f+(1-alpha)cdot g)leq alphacdot phi(f)+(1-alpha)cdot phi(f)] for $alphain (0,1)$
$endgroup$
– FuncAna09
Jan 22 at 17:59
$begingroup$
But $phi( something )$ is a function. Is $ge$ to be understood pointwise?
$endgroup$
– gerw
Jan 22 at 18:00
$begingroup$
But $phi( something )$ is a function. Is $ge$ to be understood pointwise?
$endgroup$
– gerw
Jan 22 at 18:00
$begingroup$
Sorry, the correct definition is $phi(f(x)).$ f is also a map.
$endgroup$
– FuncAna09
Jan 22 at 18:06
$begingroup$
Sorry, the correct definition is $phi(f(x)).$ f is also a map.
$endgroup$
– FuncAna09
Jan 22 at 18:06
add a comment |
1 Answer
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Waht i have so far for $phi:mathbb{R}_{>0}tomathbb{R}$ with $fmapsto phi(f)$
$begin{align*}
-phi(alphacdot f+(1-alpha)cdot g)&=-sqrt{alphacdot f+(1-alpha)cdot g}log(alphacdot f+(1-alpha)cdot g)\
&leq -(alphasqrt f+(1-alpha) sqrt{g})log(f^{alpha}g^{(1-alpha)})\
&=-left[(alphasqrt f+(1-alpha) sqrt{g})(alphalog(f)+(1-alpha)log(g))right]\
&=-[alpha^2sqrt{f}log(f)+alpha(1-alpha)sqrt{f}log(g)+alpha(1-alpha)sqrt{g}log(f)\
&+(1-alpha)^2sqrt{g}log(g)]\
&=...
end{align*}$
answered Jan 22 at 20:31
FuncAna09FuncAna09
93
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add a comment |
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Waht i have so far for $phi:mathbb{R}_{>0}tomathbb{R}$ with $fmapsto phi(f)$
$begin{align*}
-phi(alphacdot f+(1-alpha)cdot g)&=-sqrt{alphacdot f+(1-alpha)cdot g}log(alphacdot f+(1-alpha)cdot g)\
&leq -(alphasqrt f+(1-alpha) sqrt{g})log(f^{alpha}g^{(1-alpha)})\
&=-left[(alphasqrt f+(1-alpha) sqrt{g})(alphalog(f)+(1-alpha)log(g))right]\
&=-[alpha^2sqrt{f}log(f)+alpha(1-alpha)sqrt{f}log(g)+alpha(1-alpha)sqrt{g}log(f)\
&+(1-alpha)^2sqrt{g}log(g)]\
&=...
end{align*}$
answered Jan 22 at 20:31
FuncAna09FuncAna09
93
$endgroup$
add a comment |
$begingroup$
Waht i have so far for $phi:mathbb{R}_{>0}tomathbb{R}$ with $fmapsto phi(f)$
$begin{align*}
-phi(alphacdot f+(1-alpha)cdot g)&=-sqrt{alphacdot f+(1-alpha)cdot g}log(alphacdot f+(1-alpha)cdot g)\
&leq -(alphasqrt f+(1-alpha) sqrt{g})log(f^{alpha}g^{(1-alpha)})\
&=-left[(alphasqrt f+(1-alpha) sqrt{g})(alphalog(f)+(1-alpha)log(g))right]\
&=-[alpha^2sqrt{f}log(f)+alpha(1-alpha)sqrt{f}log(g)+alpha(1-alpha)sqrt{g}log(f)\
&+(1-alpha)^2sqrt{g}log(g)]\
&=...
end{align*}$
answered Jan 22 at 20:31
FuncAna09FuncAna09
93
$endgroup$
add a comment |
$begingroup$
Waht i have so far for $phi:mathbb{R}_{>0}tomathbb{R}$ with $fmapsto phi(f)$
$begin{align*}
-phi(alphacdot f+(1-alpha)cdot g)&=-sqrt{alphacdot f+(1-alpha)cdot g}log(alphacdot f+(1-alpha)cdot g)\
&leq -(alphasqrt f+(1-alpha) sqrt{g})log(f^{alpha}g^{(1-alpha)})\
&=-left[(alphasqrt f+(1-alpha) sqrt{g})(alphalog(f)+(1-alpha)log(g))right]\
&=-[alpha^2sqrt{f}log(f)+alpha(1-alpha)sqrt{f}log(g)+alpha(1-alpha)sqrt{g}log(f)\
&+(1-alpha)^2sqrt{g}log(g)]\
&=...
end{align*}$
answered Jan 22 at 20:31
FuncAna09FuncAna09
93
$endgroup$
Waht i have so far for $phi:mathbb{R}_{>0}tomathbb{R}$ with $fmapsto phi(f)$
$begin{align*}
-phi(alphacdot f+(1-alpha)cdot g)&=-sqrt{alphacdot f+(1-alpha)cdot g}log(alphacdot f+(1-alpha)cdot g)\
&leq -(alphasqrt f+(1-alpha) sqrt{g})log(f^{alpha}g^{(1-alpha)})\
&=-left[(alphasqrt f+(1-alpha) sqrt{g})(alphalog(f)+(1-alpha)log(g))right]\
&=-[alpha^2sqrt{f}log(f)+alpha(1-alpha)sqrt{f}log(g)+alpha(1-alpha)sqrt{g}log(f)\
&+(1-alpha)^2sqrt{g}log(g)]\
&=...
end{align*}$
answered Jan 22 at 20:31
FuncAna09FuncAna09
93
answered Jan 22 at 20:31
FuncAna09FuncAna09
93
answered Jan 22 at 20:31
FuncAna09FuncAna09
93
answered Jan 22 at 20:31
FuncAna09FuncAna09
93
93
add a comment |
add a comment |
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$begingroup$
How do you define convexity of $phi$?
$endgroup$
– gerw
Jan 22 at 15:20
$begingroup$
We call a map convex, if [phi(alphacdot f+(1-alpha)cdot g)leq alphacdot phi(f)+(1-alpha)cdot phi(f)] for $alphain (0,1)$
$endgroup$
– FuncAna09
Jan 22 at 17:59
$begingroup$
But $phi( something )$ is a function. Is $ge$ to be understood pointwise?
$endgroup$
– gerw
Jan 22 at 18:00
$begingroup$
Sorry, the correct definition is $phi(f(x)).$ f is also a map.
$endgroup$
– FuncAna09
Jan 22 at 18:06