Strict subadditivity of $xmapsto x^t$ for $0<t0$
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Let $f:mathbb (0,infty)tomathbb R, xmapsto x^t$ for some $0<t<1$. I want to know if $f$ is strictly subadditiv, i.e. $f(x)+f(y)>f(x+y)$.
If $t=frac1k$ for some $k>0$, it's easy to see that $f$ is strictly subadditiv. I've checked the subadditivity of $f$ for certain values $tneqfrac1k$ using the computer, so I am thinking that $f$ should be strictly subadditiv for all $t$. My question is how can you prove it for all $0<t<1$? I've tried to use the generalized binomial theorem but it didn't get me any futher.
calculus real-analysis
asked Aug 8 at 19:15
andr
84
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up vote
1
down vote
favorite
Let $f:mathbb (0,infty)tomathbb R, xmapsto x^t$ for some $0<t<1$. I want to know if $f$ is strictly subadditiv, i.e. $f(x)+f(y)>f(x+y)$.
If $t=frac1k$ for some $k>0$, it's easy to see that $f$ is strictly subadditiv. I've checked the subadditivity of $f$ for certain values $tneqfrac1k$ using the computer, so I am thinking that $f$ should be strictly subadditiv for all $t$. My question is how can you prove it for all $0<t<1$? I've tried to use the generalized binomial theorem but it didn't get me any futher.
calculus real-analysis
asked Aug 8 at 19:15
andr
84
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $f:mathbb (0,infty)tomathbb R, xmapsto x^t$ for some $0<t<1$. I want to know if $f$ is strictly subadditiv, i.e. $f(x)+f(y)>f(x+y)$.
If $t=frac1k$ for some $k>0$, it's easy to see that $f$ is strictly subadditiv. I've checked the subadditivity of $f$ for certain values $tneqfrac1k$ using the computer, so I am thinking that $f$ should be strictly subadditiv for all $t$. My question is how can you prove it for all $0<t<1$? I've tried to use the generalized binomial theorem but it didn't get me any futher.
calculus real-analysis
asked Aug 8 at 19:15
andr
84
Let $f:mathbb (0,infty)tomathbb R, xmapsto x^t$ for some $0<t<1$. I want to know if $f$ is strictly subadditiv, i.e. $f(x)+f(y)>f(x+y)$.
If $t=frac1k$ for some $k>0$, it's easy to see that $f$ is strictly subadditiv. I've checked the subadditivity of $f$ for certain values $tneqfrac1k$ using the computer, so I am thinking that $f$ should be strictly subadditiv for all $t$. My question is how can you prove it for all $0<t<1$? I've tried to use the generalized binomial theorem but it didn't get me any futher.
calculus real-analysis
asked Aug 8 at 19:15
andr
84
asked Aug 8 at 19:15
andr
84
asked Aug 8 at 19:15
andr
84
asked Aug 8 at 19:15
andr
84
84
add a comment |Â
add a comment |Â
1 Answer
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Yes, $f$ is strictly sub additive.
To prove it, fix some $x,y>0$ and observe that for $0<t<1$, we have $s^t-1>(x+s)^t-1$ for all $s>0$.
Integrating both side from $0$ to $y$, we get $int_0^ys^t-1ds>int_0^y(x+s)^t-1ds$, which gives $y^t>(x+y)^t-x^t$.
Hence the required sub additivity follows.
answered Aug 8 at 19:36
Tomath
1326
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Yes, $f$ is strictly sub additive.
To prove it, fix some $x,y>0$ and observe that for $0<t<1$, we have $s^t-1>(x+s)^t-1$ for all $s>0$.
Integrating both side from $0$ to $y$, we get $int_0^ys^t-1ds>int_0^y(x+s)^t-1ds$, which gives $y^t>(x+y)^t-x^t$.
Hence the required sub additivity follows.
answered Aug 8 at 19:36
Tomath
1326
add a comment |Â
up vote
0
down vote
accepted
Yes, $f$ is strictly sub additive.
To prove it, fix some $x,y>0$ and observe that for $0<t<1$, we have $s^t-1>(x+s)^t-1$ for all $s>0$.
Integrating both side from $0$ to $y$, we get $int_0^ys^t-1ds>int_0^y(x+s)^t-1ds$, which gives $y^t>(x+y)^t-x^t$.
Hence the required sub additivity follows.
answered Aug 8 at 19:36
Tomath
1326
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Yes, $f$ is strictly sub additive.
To prove it, fix some $x,y>0$ and observe that for $0<t<1$, we have $s^t-1>(x+s)^t-1$ for all $s>0$.
Integrating both side from $0$ to $y$, we get $int_0^ys^t-1ds>int_0^y(x+s)^t-1ds$, which gives $y^t>(x+y)^t-x^t$.
Hence the required sub additivity follows.
answered Aug 8 at 19:36
Tomath
1326
Yes, $f$ is strictly sub additive.
To prove it, fix some $x,y>0$ and observe that for $0<t<1$, we have $s^t-1>(x+s)^t-1$ for all $s>0$.
Integrating both side from $0$ to $y$, we get $int_0^ys^t-1ds>int_0^y(x+s)^t-1ds$, which gives $y^t>(x+y)^t-x^t$.
Hence the required sub additivity follows.
answered Aug 8 at 19:36
Tomath
1326
answered Aug 8 at 19:36
Tomath
1326
answered Aug 8 at 19:36
Tomath
1326
answered Aug 8 at 19:36
Tomath
1326
1326
add a comment |Â
add a comment |Â
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