Bounds on a convex function
$begingroup$
Let $f(x)$ be an increasing and infinitely differentiable strictly convex function with $f(0)=0$, it is easy to show that $f(m)-frac{m}{n}f(n)>0 $ for any $m> n> 0$ since $frac{n}{m}f(m)+frac{m-n}{m}f(0)> f(n)$ by definition. Suppose that we know $f''(0)=C>0$, is it possible to bound $f(m)-frac{m}{n}f(n)$ further?
I was hoping to show this by a Taylor expansion around zero:
$$f(x)=sum_{k=1}^infty frac{f^{(k)}(0) }{k!}x^k,$$
then we can show that:
$$ f(m)-frac{m}{n}f(n)=frac{m(m-n)}{2}C+sum_{k=3}^infty frac{f^{(k)}(0) }{k!}m(m^{k-1}-n^{k-1})>0$$
However, I'm not sure that the sum in the above equation is positive or negative, so I'm not sure if it is bounded below by sth like $frac{m(m-n)}{2}C$. Any help will be much appreciated.
convex-analysis upper-lower-bounds
asked Jan 8 at 9:14
user46666user46666
335
$endgroup$
add a comment |
$begingroup$
Let $f(x)$ be an increasing and infinitely differentiable strictly convex function with $f(0)=0$, it is easy to show that $f(m)-frac{m}{n}f(n)>0 $ for any $m> n> 0$ since $frac{n}{m}f(m)+frac{m-n}{m}f(0)> f(n)$ by definition. Suppose that we know $f''(0)=C>0$, is it possible to bound $f(m)-frac{m}{n}f(n)$ further?
I was hoping to show this by a Taylor expansion around zero:
$$f(x)=sum_{k=1}^infty frac{f^{(k)}(0) }{k!}x^k,$$
then we can show that:
$$ f(m)-frac{m}{n}f(n)=frac{m(m-n)}{2}C+sum_{k=3}^infty frac{f^{(k)}(0) }{k!}m(m^{k-1}-n^{k-1})>0$$
However, I'm not sure that the sum in the above equation is positive or negative, so I'm not sure if it is bounded below by sth like $frac{m(m-n)}{2}C$. Any help will be much appreciated.
convex-analysis upper-lower-bounds
asked Jan 8 at 9:14
user46666user46666
335
$endgroup$
add a comment |
$begingroup$
Let $f(x)$ be an increasing and infinitely differentiable strictly convex function with $f(0)=0$, it is easy to show that $f(m)-frac{m}{n}f(n)>0 $ for any $m> n> 0$ since $frac{n}{m}f(m)+frac{m-n}{m}f(0)> f(n)$ by definition. Suppose that we know $f''(0)=C>0$, is it possible to bound $f(m)-frac{m}{n}f(n)$ further?
I was hoping to show this by a Taylor expansion around zero:
$$f(x)=sum_{k=1}^infty frac{f^{(k)}(0) }{k!}x^k,$$
then we can show that:
$$ f(m)-frac{m}{n}f(n)=frac{m(m-n)}{2}C+sum_{k=3}^infty frac{f^{(k)}(0) }{k!}m(m^{k-1}-n^{k-1})>0$$
However, I'm not sure that the sum in the above equation is positive or negative, so I'm not sure if it is bounded below by sth like $frac{m(m-n)}{2}C$. Any help will be much appreciated.
convex-analysis upper-lower-bounds
asked Jan 8 at 9:14
user46666user46666
335
$endgroup$
Let $f(x)$ be an increasing and infinitely differentiable strictly convex function with $f(0)=0$, it is easy to show that $f(m)-frac{m}{n}f(n)>0 $ for any $m> n> 0$ since $frac{n}{m}f(m)+frac{m-n}{m}f(0)> f(n)$ by definition. Suppose that we know $f''(0)=C>0$, is it possible to bound $f(m)-frac{m}{n}f(n)$ further?
I was hoping to show this by a Taylor expansion around zero:
$$f(x)=sum_{k=1}^infty frac{f^{(k)}(0) }{k!}x^k,$$
then we can show that:
$$ f(m)-frac{m}{n}f(n)=frac{m(m-n)}{2}C+sum_{k=3}^infty frac{f^{(k)}(0) }{k!}m(m^{k-1}-n^{k-1})>0$$
However, I'm not sure that the sum in the above equation is positive or negative, so I'm not sure if it is bounded below by sth like $frac{m(m-n)}{2}C$. Any help will be much appreciated.
convex-analysis upper-lower-bounds
convex-analysis upper-lower-bounds
asked Jan 8 at 9:14
user46666user46666
335
asked Jan 8 at 9:14
user46666user46666
335
asked Jan 8 at 9:14
user46666user46666
335
asked Jan 8 at 9:14
user46666user46666
335
asked Jan 8 at 9:14
user46666user46666
335
335
add a comment |
add a comment |
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