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Decreasing $f:mathbbRtomathbbR$ tending to $0$ at $∞$ not convex beyond any point?
Convex function which has a limit in $-infty$ is non decreasingLocal monotonicity and differentiability a.e.Strictly decreasing function with a horizontal asymptote is convex?For which values of $x_0inmathbbR$ is $f$ differentiable$f$ convex strictly decreasing function , is $f'(x+delta)-f'(x)$ convexComputing the limit of a convex, decreasing functionSufficient condition for inflection pointStrictly convex except in a single pointSufficient condition for absolute maximumExample of convex and injective $f:I to mathbbR$ such that $f^-1$ is not concave
$begingroup$
Given a function $f:mathbbRtomathbbR$ differentiable and strictly decreasing such that $displaystyle lim_xtoinftyf(x)=0$, I am looking to find out whether or not there exists an $x_0$ such that $f$ is convex on $(x_0,infty)$. My guess is that the statement is not true but I can't find a counterexample.
real-analysis limits monotone-functions
edited yesterday
user21820
39.7k544158
asked yesterday
Andrew VAndrew V
32911
$endgroup$
add a comment |
$begingroup$
Given a function $f:mathbbRtomathbbR$ differentiable and strictly decreasing such that $displaystyle lim_xtoinftyf(x)=0$, I am looking to find out whether or not there exists an $x_0$ such that $f$ is convex on $(x_0,infty)$. My guess is that the statement is not true but I can't find a counterexample.
real-analysis limits monotone-functions
edited yesterday
user21820
39.7k544158
asked yesterday
Andrew VAndrew V
32911
$endgroup$
add a comment |
$begingroup$
Given a function $f:mathbbRtomathbbR$ differentiable and strictly decreasing such that $displaystyle lim_xtoinftyf(x)=0$, I am looking to find out whether or not there exists an $x_0$ such that $f$ is convex on $(x_0,infty)$. My guess is that the statement is not true but I can't find a counterexample.
real-analysis limits monotone-functions
edited yesterday
user21820
39.7k544158
asked yesterday
Andrew VAndrew V
32911
$endgroup$
Given a function $f:mathbbRtomathbbR$ differentiable and strictly decreasing such that $displaystyle lim_xtoinftyf(x)=0$, I am looking to find out whether or not there exists an $x_0$ such that $f$ is convex on $(x_0,infty)$. My guess is that the statement is not true but I can't find a counterexample.
real-analysis limits monotone-functions
real-analysis limits monotone-functions
edited yesterday
user21820
39.7k544158
asked yesterday
Andrew VAndrew V
32911
edited yesterday
user21820
39.7k544158
asked yesterday
Andrew VAndrew V
32911
edited yesterday
user21820
39.7k544158
edited yesterday
user21820
39.7k544158
edited yesterday
user21820
39.7k544158
39.7k544158
asked yesterday
Andrew VAndrew V
32911
asked yesterday
Andrew VAndrew V
32911
asked yesterday
Andrew VAndrew V
32911
32911
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Edit: added additional term to make the function actually decreasing. The idea remains the same.
Consider something like $$frac1x + fracsin(x)x^2+ frac4 sin^2(x/2)x^3.$$ Notice how it keeps wiggling all the way down. Here is its graph from $100$ to $120$ to give an impression.
answered yesterday
WimCWimC
24.4k23063
$endgroup$
1
$begingroup$
Is this decreasing for all reals?
$endgroup$
– marty cohen
yesterday
$begingroup$
@martycohen if it is not, you can replace it with something simple on the "prefix" e.g if x > 100, it's this function, and if x < 100 it's just linear (with coefficients to make it continious and differentiable
$endgroup$
– RiaD
yesterday
$begingroup$
@martycohen It is not. Thanks for pointing this out. I’ll fix it in a moment.
$endgroup$
– WimC
yesterday
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
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active
oldest
votes
$begingroup$
Edit: added additional term to make the function actually decreasing. The idea remains the same.
Consider something like $$frac1x + fracsin(x)x^2+ frac4 sin^2(x/2)x^3.$$ Notice how it keeps wiggling all the way down. Here is its graph from $100$ to $120$ to give an impression.
answered yesterday
WimCWimC
24.4k23063
$endgroup$
1
$begingroup$
Is this decreasing for all reals?
$endgroup$
– marty cohen
yesterday
$begingroup$
@martycohen if it is not, you can replace it with something simple on the "prefix" e.g if x > 100, it's this function, and if x < 100 it's just linear (with coefficients to make it continious and differentiable
$endgroup$
– RiaD
yesterday
$begingroup$
@martycohen It is not. Thanks for pointing this out. I’ll fix it in a moment.
$endgroup$
– WimC
yesterday
add a comment |
$begingroup$
Edit: added additional term to make the function actually decreasing. The idea remains the same.
Consider something like $$frac1x + fracsin(x)x^2+ frac4 sin^2(x/2)x^3.$$ Notice how it keeps wiggling all the way down. Here is its graph from $100$ to $120$ to give an impression.
answered yesterday
WimCWimC
24.4k23063
$endgroup$
1
$begingroup$
Is this decreasing for all reals?
$endgroup$
– marty cohen
yesterday
$begingroup$
@martycohen if it is not, you can replace it with something simple on the "prefix" e.g if x > 100, it's this function, and if x < 100 it's just linear (with coefficients to make it continious and differentiable
$endgroup$
– RiaD
yesterday
$begingroup$
@martycohen It is not. Thanks for pointing this out. I’ll fix it in a moment.
$endgroup$
– WimC
yesterday
add a comment |
$begingroup$
Edit: added additional term to make the function actually decreasing. The idea remains the same.
Consider something like $$frac1x + fracsin(x)x^2+ frac4 sin^2(x/2)x^3.$$ Notice how it keeps wiggling all the way down. Here is its graph from $100$ to $120$ to give an impression.
answered yesterday
WimCWimC
24.4k23063
$endgroup$
Edit: added additional term to make the function actually decreasing. The idea remains the same.
Consider something like $$frac1x + fracsin(x)x^2+ frac4 sin^2(x/2)x^3.$$ Notice how it keeps wiggling all the way down. Here is its graph from $100$ to $120$ to give an impression.
answered yesterday
WimCWimC
24.4k23063
edited yesterday
answered yesterday
WimCWimC
24.4k23063
answered yesterday
WimCWimC
24.4k23063
answered yesterday
WimCWimC
24.4k23063
24.4k23063
1
$begingroup$
Is this decreasing for all reals?
$endgroup$
– marty cohen
yesterday
$begingroup$
@martycohen if it is not, you can replace it with something simple on the "prefix" e.g if x > 100, it's this function, and if x < 100 it's just linear (with coefficients to make it continious and differentiable
$endgroup$
– RiaD
yesterday
$begingroup$
@martycohen It is not. Thanks for pointing this out. I’ll fix it in a moment.
$endgroup$
– WimC
yesterday
add a comment |
1
$begingroup$
Is this decreasing for all reals?
$endgroup$
– marty cohen
yesterday
$begingroup$
@martycohen if it is not, you can replace it with something simple on the "prefix" e.g if x > 100, it's this function, and if x < 100 it's just linear (with coefficients to make it continious and differentiable
$endgroup$
– RiaD
yesterday
$begingroup$
@martycohen It is not. Thanks for pointing this out. I’ll fix it in a moment.
$endgroup$
– WimC
yesterday
1
1
$begingroup$
Is this decreasing for all reals?
$endgroup$
– marty cohen
yesterday
$begingroup$
Is this decreasing for all reals?
$endgroup$
– marty cohen
yesterday
$begingroup$
@martycohen if it is not, you can replace it with something simple on the "prefix" e.g if x > 100, it's this function, and if x < 100 it's just linear (with coefficients to make it continious and differentiable
$endgroup$
– RiaD
yesterday
$begingroup$
@martycohen if it is not, you can replace it with something simple on the "prefix" e.g if x > 100, it's this function, and if x < 100 it's just linear (with coefficients to make it continious and differentiable
$endgroup$
– RiaD
yesterday
$begingroup$
@martycohen It is not. Thanks for pointing this out. I’ll fix it in a moment.
$endgroup$
– WimC
yesterday
$begingroup$
@martycohen It is not. Thanks for pointing this out. I’ll fix it in a moment.
$endgroup$
– WimC
yesterday
add a comment |
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