Function bounded by exponential functions has a bounded derivative?
$begingroup$
Let $v:mathbb{R}_{+}mapsto mathbb{R}_{+}$ and $v(t)$ is non-increasing, i.e., $dot{v}(t)le 0$. If there exist positive constants $k_{1}$, $k_{2}$, $c_{1}$, $c_{2}$, such that
begin{align}
k_{1}e^{-c_{1}t}v(0) le v(t)le k_{2}e^{-c_{2}t}v(0).
end{align}
Can I conclude that
begin{align}
dot{v}(t)le -c_{3}v(t)
end{align}
for some positive constant $c_{3}$?
Under a weaker assumption, a similar question has been asked, but with counterexamples:
function bounded by an exponential has a bounded derivative?
real-analysis ordinary-differential-equations exponential-function
asked Jan 3 at 8:02
guluzhuguluzhu
538
$endgroup$
add a comment |
$begingroup$
Let $v:mathbb{R}_{+}mapsto mathbb{R}_{+}$ and $v(t)$ is non-increasing, i.e., $dot{v}(t)le 0$. If there exist positive constants $k_{1}$, $k_{2}$, $c_{1}$, $c_{2}$, such that
begin{align}
k_{1}e^{-c_{1}t}v(0) le v(t)le k_{2}e^{-c_{2}t}v(0).
end{align}
Can I conclude that
begin{align}
dot{v}(t)le -c_{3}v(t)
end{align}
for some positive constant $c_{3}$?
Under a weaker assumption, a similar question has been asked, but with counterexamples:
function bounded by an exponential has a bounded derivative?
real-analysis ordinary-differential-equations exponential-function
asked Jan 3 at 8:02
guluzhuguluzhu
538
$endgroup$
1
$begingroup$
No. Essentially the same counter-example type holds. You can have a curve that starts near the bottom curve, but then takes a flat part (with slope 0 or as close to 0 as we want) to go over to the top curve.
$endgroup$
– Michael
Jan 3 at 8:18
add a comment |
$begingroup$
Let $v:mathbb{R}_{+}mapsto mathbb{R}_{+}$ and $v(t)$ is non-increasing, i.e., $dot{v}(t)le 0$. If there exist positive constants $k_{1}$, $k_{2}$, $c_{1}$, $c_{2}$, such that
begin{align}
k_{1}e^{-c_{1}t}v(0) le v(t)le k_{2}e^{-c_{2}t}v(0).
end{align}
Can I conclude that
begin{align}
dot{v}(t)le -c_{3}v(t)
end{align}
for some positive constant $c_{3}$?
Under a weaker assumption, a similar question has been asked, but with counterexamples:
function bounded by an exponential has a bounded derivative?
real-analysis ordinary-differential-equations exponential-function
asked Jan 3 at 8:02
guluzhuguluzhu
538
$endgroup$
Let $v:mathbb{R}_{+}mapsto mathbb{R}_{+}$ and $v(t)$ is non-increasing, i.e., $dot{v}(t)le 0$. If there exist positive constants $k_{1}$, $k_{2}$, $c_{1}$, $c_{2}$, such that
begin{align}
k_{1}e^{-c_{1}t}v(0) le v(t)le k_{2}e^{-c_{2}t}v(0).
end{align}
Can I conclude that
begin{align}
dot{v}(t)le -c_{3}v(t)
end{align}
for some positive constant $c_{3}$?
Under a weaker assumption, a similar question has been asked, but with counterexamples:
function bounded by an exponential has a bounded derivative?
real-analysis ordinary-differential-equations exponential-function
real-analysis ordinary-differential-equations exponential-function
asked Jan 3 at 8:02
guluzhuguluzhu
538
asked Jan 3 at 8:02
guluzhuguluzhu
538
asked Jan 3 at 8:02
guluzhuguluzhu
538
asked Jan 3 at 8:02
guluzhuguluzhu
538
asked Jan 3 at 8:02
guluzhuguluzhu
538
538
1
$begingroup$
No. Essentially the same counter-example type holds. You can have a curve that starts near the bottom curve, but then takes a flat part (with slope 0 or as close to 0 as we want) to go over to the top curve.
$endgroup$
– Michael
Jan 3 at 8:18
add a comment |
1
$begingroup$
No. Essentially the same counter-example type holds. You can have a curve that starts near the bottom curve, but then takes a flat part (with slope 0 or as close to 0 as we want) to go over to the top curve.
$endgroup$
– Michael
Jan 3 at 8:18
1
1
$begingroup$
No. Essentially the same counter-example type holds. You can have a curve that starts near the bottom curve, but then takes a flat part (with slope 0 or as close to 0 as we want) to go over to the top curve.
$endgroup$
– Michael
Jan 3 at 8:18
$begingroup$
No. Essentially the same counter-example type holds. You can have a curve that starts near the bottom curve, but then takes a flat part (with slope 0 or as close to 0 as we want) to go over to the top curve.
$endgroup$
– Michael
Jan 3 at 8:18
add a comment |
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$begingroup$
No. Essentially the same counter-example type holds. You can have a curve that starts near the bottom curve, but then takes a flat part (with slope 0 or as close to 0 as we want) to go over to the top curve.
$endgroup$
– Michael
Jan 3 at 8:18