How to show that $inf f(I)>0$ for an interval $I$ and a function $f$ with the following property?
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Let $f:mathbb{R}longrightarrow (0,infty)$ be a continuously differentiable function. Suppose that for an interval $Isubset mathbb{R}$ there exists a number $M>0$ such that for all $xin I$ the inequality $|f'(x)/f(x)|leq M$ is satisfied. Then it is asked to show that $inf f(I)>0$.
Now, if $I$ is a bounded interval, it is possible to prove the claim using the MVT.
However, I am unable to prove it if the interval $I$ is unbounded.
How should we proceed to show that $inf f(I)>0$, whenever $I$ is unbounded ?
real-analysis analysis
asked Jan 15 at 19:54
serenusserenus
24316
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add a comment |
$begingroup$
Let $f:mathbb{R}longrightarrow (0,infty)$ be a continuously differentiable function. Suppose that for an interval $Isubset mathbb{R}$ there exists a number $M>0$ such that for all $xin I$ the inequality $|f'(x)/f(x)|leq M$ is satisfied. Then it is asked to show that $inf f(I)>0$.
Now, if $I$ is a bounded interval, it is possible to prove the claim using the MVT.
However, I am unable to prove it if the interval $I$ is unbounded.
How should we proceed to show that $inf f(I)>0$, whenever $I$ is unbounded ?
real-analysis analysis
asked Jan 15 at 19:54
serenusserenus
24316
$endgroup$
add a comment |
$begingroup$
Let $f:mathbb{R}longrightarrow (0,infty)$ be a continuously differentiable function. Suppose that for an interval $Isubset mathbb{R}$ there exists a number $M>0$ such that for all $xin I$ the inequality $|f'(x)/f(x)|leq M$ is satisfied. Then it is asked to show that $inf f(I)>0$.
Now, if $I$ is a bounded interval, it is possible to prove the claim using the MVT.
However, I am unable to prove it if the interval $I$ is unbounded.
How should we proceed to show that $inf f(I)>0$, whenever $I$ is unbounded ?
real-analysis analysis
asked Jan 15 at 19:54
serenusserenus
24316
$endgroup$
Let $f:mathbb{R}longrightarrow (0,infty)$ be a continuously differentiable function. Suppose that for an interval $Isubset mathbb{R}$ there exists a number $M>0$ such that for all $xin I$ the inequality $|f'(x)/f(x)|leq M$ is satisfied. Then it is asked to show that $inf f(I)>0$.
Now, if $I$ is a bounded interval, it is possible to prove the claim using the MVT.
However, I am unable to prove it if the interval $I$ is unbounded.
How should we proceed to show that $inf f(I)>0$, whenever $I$ is unbounded ?
real-analysis analysis
real-analysis analysis
asked Jan 15 at 19:54
serenusserenus
24316
asked Jan 15 at 19:54
serenusserenus
24316
asked Jan 15 at 19:54
serenusserenus
24316
asked Jan 15 at 19:54
serenusserenus
24316
asked Jan 15 at 19:54
serenusserenus
24316
24316
add a comment |
add a comment |
2 Answers
2
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oldest
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Perhaps you've misstated the conditions above, but $f(x)=e^{-x}$ with $I=(0,infty)$, $M=1$, and $inf f(I)=0$ should be a counterexample to your statement.
answered Jan 15 at 20:05
smcsmc
261
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add a comment |
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This is wrong when $I$ is unbounded
Take $f(x)=e^{-x}$ and $I=(0,infty)$.
answered Jan 15 at 20:06
mathcounterexamples.netmathcounterexamples.net
27k22158
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2 Answers
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2 Answers
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$begingroup$
Perhaps you've misstated the conditions above, but $f(x)=e^{-x}$ with $I=(0,infty)$, $M=1$, and $inf f(I)=0$ should be a counterexample to your statement.
answered Jan 15 at 20:05
smcsmc
261
$endgroup$
add a comment |
$begingroup$
Perhaps you've misstated the conditions above, but $f(x)=e^{-x}$ with $I=(0,infty)$, $M=1$, and $inf f(I)=0$ should be a counterexample to your statement.
answered Jan 15 at 20:05
smcsmc
261
$endgroup$
add a comment |
$begingroup$
Perhaps you've misstated the conditions above, but $f(x)=e^{-x}$ with $I=(0,infty)$, $M=1$, and $inf f(I)=0$ should be a counterexample to your statement.
answered Jan 15 at 20:05
smcsmc
261
$endgroup$
Perhaps you've misstated the conditions above, but $f(x)=e^{-x}$ with $I=(0,infty)$, $M=1$, and $inf f(I)=0$ should be a counterexample to your statement.
answered Jan 15 at 20:05
smcsmc
261
answered Jan 15 at 20:05
smcsmc
261
answered Jan 15 at 20:05
smcsmc
261
answered Jan 15 at 20:05
smcsmc
261
261
add a comment |
add a comment |
$begingroup$
This is wrong when $I$ is unbounded
Take $f(x)=e^{-x}$ and $I=(0,infty)$.
answered Jan 15 at 20:06
mathcounterexamples.netmathcounterexamples.net
27k22158
$endgroup$
add a comment |
$begingroup$
This is wrong when $I$ is unbounded
Take $f(x)=e^{-x}$ and $I=(0,infty)$.
answered Jan 15 at 20:06
mathcounterexamples.netmathcounterexamples.net
27k22158
$endgroup$
add a comment |
$begingroup$
This is wrong when $I$ is unbounded
Take $f(x)=e^{-x}$ and $I=(0,infty)$.
answered Jan 15 at 20:06
mathcounterexamples.netmathcounterexamples.net
27k22158
$endgroup$
This is wrong when $I$ is unbounded
Take $f(x)=e^{-x}$ and $I=(0,infty)$.
answered Jan 15 at 20:06
mathcounterexamples.netmathcounterexamples.net
27k22158
answered Jan 15 at 20:06
mathcounterexamples.netmathcounterexamples.net
27k22158
answered Jan 15 at 20:06
mathcounterexamples.netmathcounterexamples.net
27k22158
answered Jan 15 at 20:06
mathcounterexamples.netmathcounterexamples.net
27k22158
27k22158
add a comment |
add a comment |
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