Proof that when f'(x) < f(x), f(x) =0
$begingroup$
Saw this problem online but have no ideas. I haven't been able to find a solution anywhere but would like to see a complete proof.
Take a real-valued function 𝑓 which is continuous and differentiable on [𝑎.𝑏] and f(x) $ge$ 0 ∀𝑥∈[𝑎,𝑏], that has the following properties:
(1) 𝑓(𝑎)=0
(2) 𝑓′(𝑥) $le$ 𝑓(𝑥) ∀𝑥∈[𝑎,𝑏]
Show that 𝑓(𝑥)=0 ∀𝑥∈[𝑎.𝑏].
What if the second property instead is that 𝑓(𝑥) $le$ $int_a^x f(t) ,dt$ ∀𝑥∈[𝑎.𝑏]? Is the conclusion still true?
real-analysis integration analysis derivatives
asked 25 mins ago
jacksonfjacksonf
15911
$endgroup$
add a comment |
$begingroup$
Saw this problem online but have no ideas. I haven't been able to find a solution anywhere but would like to see a complete proof.
Take a real-valued function 𝑓 which is continuous and differentiable on [𝑎.𝑏] and f(x) $ge$ 0 ∀𝑥∈[𝑎,𝑏], that has the following properties:
(1) 𝑓(𝑎)=0
(2) 𝑓′(𝑥) $le$ 𝑓(𝑥) ∀𝑥∈[𝑎,𝑏]
Show that 𝑓(𝑥)=0 ∀𝑥∈[𝑎.𝑏].
What if the second property instead is that 𝑓(𝑥) $le$ $int_a^x f(t) ,dt$ ∀𝑥∈[𝑎.𝑏]? Is the conclusion still true?
real-analysis integration analysis derivatives
asked 25 mins ago
jacksonfjacksonf
15911
$endgroup$
$begingroup$
Please format all math contents in MathJax. Some users may not be able to view the special charaters you used. For example, I can see no symbol between parentheses and brackets.
$endgroup$
– Trebor
14 mins ago
$begingroup$
@Trebor updated!
$endgroup$
– jacksonf
8 mins ago
add a comment |
$begingroup$
Saw this problem online but have no ideas. I haven't been able to find a solution anywhere but would like to see a complete proof.
Take a real-valued function 𝑓 which is continuous and differentiable on [𝑎.𝑏] and f(x) $ge$ 0 ∀𝑥∈[𝑎,𝑏], that has the following properties:
(1) 𝑓(𝑎)=0
(2) 𝑓′(𝑥) $le$ 𝑓(𝑥) ∀𝑥∈[𝑎,𝑏]
Show that 𝑓(𝑥)=0 ∀𝑥∈[𝑎.𝑏].
What if the second property instead is that 𝑓(𝑥) $le$ $int_a^x f(t) ,dt$ ∀𝑥∈[𝑎.𝑏]? Is the conclusion still true?
real-analysis integration analysis derivatives
asked 25 mins ago
jacksonfjacksonf
15911
$endgroup$
Saw this problem online but have no ideas. I haven't been able to find a solution anywhere but would like to see a complete proof.
Take a real-valued function 𝑓 which is continuous and differentiable on [𝑎.𝑏] and f(x) $ge$ 0 ∀𝑥∈[𝑎,𝑏], that has the following properties:
(1) 𝑓(𝑎)=0
(2) 𝑓′(𝑥) $le$ 𝑓(𝑥) ∀𝑥∈[𝑎,𝑏]
Show that 𝑓(𝑥)=0 ∀𝑥∈[𝑎.𝑏].
What if the second property instead is that 𝑓(𝑥) $le$ $int_a^x f(t) ,dt$ ∀𝑥∈[𝑎.𝑏]? Is the conclusion still true?
real-analysis integration analysis derivatives
real-analysis integration analysis derivatives
asked 25 mins ago
jacksonfjacksonf
15911
asked 25 mins ago
jacksonfjacksonf
15911
edited 8 mins ago
jacksonf
asked 25 mins ago
jacksonfjacksonf
15911
asked 25 mins ago
jacksonfjacksonf
15911
asked 25 mins ago
jacksonfjacksonf
15911
15911
$begingroup$
Please format all math contents in MathJax. Some users may not be able to view the special charaters you used. For example, I can see no symbol between parentheses and brackets.
$endgroup$
– Trebor
14 mins ago
$begingroup$
@Trebor updated!
$endgroup$
– jacksonf
8 mins ago
add a comment |
$begingroup$
Please format all math contents in MathJax. Some users may not be able to view the special charaters you used. For example, I can see no symbol between parentheses and brackets.
$endgroup$
– Trebor
14 mins ago
$begingroup$
@Trebor updated!
$endgroup$
– jacksonf
8 mins ago
$begingroup$
Please format all math contents in MathJax. Some users may not be able to view the special charaters you used. For example, I can see no symbol between parentheses and brackets.
$endgroup$
– Trebor
14 mins ago
$begingroup$
Please format all math contents in MathJax. Some users may not be able to view the special charaters you used. For example, I can see no symbol between parentheses and brackets.
$endgroup$
– Trebor
14 mins ago
$begingroup$
@Trebor updated!
$endgroup$
– jacksonf
8 mins ago
$begingroup$
@Trebor updated!
$endgroup$
– jacksonf
8 mins ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The first part is false: take $a=0,b=1$ and $f(x) =-x$. The same example shows that the second part is also false.
The condition $f(x) geq 0$ was added later. In this case we can show that $f=0$ under 2). Note that $(e^{-x}f(x))'leq 0$. So $e^{-x}f(x)$ is a decreasing non-negative function which vanishes at $a$ hence it is $0$ everywhere.
answered 9 mins ago
Kavi Rama MurthyKavi Rama Murthy
78.9k53672
$endgroup$
$begingroup$
I'm so sorry I missed a condition! It was that f(x) ≥ 0. Could you take another look?
$endgroup$
– jacksonf
6 mins ago
$begingroup$
@jacksonf I have given a proof under 2).
$endgroup$
– Kavi Rama Murthy
1 min ago
add a comment |
$begingroup$
Let us use a classical trick.
Write your inequality under the form $(f'(x)-f(x))<0$.
Multiply by the positive quantity $e^{-x}$ ; you obtain :
$(e^{-x}f(x))'<0 tag{1}$
Thus function $g(x):=e^{-x}f(x)$ is strictly decreasing.
As it is such that $g(0)=0$, it is always negative. As $e^{-x}>0$, it means that $forall x in [a,b], f(x)<0$,
which is in contradiction with the hypothesis (you have added) $f(x)geq 0$.
No answer for your second question.
answered 4 mins ago
Jean MarieJean Marie
32k42355
$endgroup$
$begingroup$
You have to replace strict inequalities by loose inequalities.
$endgroup$
– Kavi Rama Murthy
13 secs ago
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The first part is false: take $a=0,b=1$ and $f(x) =-x$. The same example shows that the second part is also false.
The condition $f(x) geq 0$ was added later. In this case we can show that $f=0$ under 2). Note that $(e^{-x}f(x))'leq 0$. So $e^{-x}f(x)$ is a decreasing non-negative function which vanishes at $a$ hence it is $0$ everywhere.
answered 9 mins ago
Kavi Rama MurthyKavi Rama Murthy
78.9k53672
$endgroup$
$begingroup$
I'm so sorry I missed a condition! It was that f(x) ≥ 0. Could you take another look?
$endgroup$
– jacksonf
6 mins ago
$begingroup$
@jacksonf I have given a proof under 2).
$endgroup$
– Kavi Rama Murthy
1 min ago
add a comment |
$begingroup$
The first part is false: take $a=0,b=1$ and $f(x) =-x$. The same example shows that the second part is also false.
The condition $f(x) geq 0$ was added later. In this case we can show that $f=0$ under 2). Note that $(e^{-x}f(x))'leq 0$. So $e^{-x}f(x)$ is a decreasing non-negative function which vanishes at $a$ hence it is $0$ everywhere.
answered 9 mins ago
Kavi Rama MurthyKavi Rama Murthy
78.9k53672
$endgroup$
$begingroup$
I'm so sorry I missed a condition! It was that f(x) ≥ 0. Could you take another look?
$endgroup$
– jacksonf
6 mins ago
$begingroup$
@jacksonf I have given a proof under 2).
$endgroup$
– Kavi Rama Murthy
1 min ago
add a comment |
$begingroup$
The first part is false: take $a=0,b=1$ and $f(x) =-x$. The same example shows that the second part is also false.
The condition $f(x) geq 0$ was added later. In this case we can show that $f=0$ under 2). Note that $(e^{-x}f(x))'leq 0$. So $e^{-x}f(x)$ is a decreasing non-negative function which vanishes at $a$ hence it is $0$ everywhere.
answered 9 mins ago
Kavi Rama MurthyKavi Rama Murthy
78.9k53672
$endgroup$
The first part is false: take $a=0,b=1$ and $f(x) =-x$. The same example shows that the second part is also false.
The condition $f(x) geq 0$ was added later. In this case we can show that $f=0$ under 2). Note that $(e^{-x}f(x))'leq 0$. So $e^{-x}f(x)$ is a decreasing non-negative function which vanishes at $a$ hence it is $0$ everywhere.
answered 9 mins ago
Kavi Rama MurthyKavi Rama Murthy
78.9k53672
edited 2 mins ago
answered 9 mins ago
Kavi Rama MurthyKavi Rama Murthy
78.9k53672
answered 9 mins ago
Kavi Rama MurthyKavi Rama Murthy
78.9k53672
answered 9 mins ago
Kavi Rama MurthyKavi Rama Murthy
78.9k53672
78.9k53672
$begingroup$
I'm so sorry I missed a condition! It was that f(x) ≥ 0. Could you take another look?
$endgroup$
– jacksonf
6 mins ago
$begingroup$
@jacksonf I have given a proof under 2).
$endgroup$
– Kavi Rama Murthy
1 min ago
add a comment |
$begingroup$
I'm so sorry I missed a condition! It was that f(x) ≥ 0. Could you take another look?
$endgroup$
– jacksonf
6 mins ago
$begingroup$
@jacksonf I have given a proof under 2).
$endgroup$
– Kavi Rama Murthy
1 min ago
$begingroup$
I'm so sorry I missed a condition! It was that f(x) ≥ 0. Could you take another look?
$endgroup$
– jacksonf
6 mins ago
$begingroup$
I'm so sorry I missed a condition! It was that f(x) ≥ 0. Could you take another look?
$endgroup$
– jacksonf
6 mins ago
$begingroup$
@jacksonf I have given a proof under 2).
$endgroup$
– Kavi Rama Murthy
1 min ago
$begingroup$
@jacksonf I have given a proof under 2).
$endgroup$
– Kavi Rama Murthy
1 min ago
add a comment |
$begingroup$
Let us use a classical trick.
Write your inequality under the form $(f'(x)-f(x))<0$.
Multiply by the positive quantity $e^{-x}$ ; you obtain :
$(e^{-x}f(x))'<0 tag{1}$
Thus function $g(x):=e^{-x}f(x)$ is strictly decreasing.
As it is such that $g(0)=0$, it is always negative. As $e^{-x}>0$, it means that $forall x in [a,b], f(x)<0$,
which is in contradiction with the hypothesis (you have added) $f(x)geq 0$.
No answer for your second question.
answered 4 mins ago
Jean MarieJean Marie
32k42355
$endgroup$
$begingroup$
You have to replace strict inequalities by loose inequalities.
$endgroup$
– Kavi Rama Murthy
13 secs ago
add a comment |
$begingroup$
Let us use a classical trick.
Write your inequality under the form $(f'(x)-f(x))<0$.
Multiply by the positive quantity $e^{-x}$ ; you obtain :
$(e^{-x}f(x))'<0 tag{1}$
Thus function $g(x):=e^{-x}f(x)$ is strictly decreasing.
As it is such that $g(0)=0$, it is always negative. As $e^{-x}>0$, it means that $forall x in [a,b], f(x)<0$,
which is in contradiction with the hypothesis (you have added) $f(x)geq 0$.
No answer for your second question.
answered 4 mins ago
Jean MarieJean Marie
32k42355
$endgroup$
$begingroup$
You have to replace strict inequalities by loose inequalities.
$endgroup$
– Kavi Rama Murthy
13 secs ago
add a comment |
$begingroup$
Let us use a classical trick.
Write your inequality under the form $(f'(x)-f(x))<0$.
Multiply by the positive quantity $e^{-x}$ ; you obtain :
$(e^{-x}f(x))'<0 tag{1}$
Thus function $g(x):=e^{-x}f(x)$ is strictly decreasing.
As it is such that $g(0)=0$, it is always negative. As $e^{-x}>0$, it means that $forall x in [a,b], f(x)<0$,
which is in contradiction with the hypothesis (you have added) $f(x)geq 0$.
No answer for your second question.
answered 4 mins ago
Jean MarieJean Marie
32k42355
$endgroup$
Let us use a classical trick.
Write your inequality under the form $(f'(x)-f(x))<0$.
Multiply by the positive quantity $e^{-x}$ ; you obtain :
$(e^{-x}f(x))'<0 tag{1}$
Thus function $g(x):=e^{-x}f(x)$ is strictly decreasing.
As it is such that $g(0)=0$, it is always negative. As $e^{-x}>0$, it means that $forall x in [a,b], f(x)<0$,
which is in contradiction with the hypothesis (you have added) $f(x)geq 0$.
No answer for your second question.
answered 4 mins ago
Jean MarieJean Marie
32k42355
answered 4 mins ago
Jean MarieJean Marie
32k42355
answered 4 mins ago
Jean MarieJean Marie
32k42355
answered 4 mins ago
Jean MarieJean Marie
32k42355
32k42355
$begingroup$
You have to replace strict inequalities by loose inequalities.
$endgroup$
– Kavi Rama Murthy
13 secs ago
add a comment |
$begingroup$
You have to replace strict inequalities by loose inequalities.
$endgroup$
– Kavi Rama Murthy
13 secs ago
$begingroup$
You have to replace strict inequalities by loose inequalities.
$endgroup$
– Kavi Rama Murthy
13 secs ago
$begingroup$
You have to replace strict inequalities by loose inequalities.
$endgroup$
– Kavi Rama Murthy
13 secs ago
add a comment |
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$begingroup$
Please format all math contents in MathJax. Some users may not be able to view the special charaters you used. For example, I can see no symbol between parentheses and brackets.
$endgroup$
– Trebor
14 mins ago
$begingroup$
@Trebor updated!
$endgroup$
– jacksonf
8 mins ago