If $|f'(x)| leq c|f(x)|$ for all $x in (0,1)$ then $f(x)=0$












3












$begingroup$


Question: Let $f:[0,1] to mathbb{R}$ be a real valued continuous function which is differentiable on $(0,1)$, and satisfies $f(0)=0$. Suppose there exists a $c in (0,1)$ such that $|f'(x)| leq c|f(x)|$ for all $x in (0,1)$. Show that $f(x)=0$.



Solution attempt:



$f$ being continuous at $x=0$, for a given $epsilon >0$ $exists$ a $delta>0$, such that $|f(x)|< epsilon$ whenever $x in [0, delta) cap [0,1]$.



Now, consider $|f'(h)|=|lim_{k to 0}frac{f(h+k)-f(h)}{k} |leq |cf(h)| implies lim_{ k to 0} |f(2h)| leq |f(h)|[1+ c|k|] $ Being continuous at $x=0$, $lim_{h to 0} f(h) = f(0) = 0 implies f(2h) = 0$ [by applying $|f(2h)| leq |f(h)|(c|k|+1) $].



In this manner, let us partition the interval $[0,1)$ into $n$ subintervals of length $h$ each. As the length $h to 0$, $n to infty$, and recursively, we get $|f(rh)| leq |f(h)|(c|k|+1)^r $. Hence, finally for all $r$, we get $f(rh)=0$. By continuity, we can safely say, for all $x$ in those respective subintervals, $f(x)=0$. Again, by continuity, we have $lim_{ x to 1} f(x) = 0$.



Hence, $f(x)=0$ for all $x$ in $[0,1]$.






real-analysis calculus derivatives proof-verification continuity







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    The third line in your attempt does not make sense.
    $endgroup$
    – Kavi Rama Murthy
    Jan 10 at 23:43










  • $begingroup$
    @KaviRamaMurthy which is?
    $endgroup$
    – Subhasis Biswas
    Jan 10 at 23:44






  • 1




    $begingroup$
    I am referring to $|f'(x)|=|lim frac {f(h+h)-f(h)} h|$
    $endgroup$
    – Kavi Rama Murthy
    Jan 10 at 23:45






  • 2




    $begingroup$
    Still makes no sense. $|f'(h)|=|lim_{kto 0} frac {f(h+k)-f(k)} k|$. Don't mix variables.
    $endgroup$
    – Kavi Rama Murthy
    Jan 10 at 23:48








  • 1




    $begingroup$
    You cannot use the same variable in your limit $(h)$ and the value of $f'$, they have to be "independent" of one another
    $endgroup$
    – symchdmath
    Jan 10 at 23:49
















3












$begingroup$


Question: Let $f:[0,1] to mathbb{R}$ be a real valued continuous function which is differentiable on $(0,1)$, and satisfies $f(0)=0$. Suppose there exists a $c in (0,1)$ such that $|f'(x)| leq c|f(x)|$ for all $x in (0,1)$. Show that $f(x)=0$.



Solution attempt:



$f$ being continuous at $x=0$, for a given $epsilon >0$ $exists$ a $delta>0$, such that $|f(x)|< epsilon$ whenever $x in [0, delta) cap [0,1]$.



Now, consider $|f'(h)|=|lim_{k to 0}frac{f(h+k)-f(h)}{k} |leq |cf(h)| implies lim_{ k to 0} |f(2h)| leq |f(h)|[1+ c|k|] $ Being continuous at $x=0$, $lim_{h to 0} f(h) = f(0) = 0 implies f(2h) = 0$ [by applying $|f(2h)| leq |f(h)|(c|k|+1) $].



In this manner, let us partition the interval $[0,1)$ into $n$ subintervals of length $h$ each. As the length $h to 0$, $n to infty$, and recursively, we get $|f(rh)| leq |f(h)|(c|k|+1)^r $. Hence, finally for all $r$, we get $f(rh)=0$. By continuity, we can safely say, for all $x$ in those respective subintervals, $f(x)=0$. Again, by continuity, we have $lim_{ x to 1} f(x) = 0$.



Hence, $f(x)=0$ for all $x$ in $[0,1]$.






real-analysis calculus derivatives proof-verification continuity







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    The third line in your attempt does not make sense.
    $endgroup$
    – Kavi Rama Murthy
    Jan 10 at 23:43










  • $begingroup$
    @KaviRamaMurthy which is?
    $endgroup$
    – Subhasis Biswas
    Jan 10 at 23:44






  • 1




    $begingroup$
    I am referring to $|f'(x)|=|lim frac {f(h+h)-f(h)} h|$
    $endgroup$
    – Kavi Rama Murthy
    Jan 10 at 23:45






  • 2




    $begingroup$
    Still makes no sense. $|f'(h)|=|lim_{kto 0} frac {f(h+k)-f(k)} k|$. Don't mix variables.
    $endgroup$
    – Kavi Rama Murthy
    Jan 10 at 23:48








  • 1




    $begingroup$
    You cannot use the same variable in your limit $(h)$ and the value of $f'$, they have to be "independent" of one another
    $endgroup$
    – symchdmath
    Jan 10 at 23:49














3












3








3


2



$begingroup$


Question: Let $f:[0,1] to mathbb{R}$ be a real valued continuous function which is differentiable on $(0,1)$, and satisfies $f(0)=0$. Suppose there exists a $c in (0,1)$ such that $|f'(x)| leq c|f(x)|$ for all $x in (0,1)$. Show that $f(x)=0$.



Solution attempt:



$f$ being continuous at $x=0$, for a given $epsilon >0$ $exists$ a $delta>0$, such that $|f(x)|< epsilon$ whenever $x in [0, delta) cap [0,1]$.



Now, consider $|f'(h)|=|lim_{k to 0}frac{f(h+k)-f(h)}{k} |leq |cf(h)| implies lim_{ k to 0} |f(2h)| leq |f(h)|[1+ c|k|] $ Being continuous at $x=0$, $lim_{h to 0} f(h) = f(0) = 0 implies f(2h) = 0$ [by applying $|f(2h)| leq |f(h)|(c|k|+1) $].



In this manner, let us partition the interval $[0,1)$ into $n$ subintervals of length $h$ each. As the length $h to 0$, $n to infty$, and recursively, we get $|f(rh)| leq |f(h)|(c|k|+1)^r $. Hence, finally for all $r$, we get $f(rh)=0$. By continuity, we can safely say, for all $x$ in those respective subintervals, $f(x)=0$. Again, by continuity, we have $lim_{ x to 1} f(x) = 0$.



Hence, $f(x)=0$ for all $x$ in $[0,1]$.






real-analysis calculus derivatives proof-verification continuity







share|cite|improve this question











$endgroup$




Question: Let $f:[0,1] to mathbb{R}$ be a real valued continuous function which is differentiable on $(0,1)$, and satisfies $f(0)=0$. Suppose there exists a $c in (0,1)$ such that $|f'(x)| leq c|f(x)|$ for all $x in (0,1)$. Show that $f(x)=0$.



Solution attempt:



$f$ being continuous at $x=0$, for a given $epsilon >0$ $exists$ a $delta>0$, such that $|f(x)|< epsilon$ whenever $x in [0, delta) cap [0,1]$.



Now, consider $|f'(h)|=|lim_{k to 0}frac{f(h+k)-f(h)}{k} |leq |cf(h)| implies lim_{ k to 0} |f(2h)| leq |f(h)|[1+ c|k|] $ Being continuous at $x=0$, $lim_{h to 0} f(h) = f(0) = 0 implies f(2h) = 0$ [by applying $|f(2h)| leq |f(h)|(c|k|+1) $].



In this manner, let us partition the interval $[0,1)$ into $n$ subintervals of length $h$ each. As the length $h to 0$, $n to infty$, and recursively, we get $|f(rh)| leq |f(h)|(c|k|+1)^r $. Hence, finally for all $r$, we get $f(rh)=0$. By continuity, we can safely say, for all $x$ in those respective subintervals, $f(x)=0$. Again, by continuity, we have $lim_{ x to 1} f(x) = 0$.



Hence, $f(x)=0$ for all $x$ in $[0,1]$.






real-analysis calculus derivatives proof-verification continuity




real-analysis calculus derivatives proof-verification continuity






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 11 at 0:16







Subhasis Biswas

















asked Jan 10 at 23:36









Subhasis BiswasSubhasis Biswas

488311




488311








  • 1




    $begingroup$
    The third line in your attempt does not make sense.
    $endgroup$
    – Kavi Rama Murthy
    Jan 10 at 23:43










  • $begingroup$
    @KaviRamaMurthy which is?
    $endgroup$
    – Subhasis Biswas
    Jan 10 at 23:44






  • 1




    $begingroup$
    I am referring to $|f'(x)|=|lim frac {f(h+h)-f(h)} h|$
    $endgroup$
    – Kavi Rama Murthy
    Jan 10 at 23:45






  • 2




    $begingroup$
    Still makes no sense. $|f'(h)|=|lim_{kto 0} frac {f(h+k)-f(k)} k|$. Don't mix variables.
    $endgroup$
    – Kavi Rama Murthy
    Jan 10 at 23:48








  • 1




    $begingroup$
    You cannot use the same variable in your limit $(h)$ and the value of $f'$, they have to be "independent" of one another
    $endgroup$
    – symchdmath
    Jan 10 at 23:49














  • 1




    $begingroup$
    The third line in your attempt does not make sense.
    $endgroup$
    – Kavi Rama Murthy
    Jan 10 at 23:43










  • $begingroup$
    @KaviRamaMurthy which is?
    $endgroup$
    – Subhasis Biswas
    Jan 10 at 23:44






  • 1




    $begingroup$
    I am referring to $|f'(x)|=|lim frac {f(h+h)-f(h)} h|$
    $endgroup$
    – Kavi Rama Murthy
    Jan 10 at 23:45






  • 2




    $begingroup$
    Still makes no sense. $|f'(h)|=|lim_{kto 0} frac {f(h+k)-f(k)} k|$. Don't mix variables.
    $endgroup$
    – Kavi Rama Murthy
    Jan 10 at 23:48








  • 1




    $begingroup$
    You cannot use the same variable in your limit $(h)$ and the value of $f'$, they have to be "independent" of one another
    $endgroup$
    – symchdmath
    Jan 10 at 23:49








1




1




$begingroup$
The third line in your attempt does not make sense.
$endgroup$
– Kavi Rama Murthy
Jan 10 at 23:43




$begingroup$
The third line in your attempt does not make sense.
$endgroup$
– Kavi Rama Murthy
Jan 10 at 23:43












$begingroup$
@KaviRamaMurthy which is?
$endgroup$
– Subhasis Biswas
Jan 10 at 23:44




$begingroup$
@KaviRamaMurthy which is?
$endgroup$
– Subhasis Biswas
Jan 10 at 23:44




1




1




$begingroup$
I am referring to $|f'(x)|=|lim frac {f(h+h)-f(h)} h|$
$endgroup$
– Kavi Rama Murthy
Jan 10 at 23:45




$begingroup$
I am referring to $|f'(x)|=|lim frac {f(h+h)-f(h)} h|$
$endgroup$
– Kavi Rama Murthy
Jan 10 at 23:45




2




2




$begingroup$
Still makes no sense. $|f'(h)|=|lim_{kto 0} frac {f(h+k)-f(k)} k|$. Don't mix variables.
$endgroup$
– Kavi Rama Murthy
Jan 10 at 23:48






$begingroup$
Still makes no sense. $|f'(h)|=|lim_{kto 0} frac {f(h+k)-f(k)} k|$. Don't mix variables.
$endgroup$
– Kavi Rama Murthy
Jan 10 at 23:48






1




1




$begingroup$
You cannot use the same variable in your limit $(h)$ and the value of $f'$, they have to be "independent" of one another
$endgroup$
– symchdmath
Jan 10 at 23:49




$begingroup$
You cannot use the same variable in your limit $(h)$ and the value of $f'$, they have to be "independent" of one another
$endgroup$
– symchdmath
Jan 10 at 23:49










1 Answer
1






active

oldest

votes


















4












$begingroup$

Let $M$ be the sup norm of $f$. Let us prove by induction that $|f(x)| leq frac{c^nMx^n}{n!}$.
If $n=0$, this is obvious.



Let $n geq 0$ be such that $|f(x)| leq frac{c^nMx^n}{n!}$ for each $x$.



Then $|f’(x)| leq frac{c^{n+1}Mx^n}{n!}$ for each $x$.



Since $f(0)=0$, integration yields the desired inequality.



Now, the $RHS$ goes to $0$ as $n$ goes to $infty$, thus $f=0$ (note that $c<1$ is unimportant).






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Can you please check out mine?! Correcting the mistakes will help me immensely.
    $endgroup$
    – Subhasis Biswas
    Jan 11 at 0:05










  • $begingroup$
    Ok: what does the inequality just after the first $Rightarrow$ mean?
    $endgroup$
    – Mindlack
    Jan 11 at 0:08










  • $begingroup$
    I used the fact that $||a|-|b||≤|a-b|$ and $k$ approaches $0$
    $endgroup$
    – Subhasis Biswas
    Jan 11 at 0:10












  • $begingroup$
    No, you did not. Because this inequality does not mean anything. $k$ is defined on the right hand side but not on the left hand side, for instance.
    $endgroup$
    – Mindlack
    Jan 11 at 0:11






  • 1




    $begingroup$
    To be brutally honest, I don’t think so. It seems to me to be very deeply flawed owing to it not manipulating limits with the necessary rigor. However, I cannot prove that there is no proof that borrows some part of your ideas, so ... ;)
    $endgroup$
    – Mindlack
    Jan 11 at 0:32











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









4












$begingroup$

Let $M$ be the sup norm of $f$. Let us prove by induction that $|f(x)| leq frac{c^nMx^n}{n!}$.
If $n=0$, this is obvious.



Let $n geq 0$ be such that $|f(x)| leq frac{c^nMx^n}{n!}$ for each $x$.



Then $|f’(x)| leq frac{c^{n+1}Mx^n}{n!}$ for each $x$.



Since $f(0)=0$, integration yields the desired inequality.



Now, the $RHS$ goes to $0$ as $n$ goes to $infty$, thus $f=0$ (note that $c<1$ is unimportant).






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Can you please check out mine?! Correcting the mistakes will help me immensely.
    $endgroup$
    – Subhasis Biswas
    Jan 11 at 0:05










  • $begingroup$
    Ok: what does the inequality just after the first $Rightarrow$ mean?
    $endgroup$
    – Mindlack
    Jan 11 at 0:08










  • $begingroup$
    I used the fact that $||a|-|b||≤|a-b|$ and $k$ approaches $0$
    $endgroup$
    – Subhasis Biswas
    Jan 11 at 0:10












  • $begingroup$
    No, you did not. Because this inequality does not mean anything. $k$ is defined on the right hand side but not on the left hand side, for instance.
    $endgroup$
    – Mindlack
    Jan 11 at 0:11






  • 1




    $begingroup$
    To be brutally honest, I don’t think so. It seems to me to be very deeply flawed owing to it not manipulating limits with the necessary rigor. However, I cannot prove that there is no proof that borrows some part of your ideas, so ... ;)
    $endgroup$
    – Mindlack
    Jan 11 at 0:32
















4












$begingroup$

Let $M$ be the sup norm of $f$. Let us prove by induction that $|f(x)| leq frac{c^nMx^n}{n!}$.
If $n=0$, this is obvious.



Let $n geq 0$ be such that $|f(x)| leq frac{c^nMx^n}{n!}$ for each $x$.



Then $|f’(x)| leq frac{c^{n+1}Mx^n}{n!}$ for each $x$.



Since $f(0)=0$, integration yields the desired inequality.



Now, the $RHS$ goes to $0$ as $n$ goes to $infty$, thus $f=0$ (note that $c<1$ is unimportant).






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Can you please check out mine?! Correcting the mistakes will help me immensely.
    $endgroup$
    – Subhasis Biswas
    Jan 11 at 0:05










  • $begingroup$
    Ok: what does the inequality just after the first $Rightarrow$ mean?
    $endgroup$
    – Mindlack
    Jan 11 at 0:08










  • $begingroup$
    I used the fact that $||a|-|b||≤|a-b|$ and $k$ approaches $0$
    $endgroup$
    – Subhasis Biswas
    Jan 11 at 0:10












  • $begingroup$
    No, you did not. Because this inequality does not mean anything. $k$ is defined on the right hand side but not on the left hand side, for instance.
    $endgroup$
    – Mindlack
    Jan 11 at 0:11






  • 1




    $begingroup$
    To be brutally honest, I don’t think so. It seems to me to be very deeply flawed owing to it not manipulating limits with the necessary rigor. However, I cannot prove that there is no proof that borrows some part of your ideas, so ... ;)
    $endgroup$
    – Mindlack
    Jan 11 at 0:32














4












4








4





$begingroup$

Let $M$ be the sup norm of $f$. Let us prove by induction that $|f(x)| leq frac{c^nMx^n}{n!}$.
If $n=0$, this is obvious.



Let $n geq 0$ be such that $|f(x)| leq frac{c^nMx^n}{n!}$ for each $x$.



Then $|f’(x)| leq frac{c^{n+1}Mx^n}{n!}$ for each $x$.



Since $f(0)=0$, integration yields the desired inequality.



Now, the $RHS$ goes to $0$ as $n$ goes to $infty$, thus $f=0$ (note that $c<1$ is unimportant).






share|cite|improve this answer











$endgroup$



Let $M$ be the sup norm of $f$. Let us prove by induction that $|f(x)| leq frac{c^nMx^n}{n!}$.
If $n=0$, this is obvious.



Let $n geq 0$ be such that $|f(x)| leq frac{c^nMx^n}{n!}$ for each $x$.



Then $|f’(x)| leq frac{c^{n+1}Mx^n}{n!}$ for each $x$.



Since $f(0)=0$, integration yields the desired inequality.



Now, the $RHS$ goes to $0$ as $n$ goes to $infty$, thus $f=0$ (note that $c<1$ is unimportant).







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 11 at 0:05

























answered Jan 11 at 0:04









MindlackMindlack

3,11717




3,11717












  • $begingroup$
    Can you please check out mine?! Correcting the mistakes will help me immensely.
    $endgroup$
    – Subhasis Biswas
    Jan 11 at 0:05










  • $begingroup$
    Ok: what does the inequality just after the first $Rightarrow$ mean?
    $endgroup$
    – Mindlack
    Jan 11 at 0:08










  • $begingroup$
    I used the fact that $||a|-|b||≤|a-b|$ and $k$ approaches $0$
    $endgroup$
    – Subhasis Biswas
    Jan 11 at 0:10












  • $begingroup$
    No, you did not. Because this inequality does not mean anything. $k$ is defined on the right hand side but not on the left hand side, for instance.
    $endgroup$
    – Mindlack
    Jan 11 at 0:11






  • 1




    $begingroup$
    To be brutally honest, I don’t think so. It seems to me to be very deeply flawed owing to it not manipulating limits with the necessary rigor. However, I cannot prove that there is no proof that borrows some part of your ideas, so ... ;)
    $endgroup$
    – Mindlack
    Jan 11 at 0:32


















  • $begingroup$
    Can you please check out mine?! Correcting the mistakes will help me immensely.
    $endgroup$
    – Subhasis Biswas
    Jan 11 at 0:05










  • $begingroup$
    Ok: what does the inequality just after the first $Rightarrow$ mean?
    $endgroup$
    – Mindlack
    Jan 11 at 0:08










  • $begingroup$
    I used the fact that $||a|-|b||≤|a-b|$ and $k$ approaches $0$
    $endgroup$
    – Subhasis Biswas
    Jan 11 at 0:10












  • $begingroup$
    No, you did not. Because this inequality does not mean anything. $k$ is defined on the right hand side but not on the left hand side, for instance.
    $endgroup$
    – Mindlack
    Jan 11 at 0:11






  • 1




    $begingroup$
    To be brutally honest, I don’t think so. It seems to me to be very deeply flawed owing to it not manipulating limits with the necessary rigor. However, I cannot prove that there is no proof that borrows some part of your ideas, so ... ;)
    $endgroup$
    – Mindlack
    Jan 11 at 0:32
















$begingroup$
Can you please check out mine?! Correcting the mistakes will help me immensely.
$endgroup$
– Subhasis Biswas
Jan 11 at 0:05




$begingroup$
Can you please check out mine?! Correcting the mistakes will help me immensely.
$endgroup$
– Subhasis Biswas
Jan 11 at 0:05












$begingroup$
Ok: what does the inequality just after the first $Rightarrow$ mean?
$endgroup$
– Mindlack
Jan 11 at 0:08




$begingroup$
Ok: what does the inequality just after the first $Rightarrow$ mean?
$endgroup$
– Mindlack
Jan 11 at 0:08












$begingroup$
I used the fact that $||a|-|b||≤|a-b|$ and $k$ approaches $0$
$endgroup$
– Subhasis Biswas
Jan 11 at 0:10






$begingroup$
I used the fact that $||a|-|b||≤|a-b|$ and $k$ approaches $0$
$endgroup$
– Subhasis Biswas
Jan 11 at 0:10














$begingroup$
No, you did not. Because this inequality does not mean anything. $k$ is defined on the right hand side but not on the left hand side, for instance.
$endgroup$
– Mindlack
Jan 11 at 0:11




$begingroup$
No, you did not. Because this inequality does not mean anything. $k$ is defined on the right hand side but not on the left hand side, for instance.
$endgroup$
– Mindlack
Jan 11 at 0:11




1




1




$begingroup$
To be brutally honest, I don’t think so. It seems to me to be very deeply flawed owing to it not manipulating limits with the necessary rigor. However, I cannot prove that there is no proof that borrows some part of your ideas, so ... ;)
$endgroup$
– Mindlack
Jan 11 at 0:32




$begingroup$
To be brutally honest, I don’t think so. It seems to me to be very deeply flawed owing to it not manipulating limits with the necessary rigor. However, I cannot prove that there is no proof that borrows some part of your ideas, so ... ;)
$endgroup$
– Mindlack
Jan 11 at 0:32


















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