Suppose $f: [0,1] rightarrow [0,1] $ and $f(x) leq int_0^x sqrt{f(t)}dt$. Show that $f(x) leq x^2$ for all $x...
Let $f(x): [0,1] rightarrow [0,1] $ such that $f(x) leq int_0^x sqrt{f(t)},dt$. Show that $f(x) leq x^2$ for all $x in [0,1]$.
I tried reiterating the inequality, obtaining $f(x) leq int_0^x1dt = x; f(x) leq int_0^x sqrt{t}dt = frac{2}{3}x^{3/2}$ etc... While it's easy to see that the exponent of $x$ tends to $2$, it's more difficult to show that the coefficient is $1$. Can somebody help me?
calculus real-analysis integration inequality
edited Nov 30 at 21:44
zhw.
71.6k43075
asked Nov 30 at 18:55
Lance
7712
add a comment |
Let $f(x): [0,1] rightarrow [0,1] $ such that $f(x) leq int_0^x sqrt{f(t)},dt$. Show that $f(x) leq x^2$ for all $x in [0,1]$.
I tried reiterating the inequality, obtaining $f(x) leq int_0^x1dt = x; f(x) leq int_0^x sqrt{t}dt = frac{2}{3}x^{3/2}$ etc... While it's easy to see that the exponent of $x$ tends to $2$, it's more difficult to show that the coefficient is $1$. Can somebody help me?
calculus real-analysis integration inequality
edited Nov 30 at 21:44
zhw.
71.6k43075
asked Nov 30 at 18:55
Lance
7712
This is a special case of math.stackexchange.com/q/2050893/42969 – choose $C=1$ and $alpha = 1/2$. It follows that $f(x) le x^2/4$.
– Martin R
Nov 30 at 20:31
@MartinR $f$ is not necessarily conitnuous here
– Thinking
Nov 30 at 20:38
@Thinking: You are right. – I assume that a similar argument works if $f$ is only integrable, but I am not 100% sure about that.
– Martin R
Nov 30 at 20:43
@MartinR maybe... but in this case $v$ is not necessarily differentiable so there might be several problems to fix.
– Thinking
Nov 30 at 20:46
add a comment |
Let $f(x): [0,1] rightarrow [0,1] $ such that $f(x) leq int_0^x sqrt{f(t)},dt$. Show that $f(x) leq x^2$ for all $x in [0,1]$.
I tried reiterating the inequality, obtaining $f(x) leq int_0^x1dt = x; f(x) leq int_0^x sqrt{t}dt = frac{2}{3}x^{3/2}$ etc... While it's easy to see that the exponent of $x$ tends to $2$, it's more difficult to show that the coefficient is $1$. Can somebody help me?
calculus real-analysis integration inequality
edited Nov 30 at 21:44
zhw.
71.6k43075
asked Nov 30 at 18:55
Lance
7712
Let $f(x): [0,1] rightarrow [0,1] $ such that $f(x) leq int_0^x sqrt{f(t)},dt$. Show that $f(x) leq x^2$ for all $x in [0,1]$.
I tried reiterating the inequality, obtaining $f(x) leq int_0^x1dt = x; f(x) leq int_0^x sqrt{t}dt = frac{2}{3}x^{3/2}$ etc... While it's easy to see that the exponent of $x$ tends to $2$, it's more difficult to show that the coefficient is $1$. Can somebody help me?
calculus real-analysis integration inequality
calculus real-analysis integration inequality
edited Nov 30 at 21:44
zhw.
71.6k43075
asked Nov 30 at 18:55
Lance
7712
edited Nov 30 at 21:44
zhw.
71.6k43075
asked Nov 30 at 18:55
Lance
7712
edited Nov 30 at 21:44
zhw.
71.6k43075
edited Nov 30 at 21:44
zhw.
71.6k43075
edited Nov 30 at 21:44
zhw.
71.6k43075
71.6k43075
asked Nov 30 at 18:55
Lance
7712
asked Nov 30 at 18:55
Lance
7712
asked Nov 30 at 18:55
Lance
7712
7712
This is a special case of math.stackexchange.com/q/2050893/42969 – choose $C=1$ and $alpha = 1/2$. It follows that $f(x) le x^2/4$.
– Martin R
Nov 30 at 20:31
@MartinR $f$ is not necessarily conitnuous here
– Thinking
Nov 30 at 20:38
@Thinking: You are right. – I assume that a similar argument works if $f$ is only integrable, but I am not 100% sure about that.
– Martin R
Nov 30 at 20:43
@MartinR maybe... but in this case $v$ is not necessarily differentiable so there might be several problems to fix.
– Thinking
Nov 30 at 20:46
add a comment |
This is a special case of math.stackexchange.com/q/2050893/42969 – choose $C=1$ and $alpha = 1/2$. It follows that $f(x) le x^2/4$.
– Martin R
Nov 30 at 20:31
@MartinR $f$ is not necessarily conitnuous here
– Thinking
Nov 30 at 20:38
@Thinking: You are right. – I assume that a similar argument works if $f$ is only integrable, but I am not 100% sure about that.
– Martin R
Nov 30 at 20:43
@MartinR maybe... but in this case $v$ is not necessarily differentiable so there might be several problems to fix.
– Thinking
Nov 30 at 20:46
This is a special case of math.stackexchange.com/q/2050893/42969 – choose $C=1$ and $alpha = 1/2$. It follows that $f(x) le x^2/4$.
– Martin R
Nov 30 at 20:31
This is a special case of math.stackexchange.com/q/2050893/42969 – choose $C=1$ and $alpha = 1/2$. It follows that $f(x) le x^2/4$.
– Martin R
Nov 30 at 20:31
@MartinR $f$ is not necessarily conitnuous here
– Thinking
Nov 30 at 20:38
@MartinR $f$ is not necessarily conitnuous here
– Thinking
Nov 30 at 20:38
@Thinking: You are right. – I assume that a similar argument works if $f$ is only integrable, but I am not 100% sure about that.
– Martin R
Nov 30 at 20:43
@Thinking: You are right. – I assume that a similar argument works if $f$ is only integrable, but I am not 100% sure about that.
– Martin R
Nov 30 at 20:43
@MartinR maybe... but in this case $v$ is not necessarily differentiable so there might be several problems to fix.
– Thinking
Nov 30 at 20:46
@MartinR maybe... but in this case $v$ is not necessarily differentiable so there might be several problems to fix.
– Thinking
Nov 30 at 20:46
add a comment |
2 Answers
2
active
oldest
votes
I assume that $f$ is Lebesgue integrable along with the other hypotheses. Let $xin [0,1].$ Let $M_x$ be the supremum of $f$ over $[0,x].$ If $M_x=0,$ there is nothing to prove. So assume $M_x>0.$ Let $0<epsilon< M_x.$ Choose $x_epsilonin [0,x]$ such that $f(x_epsilon) > M_x-epsilon.$ Then
$$M_x-epsilon < f(x_epsilon) le int_0^{x_epsilon}sqrt {f(t)},dt le xsqrt{M_x}.$$
Now let $epsilonto 0^+$ to see $M_xle xsqrt{M_x},$ which implies $sqrt{M_x}le x.$ Squaring, we see $f(x)le M_xle x^2,$ giving the result.
answered Nov 30 at 20:50
zhw.
71.6k43075
There's a problem, if the number of local maxima is dense in $[0,1]$ no ? $x_0$ can be really close to $0$, then the next one $x_1$ can be really close to $x_0$... and we never reach the whole interval $[0,1]$
– Thinking
Nov 30 at 20:54
1
@Thinking I don't see the problem. Where exactly do you think my proof breaks down?
– zhw.
Nov 30 at 21:04
You are right, I completely missounderstood the proof. It's actually clever. (+1)
– Thinking
Nov 30 at 21:08
@Thinking I've edited to remove the continuity hypothesis.
– zhw.
Nov 30 at 21:24
@Thinking Yes of course. Otherwise $M_x-epsilon$ is an upper bound.
– zhw.
Nov 30 at 21:52
add a comment |
Suppose $b_{n+1}=frac 12(b_n+2)$ and $a_{n+1}=frac{a_n^{1/2}}{b_{n+1}}$, then $f(x) le a_nx^{b_n}Rightarrow f(x)le a_{n+1}x^{b_{n+1}}$. As you showed, $f(x)le a_0x^{b_0}$ where $a_0=b_0=1$. To solve the recurrence for $b_n$ put $B_n=2^{n}b_n$ and notice $b_{n+1}-frac 12b_n=1Rightarrow 2^{n+1}b_{n+1}-2^{n}b_n=B_{n+1}-B_n=2^{n+1}$, summing over $n$ gives $B_{N}-B_0=sum_{n=0}^{N-1} 2^{n+1}=2^{N+1}-2$ and therefore $b_{N}=2^{-N}(2^{N+1}-2+b_0)=2-2^{-N}$. The $b_n$'s are increasing, so $frac 1{b_n}$ is a decreasing sequence. Assume $a_{n+1}le a_n$ so $a_{n+1}^{1/2}le a_n^{1/2}$ and hence $a_{n+2}=frac{a_{n+1}^{1/2}}{b_{n+2}} le frac{a_{n}^{1/2}}{b_{n+1}}=a_{n+1}$. Because $a_1=frac 23le 1 =a_0$, this holds for $n=0$ and induction shows $a_n$ is a decreasing sequence bounded below by $0$, thus has a limit $a$. Taking this limit shows $a=a^{1/2}/2$ because $b_nto 2$, and therefore $a=1/4$. Finally, $f(x)le a_nx^{b_n}to frac 14x^2$.
answered Nov 30 at 20:12
Guacho Perez
3,88911131
add a comment |
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2 Answers
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2 Answers
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I assume that $f$ is Lebesgue integrable along with the other hypotheses. Let $xin [0,1].$ Let $M_x$ be the supremum of $f$ over $[0,x].$ If $M_x=0,$ there is nothing to prove. So assume $M_x>0.$ Let $0<epsilon< M_x.$ Choose $x_epsilonin [0,x]$ such that $f(x_epsilon) > M_x-epsilon.$ Then
$$M_x-epsilon < f(x_epsilon) le int_0^{x_epsilon}sqrt {f(t)},dt le xsqrt{M_x}.$$
Now let $epsilonto 0^+$ to see $M_xle xsqrt{M_x},$ which implies $sqrt{M_x}le x.$ Squaring, we see $f(x)le M_xle x^2,$ giving the result.
answered Nov 30 at 20:50
zhw.
71.6k43075
There's a problem, if the number of local maxima is dense in $[0,1]$ no ? $x_0$ can be really close to $0$, then the next one $x_1$ can be really close to $x_0$... and we never reach the whole interval $[0,1]$
– Thinking
Nov 30 at 20:54
1
@Thinking I don't see the problem. Where exactly do you think my proof breaks down?
– zhw.
Nov 30 at 21:04
You are right, I completely missounderstood the proof. It's actually clever. (+1)
– Thinking
Nov 30 at 21:08
@Thinking I've edited to remove the continuity hypothesis.
– zhw.
Nov 30 at 21:24
@Thinking Yes of course. Otherwise $M_x-epsilon$ is an upper bound.
– zhw.
Nov 30 at 21:52
add a comment |
I assume that $f$ is Lebesgue integrable along with the other hypotheses. Let $xin [0,1].$ Let $M_x$ be the supremum of $f$ over $[0,x].$ If $M_x=0,$ there is nothing to prove. So assume $M_x>0.$ Let $0<epsilon< M_x.$ Choose $x_epsilonin [0,x]$ such that $f(x_epsilon) > M_x-epsilon.$ Then
$$M_x-epsilon < f(x_epsilon) le int_0^{x_epsilon}sqrt {f(t)},dt le xsqrt{M_x}.$$
Now let $epsilonto 0^+$ to see $M_xle xsqrt{M_x},$ which implies $sqrt{M_x}le x.$ Squaring, we see $f(x)le M_xle x^2,$ giving the result.
answered Nov 30 at 20:50
zhw.
71.6k43075
There's a problem, if the number of local maxima is dense in $[0,1]$ no ? $x_0$ can be really close to $0$, then the next one $x_1$ can be really close to $x_0$... and we never reach the whole interval $[0,1]$
– Thinking
Nov 30 at 20:54
1
@Thinking I don't see the problem. Where exactly do you think my proof breaks down?
– zhw.
Nov 30 at 21:04
You are right, I completely missounderstood the proof. It's actually clever. (+1)
– Thinking
Nov 30 at 21:08
@Thinking I've edited to remove the continuity hypothesis.
– zhw.
Nov 30 at 21:24
@Thinking Yes of course. Otherwise $M_x-epsilon$ is an upper bound.
– zhw.
Nov 30 at 21:52
add a comment |
I assume that $f$ is Lebesgue integrable along with the other hypotheses. Let $xin [0,1].$ Let $M_x$ be the supremum of $f$ over $[0,x].$ If $M_x=0,$ there is nothing to prove. So assume $M_x>0.$ Let $0<epsilon< M_x.$ Choose $x_epsilonin [0,x]$ such that $f(x_epsilon) > M_x-epsilon.$ Then
$$M_x-epsilon < f(x_epsilon) le int_0^{x_epsilon}sqrt {f(t)},dt le xsqrt{M_x}.$$
Now let $epsilonto 0^+$ to see $M_xle xsqrt{M_x},$ which implies $sqrt{M_x}le x.$ Squaring, we see $f(x)le M_xle x^2,$ giving the result.
answered Nov 30 at 20:50
zhw.
71.6k43075
I assume that $f$ is Lebesgue integrable along with the other hypotheses. Let $xin [0,1].$ Let $M_x$ be the supremum of $f$ over $[0,x].$ If $M_x=0,$ there is nothing to prove. So assume $M_x>0.$ Let $0<epsilon< M_x.$ Choose $x_epsilonin [0,x]$ such that $f(x_epsilon) > M_x-epsilon.$ Then
$$M_x-epsilon < f(x_epsilon) le int_0^{x_epsilon}sqrt {f(t)},dt le xsqrt{M_x}.$$
Now let $epsilonto 0^+$ to see $M_xle xsqrt{M_x},$ which implies $sqrt{M_x}le x.$ Squaring, we see $f(x)le M_xle x^2,$ giving the result.
answered Nov 30 at 20:50
zhw.
71.6k43075
edited Dec 1 at 16:42
answered Nov 30 at 20:50
zhw.
71.6k43075
answered Nov 30 at 20:50
zhw.
71.6k43075
answered Nov 30 at 20:50
zhw.
71.6k43075
71.6k43075
There's a problem, if the number of local maxima is dense in $[0,1]$ no ? $x_0$ can be really close to $0$, then the next one $x_1$ can be really close to $x_0$... and we never reach the whole interval $[0,1]$
– Thinking
Nov 30 at 20:54
1
@Thinking I don't see the problem. Where exactly do you think my proof breaks down?
– zhw.
Nov 30 at 21:04
You are right, I completely missounderstood the proof. It's actually clever. (+1)
– Thinking
Nov 30 at 21:08
@Thinking I've edited to remove the continuity hypothesis.
– zhw.
Nov 30 at 21:24
@Thinking Yes of course. Otherwise $M_x-epsilon$ is an upper bound.
– zhw.
Nov 30 at 21:52
add a comment |
There's a problem, if the number of local maxima is dense in $[0,1]$ no ? $x_0$ can be really close to $0$, then the next one $x_1$ can be really close to $x_0$... and we never reach the whole interval $[0,1]$
– Thinking
Nov 30 at 20:54
1
@Thinking I don't see the problem. Where exactly do you think my proof breaks down?
– zhw.
Nov 30 at 21:04
You are right, I completely missounderstood the proof. It's actually clever. (+1)
– Thinking
Nov 30 at 21:08
@Thinking I've edited to remove the continuity hypothesis.
– zhw.
Nov 30 at 21:24
@Thinking Yes of course. Otherwise $M_x-epsilon$ is an upper bound.
– zhw.
Nov 30 at 21:52
There's a problem, if the number of local maxima is dense in $[0,1]$ no ? $x_0$ can be really close to $0$, then the next one $x_1$ can be really close to $x_0$... and we never reach the whole interval $[0,1]$
– Thinking
Nov 30 at 20:54
There's a problem, if the number of local maxima is dense in $[0,1]$ no ? $x_0$ can be really close to $0$, then the next one $x_1$ can be really close to $x_0$... and we never reach the whole interval $[0,1]$
– Thinking
Nov 30 at 20:54
1
1
@Thinking I don't see the problem. Where exactly do you think my proof breaks down?
– zhw.
Nov 30 at 21:04
@Thinking I don't see the problem. Where exactly do you think my proof breaks down?
– zhw.
Nov 30 at 21:04
You are right, I completely missounderstood the proof. It's actually clever. (+1)
– Thinking
Nov 30 at 21:08
You are right, I completely missounderstood the proof. It's actually clever. (+1)
– Thinking
Nov 30 at 21:08
@Thinking I've edited to remove the continuity hypothesis.
– zhw.
Nov 30 at 21:24
@Thinking I've edited to remove the continuity hypothesis.
– zhw.
Nov 30 at 21:24
@Thinking Yes of course. Otherwise $M_x-epsilon$ is an upper bound.
– zhw.
Nov 30 at 21:52
@Thinking Yes of course. Otherwise $M_x-epsilon$ is an upper bound.
– zhw.
Nov 30 at 21:52
add a comment |
Suppose $b_{n+1}=frac 12(b_n+2)$ and $a_{n+1}=frac{a_n^{1/2}}{b_{n+1}}$, then $f(x) le a_nx^{b_n}Rightarrow f(x)le a_{n+1}x^{b_{n+1}}$. As you showed, $f(x)le a_0x^{b_0}$ where $a_0=b_0=1$. To solve the recurrence for $b_n$ put $B_n=2^{n}b_n$ and notice $b_{n+1}-frac 12b_n=1Rightarrow 2^{n+1}b_{n+1}-2^{n}b_n=B_{n+1}-B_n=2^{n+1}$, summing over $n$ gives $B_{N}-B_0=sum_{n=0}^{N-1} 2^{n+1}=2^{N+1}-2$ and therefore $b_{N}=2^{-N}(2^{N+1}-2+b_0)=2-2^{-N}$. The $b_n$'s are increasing, so $frac 1{b_n}$ is a decreasing sequence. Assume $a_{n+1}le a_n$ so $a_{n+1}^{1/2}le a_n^{1/2}$ and hence $a_{n+2}=frac{a_{n+1}^{1/2}}{b_{n+2}} le frac{a_{n}^{1/2}}{b_{n+1}}=a_{n+1}$. Because $a_1=frac 23le 1 =a_0$, this holds for $n=0$ and induction shows $a_n$ is a decreasing sequence bounded below by $0$, thus has a limit $a$. Taking this limit shows $a=a^{1/2}/2$ because $b_nto 2$, and therefore $a=1/4$. Finally, $f(x)le a_nx^{b_n}to frac 14x^2$.
answered Nov 30 at 20:12
Guacho Perez
3,88911131
add a comment |
Suppose $b_{n+1}=frac 12(b_n+2)$ and $a_{n+1}=frac{a_n^{1/2}}{b_{n+1}}$, then $f(x) le a_nx^{b_n}Rightarrow f(x)le a_{n+1}x^{b_{n+1}}$. As you showed, $f(x)le a_0x^{b_0}$ where $a_0=b_0=1$. To solve the recurrence for $b_n$ put $B_n=2^{n}b_n$ and notice $b_{n+1}-frac 12b_n=1Rightarrow 2^{n+1}b_{n+1}-2^{n}b_n=B_{n+1}-B_n=2^{n+1}$, summing over $n$ gives $B_{N}-B_0=sum_{n=0}^{N-1} 2^{n+1}=2^{N+1}-2$ and therefore $b_{N}=2^{-N}(2^{N+1}-2+b_0)=2-2^{-N}$. The $b_n$'s are increasing, so $frac 1{b_n}$ is a decreasing sequence. Assume $a_{n+1}le a_n$ so $a_{n+1}^{1/2}le a_n^{1/2}$ and hence $a_{n+2}=frac{a_{n+1}^{1/2}}{b_{n+2}} le frac{a_{n}^{1/2}}{b_{n+1}}=a_{n+1}$. Because $a_1=frac 23le 1 =a_0$, this holds for $n=0$ and induction shows $a_n$ is a decreasing sequence bounded below by $0$, thus has a limit $a$. Taking this limit shows $a=a^{1/2}/2$ because $b_nto 2$, and therefore $a=1/4$. Finally, $f(x)le a_nx^{b_n}to frac 14x^2$.
answered Nov 30 at 20:12
Guacho Perez
3,88911131
add a comment |
Suppose $b_{n+1}=frac 12(b_n+2)$ and $a_{n+1}=frac{a_n^{1/2}}{b_{n+1}}$, then $f(x) le a_nx^{b_n}Rightarrow f(x)le a_{n+1}x^{b_{n+1}}$. As you showed, $f(x)le a_0x^{b_0}$ where $a_0=b_0=1$. To solve the recurrence for $b_n$ put $B_n=2^{n}b_n$ and notice $b_{n+1}-frac 12b_n=1Rightarrow 2^{n+1}b_{n+1}-2^{n}b_n=B_{n+1}-B_n=2^{n+1}$, summing over $n$ gives $B_{N}-B_0=sum_{n=0}^{N-1} 2^{n+1}=2^{N+1}-2$ and therefore $b_{N}=2^{-N}(2^{N+1}-2+b_0)=2-2^{-N}$. The $b_n$'s are increasing, so $frac 1{b_n}$ is a decreasing sequence. Assume $a_{n+1}le a_n$ so $a_{n+1}^{1/2}le a_n^{1/2}$ and hence $a_{n+2}=frac{a_{n+1}^{1/2}}{b_{n+2}} le frac{a_{n}^{1/2}}{b_{n+1}}=a_{n+1}$. Because $a_1=frac 23le 1 =a_0$, this holds for $n=0$ and induction shows $a_n$ is a decreasing sequence bounded below by $0$, thus has a limit $a$. Taking this limit shows $a=a^{1/2}/2$ because $b_nto 2$, and therefore $a=1/4$. Finally, $f(x)le a_nx^{b_n}to frac 14x^2$.
answered Nov 30 at 20:12
Guacho Perez
3,88911131
Suppose $b_{n+1}=frac 12(b_n+2)$ and $a_{n+1}=frac{a_n^{1/2}}{b_{n+1}}$, then $f(x) le a_nx^{b_n}Rightarrow f(x)le a_{n+1}x^{b_{n+1}}$. As you showed, $f(x)le a_0x^{b_0}$ where $a_0=b_0=1$. To solve the recurrence for $b_n$ put $B_n=2^{n}b_n$ and notice $b_{n+1}-frac 12b_n=1Rightarrow 2^{n+1}b_{n+1}-2^{n}b_n=B_{n+1}-B_n=2^{n+1}$, summing over $n$ gives $B_{N}-B_0=sum_{n=0}^{N-1} 2^{n+1}=2^{N+1}-2$ and therefore $b_{N}=2^{-N}(2^{N+1}-2+b_0)=2-2^{-N}$. The $b_n$'s are increasing, so $frac 1{b_n}$ is a decreasing sequence. Assume $a_{n+1}le a_n$ so $a_{n+1}^{1/2}le a_n^{1/2}$ and hence $a_{n+2}=frac{a_{n+1}^{1/2}}{b_{n+2}} le frac{a_{n}^{1/2}}{b_{n+1}}=a_{n+1}$. Because $a_1=frac 23le 1 =a_0$, this holds for $n=0$ and induction shows $a_n$ is a decreasing sequence bounded below by $0$, thus has a limit $a$. Taking this limit shows $a=a^{1/2}/2$ because $b_nto 2$, and therefore $a=1/4$. Finally, $f(x)le a_nx^{b_n}to frac 14x^2$.
answered Nov 30 at 20:12
Guacho Perez
3,88911131
answered Nov 30 at 20:12
Guacho Perez
3,88911131
answered Nov 30 at 20:12
Guacho Perez
3,88911131
answered Nov 30 at 20:12
Guacho Perez
3,88911131
3,88911131
add a comment |
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This is a special case of math.stackexchange.com/q/2050893/42969 – choose $C=1$ and $alpha = 1/2$. It follows that $f(x) le x^2/4$.
– Martin R
Nov 30 at 20:31
@MartinR $f$ is not necessarily conitnuous here
– Thinking
Nov 30 at 20:38
@Thinking: You are right. – I assume that a similar argument works if $f$ is only integrable, but I am not 100% sure about that.
– Martin R
Nov 30 at 20:43
@MartinR maybe... but in this case $v$ is not necessarily differentiable so there might be several problems to fix.
– Thinking
Nov 30 at 20:46