Bounded function with bounded second derivative imply bounded first derivative
Let $f$ be a $C^2$ function from $(t_1,t_2)$ to $R^n$ such that $Vert f(t)Vertleq A$ and $Vert f''(t)Vert leq B$ for all $tin (t_1,t_2)$, where $A$ and $B$ are nonnegative reals.
Let $t_0in (t_1,t_2)$ and $alpha>0$ be such that $(t_0-alpha,t_0+alpha)subset (t_1,t_2)$.
I would like to prove that $Vert f'(t_0)Vertleq 2A/alpha + Balpha/2$.
I thought of using a Taylor expansion of $f$ but cannot manage to get the inequality.
real-analysis derivatives taylor-expansion
asked Nov 27 '18 at 15:27
filip
214
add a comment |
Let $f$ be a $C^2$ function from $(t_1,t_2)$ to $R^n$ such that $Vert f(t)Vertleq A$ and $Vert f''(t)Vert leq B$ for all $tin (t_1,t_2)$, where $A$ and $B$ are nonnegative reals.
Let $t_0in (t_1,t_2)$ and $alpha>0$ be such that $(t_0-alpha,t_0+alpha)subset (t_1,t_2)$.
I would like to prove that $Vert f'(t_0)Vertleq 2A/alpha + Balpha/2$.
I thought of using a Taylor expansion of $f$ but cannot manage to get the inequality.
real-analysis derivatives taylor-expansion
asked Nov 27 '18 at 15:27
filip
214
Are you sure about the inequality? I can obtain that the left hand side is smaller than $2A/alpha+Balpha/2.$
– Uskebasi
Nov 27 '18 at 16:30
In my inequality there is a factor $2$ that multiplies the term $A/alpha$
– Uskebasi
Nov 27 '18 at 16:32
add a comment |
Let $f$ be a $C^2$ function from $(t_1,t_2)$ to $R^n$ such that $Vert f(t)Vertleq A$ and $Vert f''(t)Vert leq B$ for all $tin (t_1,t_2)$, where $A$ and $B$ are nonnegative reals.
Let $t_0in (t_1,t_2)$ and $alpha>0$ be such that $(t_0-alpha,t_0+alpha)subset (t_1,t_2)$.
I would like to prove that $Vert f'(t_0)Vertleq 2A/alpha + Balpha/2$.
I thought of using a Taylor expansion of $f$ but cannot manage to get the inequality.
real-analysis derivatives taylor-expansion
asked Nov 27 '18 at 15:27
filip
214
Let $f$ be a $C^2$ function from $(t_1,t_2)$ to $R^n$ such that $Vert f(t)Vertleq A$ and $Vert f''(t)Vert leq B$ for all $tin (t_1,t_2)$, where $A$ and $B$ are nonnegative reals.
Let $t_0in (t_1,t_2)$ and $alpha>0$ be such that $(t_0-alpha,t_0+alpha)subset (t_1,t_2)$.
I would like to prove that $Vert f'(t_0)Vertleq 2A/alpha + Balpha/2$.
I thought of using a Taylor expansion of $f$ but cannot manage to get the inequality.
real-analysis derivatives taylor-expansion
real-analysis derivatives taylor-expansion
asked Nov 27 '18 at 15:27
filip
214
asked Nov 27 '18 at 15:27
filip
214
edited Nov 27 '18 at 17:01
asked Nov 27 '18 at 15:27
filip
214
asked Nov 27 '18 at 15:27
filip
214
asked Nov 27 '18 at 15:27
filip
214
214
Are you sure about the inequality? I can obtain that the left hand side is smaller than $2A/alpha+Balpha/2.$
– Uskebasi
Nov 27 '18 at 16:30
In my inequality there is a factor $2$ that multiplies the term $A/alpha$
– Uskebasi
Nov 27 '18 at 16:32
add a comment |
Are you sure about the inequality? I can obtain that the left hand side is smaller than $2A/alpha+Balpha/2.$
– Uskebasi
Nov 27 '18 at 16:30
In my inequality there is a factor $2$ that multiplies the term $A/alpha$
– Uskebasi
Nov 27 '18 at 16:32
Are you sure about the inequality? I can obtain that the left hand side is smaller than $2A/alpha+Balpha/2.$
– Uskebasi
Nov 27 '18 at 16:30
Are you sure about the inequality? I can obtain that the left hand side is smaller than $2A/alpha+Balpha/2.$
– Uskebasi
Nov 27 '18 at 16:30
In my inequality there is a factor $2$ that multiplies the term $A/alpha$
– Uskebasi
Nov 27 '18 at 16:32
In my inequality there is a factor $2$ that multiplies the term $A/alpha$
– Uskebasi
Nov 27 '18 at 16:32
add a comment |
1 Answer
1
active
oldest
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First, it's enough to consider the case $n=1$: Say $vinBbb R^n$ with $||v||=1$ and let $g(t)=f(t)cdot v$. Then $|g'|le A$ and $|g''(t)|le B$; if we show that $|g'(t)|le C$ then $|f'(t)cdot v|le C$ for all $v$ with $||v||=1$, hence $|f'(t)|le C$.
Taylor's Theorem shows that for $tin (t_0-alpha,t_0+alpha)$ we have $$f(t)=P(t)+R(t)$$where $$P(t)=f(t_0)+(t-t_0)f'(t_0)$$and $$|R(t)|lefrac12 B(t-t_0)^2<frac12 Balpha^2.$$
So $$2Age|f(t)-f(t_0)|=|(t-t_0)f'(t_0)+R(t)|
ge|t-t_0||f'(t_0)|-frac12 Balpha^2$$Chooing $t$ close to $t_0pmalpha$ now gives $$2Agealpha|f'(t_0)|-frac12Balpha^2,$$or$$|f'(t_0)|le2A/alpha+frac12 Balpha.$$
answered Nov 27 '18 at 16:49
David C. Ullrich
58.5k43892
I am not sure I understand how you restrict to $n=1$, since the hypothesis is on $f$ and not on $g$.
– filip
Nov 27 '18 at 17:48
@filip We're given $||f||le A$ and $||f''||le B$. It follows that $|g|le A$ and $|g''|le B$.
– David C. Ullrich
Nov 27 '18 at 17:58
Yes but then how do you go back to the inequality on $f$ ? I agree that it is straight forward for the $L^1$ norm but what about the $L^2$ norm?
– filip
Nov 27 '18 at 19:31
@filip If $|f'(t)cdot v|le C$ for every $v$ with $||v||=1$ then $||f'(t)||le C$. (Let $v=f'(t)/||f'(t)||$.)
– David C. Ullrich
Nov 27 '18 at 20:20
Something still bugs me. Suppose $n=2$. We have $vert f'_i(t) vertleq C$ for $i=1,2$. Then we take the scalar product with $(vert f'_1(t) vert/Vert f'(t) Vert,vert f'_2(t) vert/Vert f'(t) Vert)$ we get $Vert f'(t) Vertleq C((vert f'_1(t) vert+vert f'_2(t) vert)/Vert f'(t) Vert)$.
– filip
Nov 28 '18 at 9:13
|
show 4 more comments
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1 Answer
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First, it's enough to consider the case $n=1$: Say $vinBbb R^n$ with $||v||=1$ and let $g(t)=f(t)cdot v$. Then $|g'|le A$ and $|g''(t)|le B$; if we show that $|g'(t)|le C$ then $|f'(t)cdot v|le C$ for all $v$ with $||v||=1$, hence $|f'(t)|le C$.
Taylor's Theorem shows that for $tin (t_0-alpha,t_0+alpha)$ we have $$f(t)=P(t)+R(t)$$where $$P(t)=f(t_0)+(t-t_0)f'(t_0)$$and $$|R(t)|lefrac12 B(t-t_0)^2<frac12 Balpha^2.$$
So $$2Age|f(t)-f(t_0)|=|(t-t_0)f'(t_0)+R(t)|
ge|t-t_0||f'(t_0)|-frac12 Balpha^2$$Chooing $t$ close to $t_0pmalpha$ now gives $$2Agealpha|f'(t_0)|-frac12Balpha^2,$$or$$|f'(t_0)|le2A/alpha+frac12 Balpha.$$
answered Nov 27 '18 at 16:49
David C. Ullrich
58.5k43892
I am not sure I understand how you restrict to $n=1$, since the hypothesis is on $f$ and not on $g$.
– filip
Nov 27 '18 at 17:48
@filip We're given $||f||le A$ and $||f''||le B$. It follows that $|g|le A$ and $|g''|le B$.
– David C. Ullrich
Nov 27 '18 at 17:58
Yes but then how do you go back to the inequality on $f$ ? I agree that it is straight forward for the $L^1$ norm but what about the $L^2$ norm?
– filip
Nov 27 '18 at 19:31
@filip If $|f'(t)cdot v|le C$ for every $v$ with $||v||=1$ then $||f'(t)||le C$. (Let $v=f'(t)/||f'(t)||$.)
– David C. Ullrich
Nov 27 '18 at 20:20
Something still bugs me. Suppose $n=2$. We have $vert f'_i(t) vertleq C$ for $i=1,2$. Then we take the scalar product with $(vert f'_1(t) vert/Vert f'(t) Vert,vert f'_2(t) vert/Vert f'(t) Vert)$ we get $Vert f'(t) Vertleq C((vert f'_1(t) vert+vert f'_2(t) vert)/Vert f'(t) Vert)$.
– filip
Nov 28 '18 at 9:13
|
show 4 more comments
First, it's enough to consider the case $n=1$: Say $vinBbb R^n$ with $||v||=1$ and let $g(t)=f(t)cdot v$. Then $|g'|le A$ and $|g''(t)|le B$; if we show that $|g'(t)|le C$ then $|f'(t)cdot v|le C$ for all $v$ with $||v||=1$, hence $|f'(t)|le C$.
Taylor's Theorem shows that for $tin (t_0-alpha,t_0+alpha)$ we have $$f(t)=P(t)+R(t)$$where $$P(t)=f(t_0)+(t-t_0)f'(t_0)$$and $$|R(t)|lefrac12 B(t-t_0)^2<frac12 Balpha^2.$$
So $$2Age|f(t)-f(t_0)|=|(t-t_0)f'(t_0)+R(t)|
ge|t-t_0||f'(t_0)|-frac12 Balpha^2$$Chooing $t$ close to $t_0pmalpha$ now gives $$2Agealpha|f'(t_0)|-frac12Balpha^2,$$or$$|f'(t_0)|le2A/alpha+frac12 Balpha.$$
answered Nov 27 '18 at 16:49
David C. Ullrich
58.5k43892
I am not sure I understand how you restrict to $n=1$, since the hypothesis is on $f$ and not on $g$.
– filip
Nov 27 '18 at 17:48
@filip We're given $||f||le A$ and $||f''||le B$. It follows that $|g|le A$ and $|g''|le B$.
– David C. Ullrich
Nov 27 '18 at 17:58
Yes but then how do you go back to the inequality on $f$ ? I agree that it is straight forward for the $L^1$ norm but what about the $L^2$ norm?
– filip
Nov 27 '18 at 19:31
@filip If $|f'(t)cdot v|le C$ for every $v$ with $||v||=1$ then $||f'(t)||le C$. (Let $v=f'(t)/||f'(t)||$.)
– David C. Ullrich
Nov 27 '18 at 20:20
Something still bugs me. Suppose $n=2$. We have $vert f'_i(t) vertleq C$ for $i=1,2$. Then we take the scalar product with $(vert f'_1(t) vert/Vert f'(t) Vert,vert f'_2(t) vert/Vert f'(t) Vert)$ we get $Vert f'(t) Vertleq C((vert f'_1(t) vert+vert f'_2(t) vert)/Vert f'(t) Vert)$.
– filip
Nov 28 '18 at 9:13
|
show 4 more comments
First, it's enough to consider the case $n=1$: Say $vinBbb R^n$ with $||v||=1$ and let $g(t)=f(t)cdot v$. Then $|g'|le A$ and $|g''(t)|le B$; if we show that $|g'(t)|le C$ then $|f'(t)cdot v|le C$ for all $v$ with $||v||=1$, hence $|f'(t)|le C$.
Taylor's Theorem shows that for $tin (t_0-alpha,t_0+alpha)$ we have $$f(t)=P(t)+R(t)$$where $$P(t)=f(t_0)+(t-t_0)f'(t_0)$$and $$|R(t)|lefrac12 B(t-t_0)^2<frac12 Balpha^2.$$
So $$2Age|f(t)-f(t_0)|=|(t-t_0)f'(t_0)+R(t)|
ge|t-t_0||f'(t_0)|-frac12 Balpha^2$$Chooing $t$ close to $t_0pmalpha$ now gives $$2Agealpha|f'(t_0)|-frac12Balpha^2,$$or$$|f'(t_0)|le2A/alpha+frac12 Balpha.$$
answered Nov 27 '18 at 16:49
David C. Ullrich
58.5k43892
First, it's enough to consider the case $n=1$: Say $vinBbb R^n$ with $||v||=1$ and let $g(t)=f(t)cdot v$. Then $|g'|le A$ and $|g''(t)|le B$; if we show that $|g'(t)|le C$ then $|f'(t)cdot v|le C$ for all $v$ with $||v||=1$, hence $|f'(t)|le C$.
Taylor's Theorem shows that for $tin (t_0-alpha,t_0+alpha)$ we have $$f(t)=P(t)+R(t)$$where $$P(t)=f(t_0)+(t-t_0)f'(t_0)$$and $$|R(t)|lefrac12 B(t-t_0)^2<frac12 Balpha^2.$$
So $$2Age|f(t)-f(t_0)|=|(t-t_0)f'(t_0)+R(t)|
ge|t-t_0||f'(t_0)|-frac12 Balpha^2$$Chooing $t$ close to $t_0pmalpha$ now gives $$2Agealpha|f'(t_0)|-frac12Balpha^2,$$or$$|f'(t_0)|le2A/alpha+frac12 Balpha.$$
answered Nov 27 '18 at 16:49
David C. Ullrich
58.5k43892
edited Nov 27 '18 at 18:00
answered Nov 27 '18 at 16:49
David C. Ullrich
58.5k43892
answered Nov 27 '18 at 16:49
David C. Ullrich
58.5k43892
answered Nov 27 '18 at 16:49
David C. Ullrich
58.5k43892
58.5k43892
I am not sure I understand how you restrict to $n=1$, since the hypothesis is on $f$ and not on $g$.
– filip
Nov 27 '18 at 17:48
@filip We're given $||f||le A$ and $||f''||le B$. It follows that $|g|le A$ and $|g''|le B$.
– David C. Ullrich
Nov 27 '18 at 17:58
Yes but then how do you go back to the inequality on $f$ ? I agree that it is straight forward for the $L^1$ norm but what about the $L^2$ norm?
– filip
Nov 27 '18 at 19:31
@filip If $|f'(t)cdot v|le C$ for every $v$ with $||v||=1$ then $||f'(t)||le C$. (Let $v=f'(t)/||f'(t)||$.)
– David C. Ullrich
Nov 27 '18 at 20:20
Something still bugs me. Suppose $n=2$. We have $vert f'_i(t) vertleq C$ for $i=1,2$. Then we take the scalar product with $(vert f'_1(t) vert/Vert f'(t) Vert,vert f'_2(t) vert/Vert f'(t) Vert)$ we get $Vert f'(t) Vertleq C((vert f'_1(t) vert+vert f'_2(t) vert)/Vert f'(t) Vert)$.
– filip
Nov 28 '18 at 9:13
|
show 4 more comments
I am not sure I understand how you restrict to $n=1$, since the hypothesis is on $f$ and not on $g$.
– filip
Nov 27 '18 at 17:48
@filip We're given $||f||le A$ and $||f''||le B$. It follows that $|g|le A$ and $|g''|le B$.
– David C. Ullrich
Nov 27 '18 at 17:58
Yes but then how do you go back to the inequality on $f$ ? I agree that it is straight forward for the $L^1$ norm but what about the $L^2$ norm?
– filip
Nov 27 '18 at 19:31
@filip If $|f'(t)cdot v|le C$ for every $v$ with $||v||=1$ then $||f'(t)||le C$. (Let $v=f'(t)/||f'(t)||$.)
– David C. Ullrich
Nov 27 '18 at 20:20
Something still bugs me. Suppose $n=2$. We have $vert f'_i(t) vertleq C$ for $i=1,2$. Then we take the scalar product with $(vert f'_1(t) vert/Vert f'(t) Vert,vert f'_2(t) vert/Vert f'(t) Vert)$ we get $Vert f'(t) Vertleq C((vert f'_1(t) vert+vert f'_2(t) vert)/Vert f'(t) Vert)$.
– filip
Nov 28 '18 at 9:13
I am not sure I understand how you restrict to $n=1$, since the hypothesis is on $f$ and not on $g$.
– filip
Nov 27 '18 at 17:48
I am not sure I understand how you restrict to $n=1$, since the hypothesis is on $f$ and not on $g$.
– filip
Nov 27 '18 at 17:48
@filip We're given $||f||le A$ and $||f''||le B$. It follows that $|g|le A$ and $|g''|le B$.
– David C. Ullrich
Nov 27 '18 at 17:58
@filip We're given $||f||le A$ and $||f''||le B$. It follows that $|g|le A$ and $|g''|le B$.
– David C. Ullrich
Nov 27 '18 at 17:58
Yes but then how do you go back to the inequality on $f$ ? I agree that it is straight forward for the $L^1$ norm but what about the $L^2$ norm?
– filip
Nov 27 '18 at 19:31
Yes but then how do you go back to the inequality on $f$ ? I agree that it is straight forward for the $L^1$ norm but what about the $L^2$ norm?
– filip
Nov 27 '18 at 19:31
@filip If $|f'(t)cdot v|le C$ for every $v$ with $||v||=1$ then $||f'(t)||le C$. (Let $v=f'(t)/||f'(t)||$.)
– David C. Ullrich
Nov 27 '18 at 20:20
@filip If $|f'(t)cdot v|le C$ for every $v$ with $||v||=1$ then $||f'(t)||le C$. (Let $v=f'(t)/||f'(t)||$.)
– David C. Ullrich
Nov 27 '18 at 20:20
Something still bugs me. Suppose $n=2$. We have $vert f'_i(t) vertleq C$ for $i=1,2$. Then we take the scalar product with $(vert f'_1(t) vert/Vert f'(t) Vert,vert f'_2(t) vert/Vert f'(t) Vert)$ we get $Vert f'(t) Vertleq C((vert f'_1(t) vert+vert f'_2(t) vert)/Vert f'(t) Vert)$.
– filip
Nov 28 '18 at 9:13
Something still bugs me. Suppose $n=2$. We have $vert f'_i(t) vertleq C$ for $i=1,2$. Then we take the scalar product with $(vert f'_1(t) vert/Vert f'(t) Vert,vert f'_2(t) vert/Vert f'(t) Vert)$ we get $Vert f'(t) Vertleq C((vert f'_1(t) vert+vert f'_2(t) vert)/Vert f'(t) Vert)$.
– filip
Nov 28 '18 at 9:13
|
show 4 more comments
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Are you sure about the inequality? I can obtain that the left hand side is smaller than $2A/alpha+Balpha/2.$
– Uskebasi
Nov 27 '18 at 16:30
In my inequality there is a factor $2$ that multiplies the term $A/alpha$
– Uskebasi
Nov 27 '18 at 16:32