Proving that $|x , g(y) - y , g(x)| leq C$ for $x,y in [-1,1]$, where $|g''(t)| leq C$
Let $f : mathbb{R}^2 rightarrow mathbb{R}^2$ given by $f(x,y) = x , g(y) - y , g(x)$, where $g:mathbb{R} rightarrow mathbb{R}$ such that $g in C^2$, $g(0)=0$ and $|g''(x)| leq C $ $forall x in mathbb{R}$. I Want to prove:
a) If $(x,y) in [-1,1]times[-1,1]$ then $|f(x,y)| leq C$.
b) Evaluate $int^1_0 int^1_0 f(x,y) , dx , dy$.
For the b part i guess that a simple application of green theorem will solve it, the value will be in function of $g$.
But, for the a part I'm getting trouble.
I tried to use the mean value theorem for the derivative of $g$ and then obtain that $|g'(x)|leq$ sup $ g^{''}$ $|x|$ on $x in [-1,1]$ and then conclude $ |g'(x)| leq C$ either.
So, applying again to $f$ I would get $|f(x,y)|$ is less or equal to a multiple of $C$.
calculus derivatives absolute-value
edited Nov 28 '18 at 10:48
PhoemueX
27.3k22456
asked Nov 23 '18 at 18:08
Eduardo Silva
68239
add a comment |
Let $f : mathbb{R}^2 rightarrow mathbb{R}^2$ given by $f(x,y) = x , g(y) - y , g(x)$, where $g:mathbb{R} rightarrow mathbb{R}$ such that $g in C^2$, $g(0)=0$ and $|g''(x)| leq C $ $forall x in mathbb{R}$. I Want to prove:
a) If $(x,y) in [-1,1]times[-1,1]$ then $|f(x,y)| leq C$.
b) Evaluate $int^1_0 int^1_0 f(x,y) , dx , dy$.
For the b part i guess that a simple application of green theorem will solve it, the value will be in function of $g$.
But, for the a part I'm getting trouble.
I tried to use the mean value theorem for the derivative of $g$ and then obtain that $|g'(x)|leq$ sup $ g^{''}$ $|x|$ on $x in [-1,1]$ and then conclude $ |g'(x)| leq C$ either.
So, applying again to $f$ I would get $|f(x,y)|$ is less or equal to a multiple of $C$.
calculus derivatives absolute-value
edited Nov 28 '18 at 10:48
PhoemueX
27.3k22456
asked Nov 23 '18 at 18:08
Eduardo Silva
68239
Hint: How does $f$ change if you change $g$ to $widetilde{g}(x) = ax + g(x)$?
– PhoemueX
Nov 23 '18 at 19:03
@PhoemueX, there is no change, but I still can't see for where it is going
– Eduardo Silva
Nov 23 '18 at 19:20
I now also added an answer for part b)
– PhoemueX
Nov 28 '18 at 10:48
add a comment |
Let $f : mathbb{R}^2 rightarrow mathbb{R}^2$ given by $f(x,y) = x , g(y) - y , g(x)$, where $g:mathbb{R} rightarrow mathbb{R}$ such that $g in C^2$, $g(0)=0$ and $|g''(x)| leq C $ $forall x in mathbb{R}$. I Want to prove:
a) If $(x,y) in [-1,1]times[-1,1]$ then $|f(x,y)| leq C$.
b) Evaluate $int^1_0 int^1_0 f(x,y) , dx , dy$.
For the b part i guess that a simple application of green theorem will solve it, the value will be in function of $g$.
But, for the a part I'm getting trouble.
I tried to use the mean value theorem for the derivative of $g$ and then obtain that $|g'(x)|leq$ sup $ g^{''}$ $|x|$ on $x in [-1,1]$ and then conclude $ |g'(x)| leq C$ either.
So, applying again to $f$ I would get $|f(x,y)|$ is less or equal to a multiple of $C$.
calculus derivatives absolute-value
edited Nov 28 '18 at 10:48
PhoemueX
27.3k22456
asked Nov 23 '18 at 18:08
Eduardo Silva
68239
Let $f : mathbb{R}^2 rightarrow mathbb{R}^2$ given by $f(x,y) = x , g(y) - y , g(x)$, where $g:mathbb{R} rightarrow mathbb{R}$ such that $g in C^2$, $g(0)=0$ and $|g''(x)| leq C $ $forall x in mathbb{R}$. I Want to prove:
a) If $(x,y) in [-1,1]times[-1,1]$ then $|f(x,y)| leq C$.
b) Evaluate $int^1_0 int^1_0 f(x,y) , dx , dy$.
For the b part i guess that a simple application of green theorem will solve it, the value will be in function of $g$.
But, for the a part I'm getting trouble.
I tried to use the mean value theorem for the derivative of $g$ and then obtain that $|g'(x)|leq$ sup $ g^{''}$ $|x|$ on $x in [-1,1]$ and then conclude $ |g'(x)| leq C$ either.
So, applying again to $f$ I would get $|f(x,y)|$ is less or equal to a multiple of $C$.
calculus derivatives absolute-value
calculus derivatives absolute-value
edited Nov 28 '18 at 10:48
PhoemueX
27.3k22456
asked Nov 23 '18 at 18:08
Eduardo Silva
68239
edited Nov 28 '18 at 10:48
PhoemueX
27.3k22456
asked Nov 23 '18 at 18:08
Eduardo Silva
68239
edited Nov 28 '18 at 10:48
PhoemueX
27.3k22456
edited Nov 28 '18 at 10:48
PhoemueX
27.3k22456
edited Nov 28 '18 at 10:48
PhoemueX
27.3k22456
27.3k22456
asked Nov 23 '18 at 18:08
Eduardo Silva
68239
asked Nov 23 '18 at 18:08
Eduardo Silva
68239
asked Nov 23 '18 at 18:08
Eduardo Silva
68239
68239
Hint: How does $f$ change if you change $g$ to $widetilde{g}(x) = ax + g(x)$?
– PhoemueX
Nov 23 '18 at 19:03
@PhoemueX, there is no change, but I still can't see for where it is going
– Eduardo Silva
Nov 23 '18 at 19:20
I now also added an answer for part b)
– PhoemueX
Nov 28 '18 at 10:48
add a comment |
Hint: How does $f$ change if you change $g$ to $widetilde{g}(x) = ax + g(x)$?
– PhoemueX
Nov 23 '18 at 19:03
@PhoemueX, there is no change, but I still can't see for where it is going
– Eduardo Silva
Nov 23 '18 at 19:20
I now also added an answer for part b)
– PhoemueX
Nov 28 '18 at 10:48
Hint: How does $f$ change if you change $g$ to $widetilde{g}(x) = ax + g(x)$?
– PhoemueX
Nov 23 '18 at 19:03
Hint: How does $f$ change if you change $g$ to $widetilde{g}(x) = ax + g(x)$?
– PhoemueX
Nov 23 '18 at 19:03
@PhoemueX, there is no change, but I still can't see for where it is going
– Eduardo Silva
Nov 23 '18 at 19:20
@PhoemueX, there is no change, but I still can't see for where it is going
– Eduardo Silva
Nov 23 '18 at 19:20
I now also added an answer for part b)
– PhoemueX
Nov 28 '18 at 10:48
I now also added an answer for part b)
– PhoemueX
Nov 28 '18 at 10:48
add a comment |
1 Answer
1
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oldest
votes
a) If we change from $g$ to $widetilde{g}(x) = g(x) + a x$ for any $a in Bbb{R}$, this does not change the function $f$. Also, we will still have $widetilde{g}(0) = 0$ and $|widetilde{g}''(x)| leq C$. Therefore, we can without loss of generality assume that $g'(0) = 0$.
Now, Taylor's theorem, with the Lagrange form of the remainder shows that for each $x in [-1,1]$, there is some $xi in [0,x] cup [x,0] subset [-1,1]$ satisfying
$$
g(x) = g(0) + g'(0) cdot x + frac{g''(xi)}{2} (x - 0)^2 , ,
$$
and hence $|g(x)| leq frac{C}{2} cdot x^2 leq frac{C}{2}$ for all $x in [-1,1]$.
Now, we easily see
$$
|f(x,y)| leq |x| cdot |g(y)| + |y| cdot |g(x)| leq frac{C}{2} + frac{C}{2} = C
$$
for all $x,y in [-1,1]$.
Final comment: This is more or less what you were also trying to do (using the estimate for $g''$ to estimate $g'$, and then $g$). The main points was to see that one can assume $g'(x) = 0$. Finally, instead of using the mean-value theorem twice, we used Taylor's theorem, which got rid of the factor two that you would otherwise get if you naively apply the mean-value theorem.
b) This part of the question is actually pretty easy; one does not need anything fancy like Green's theorem. Simply note that if we set $a := int_0^1 t , d t$ (so that $a = 1/2$) and $b := int_0^1 g(t) , dt$, then
begin{align*}
int_0^1 int_0^1 f(x,y) , d x , d y
& = int_0^1 int_0^1 x , g(y) , d x , d y
- int_0^1 int_0^1 y , g(x) , d x , d y \
& = int_0^1 g(y) cdot a , d y
- int_0^1 y , b , d y \
& = a cdot b - a cdot b
= 0 , .
end{align*}
answered Nov 25 '18 at 18:17
PhoemueX
27.3k22456
add a comment |
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1 Answer
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a) If we change from $g$ to $widetilde{g}(x) = g(x) + a x$ for any $a in Bbb{R}$, this does not change the function $f$. Also, we will still have $widetilde{g}(0) = 0$ and $|widetilde{g}''(x)| leq C$. Therefore, we can without loss of generality assume that $g'(0) = 0$.
Now, Taylor's theorem, with the Lagrange form of the remainder shows that for each $x in [-1,1]$, there is some $xi in [0,x] cup [x,0] subset [-1,1]$ satisfying
$$
g(x) = g(0) + g'(0) cdot x + frac{g''(xi)}{2} (x - 0)^2 , ,
$$
and hence $|g(x)| leq frac{C}{2} cdot x^2 leq frac{C}{2}$ for all $x in [-1,1]$.
Now, we easily see
$$
|f(x,y)| leq |x| cdot |g(y)| + |y| cdot |g(x)| leq frac{C}{2} + frac{C}{2} = C
$$
for all $x,y in [-1,1]$.
Final comment: This is more or less what you were also trying to do (using the estimate for $g''$ to estimate $g'$, and then $g$). The main points was to see that one can assume $g'(x) = 0$. Finally, instead of using the mean-value theorem twice, we used Taylor's theorem, which got rid of the factor two that you would otherwise get if you naively apply the mean-value theorem.
b) This part of the question is actually pretty easy; one does not need anything fancy like Green's theorem. Simply note that if we set $a := int_0^1 t , d t$ (so that $a = 1/2$) and $b := int_0^1 g(t) , dt$, then
begin{align*}
int_0^1 int_0^1 f(x,y) , d x , d y
& = int_0^1 int_0^1 x , g(y) , d x , d y
- int_0^1 int_0^1 y , g(x) , d x , d y \
& = int_0^1 g(y) cdot a , d y
- int_0^1 y , b , d y \
& = a cdot b - a cdot b
= 0 , .
end{align*}
answered Nov 25 '18 at 18:17
PhoemueX
27.3k22456
add a comment |
a) If we change from $g$ to $widetilde{g}(x) = g(x) + a x$ for any $a in Bbb{R}$, this does not change the function $f$. Also, we will still have $widetilde{g}(0) = 0$ and $|widetilde{g}''(x)| leq C$. Therefore, we can without loss of generality assume that $g'(0) = 0$.
Now, Taylor's theorem, with the Lagrange form of the remainder shows that for each $x in [-1,1]$, there is some $xi in [0,x] cup [x,0] subset [-1,1]$ satisfying
$$
g(x) = g(0) + g'(0) cdot x + frac{g''(xi)}{2} (x - 0)^2 , ,
$$
and hence $|g(x)| leq frac{C}{2} cdot x^2 leq frac{C}{2}$ for all $x in [-1,1]$.
Now, we easily see
$$
|f(x,y)| leq |x| cdot |g(y)| + |y| cdot |g(x)| leq frac{C}{2} + frac{C}{2} = C
$$
for all $x,y in [-1,1]$.
Final comment: This is more or less what you were also trying to do (using the estimate for $g''$ to estimate $g'$, and then $g$). The main points was to see that one can assume $g'(x) = 0$. Finally, instead of using the mean-value theorem twice, we used Taylor's theorem, which got rid of the factor two that you would otherwise get if you naively apply the mean-value theorem.
b) This part of the question is actually pretty easy; one does not need anything fancy like Green's theorem. Simply note that if we set $a := int_0^1 t , d t$ (so that $a = 1/2$) and $b := int_0^1 g(t) , dt$, then
begin{align*}
int_0^1 int_0^1 f(x,y) , d x , d y
& = int_0^1 int_0^1 x , g(y) , d x , d y
- int_0^1 int_0^1 y , g(x) , d x , d y \
& = int_0^1 g(y) cdot a , d y
- int_0^1 y , b , d y \
& = a cdot b - a cdot b
= 0 , .
end{align*}
answered Nov 25 '18 at 18:17
PhoemueX
27.3k22456
add a comment |
a) If we change from $g$ to $widetilde{g}(x) = g(x) + a x$ for any $a in Bbb{R}$, this does not change the function $f$. Also, we will still have $widetilde{g}(0) = 0$ and $|widetilde{g}''(x)| leq C$. Therefore, we can without loss of generality assume that $g'(0) = 0$.
Now, Taylor's theorem, with the Lagrange form of the remainder shows that for each $x in [-1,1]$, there is some $xi in [0,x] cup [x,0] subset [-1,1]$ satisfying
$$
g(x) = g(0) + g'(0) cdot x + frac{g''(xi)}{2} (x - 0)^2 , ,
$$
and hence $|g(x)| leq frac{C}{2} cdot x^2 leq frac{C}{2}$ for all $x in [-1,1]$.
Now, we easily see
$$
|f(x,y)| leq |x| cdot |g(y)| + |y| cdot |g(x)| leq frac{C}{2} + frac{C}{2} = C
$$
for all $x,y in [-1,1]$.
Final comment: This is more or less what you were also trying to do (using the estimate for $g''$ to estimate $g'$, and then $g$). The main points was to see that one can assume $g'(x) = 0$. Finally, instead of using the mean-value theorem twice, we used Taylor's theorem, which got rid of the factor two that you would otherwise get if you naively apply the mean-value theorem.
b) This part of the question is actually pretty easy; one does not need anything fancy like Green's theorem. Simply note that if we set $a := int_0^1 t , d t$ (so that $a = 1/2$) and $b := int_0^1 g(t) , dt$, then
begin{align*}
int_0^1 int_0^1 f(x,y) , d x , d y
& = int_0^1 int_0^1 x , g(y) , d x , d y
- int_0^1 int_0^1 y , g(x) , d x , d y \
& = int_0^1 g(y) cdot a , d y
- int_0^1 y , b , d y \
& = a cdot b - a cdot b
= 0 , .
end{align*}
answered Nov 25 '18 at 18:17
PhoemueX
27.3k22456
a) If we change from $g$ to $widetilde{g}(x) = g(x) + a x$ for any $a in Bbb{R}$, this does not change the function $f$. Also, we will still have $widetilde{g}(0) = 0$ and $|widetilde{g}''(x)| leq C$. Therefore, we can without loss of generality assume that $g'(0) = 0$.
Now, Taylor's theorem, with the Lagrange form of the remainder shows that for each $x in [-1,1]$, there is some $xi in [0,x] cup [x,0] subset [-1,1]$ satisfying
$$
g(x) = g(0) + g'(0) cdot x + frac{g''(xi)}{2} (x - 0)^2 , ,
$$
and hence $|g(x)| leq frac{C}{2} cdot x^2 leq frac{C}{2}$ for all $x in [-1,1]$.
Now, we easily see
$$
|f(x,y)| leq |x| cdot |g(y)| + |y| cdot |g(x)| leq frac{C}{2} + frac{C}{2} = C
$$
for all $x,y in [-1,1]$.
Final comment: This is more or less what you were also trying to do (using the estimate for $g''$ to estimate $g'$, and then $g$). The main points was to see that one can assume $g'(x) = 0$. Finally, instead of using the mean-value theorem twice, we used Taylor's theorem, which got rid of the factor two that you would otherwise get if you naively apply the mean-value theorem.
b) This part of the question is actually pretty easy; one does not need anything fancy like Green's theorem. Simply note that if we set $a := int_0^1 t , d t$ (so that $a = 1/2$) and $b := int_0^1 g(t) , dt$, then
begin{align*}
int_0^1 int_0^1 f(x,y) , d x , d y
& = int_0^1 int_0^1 x , g(y) , d x , d y
- int_0^1 int_0^1 y , g(x) , d x , d y \
& = int_0^1 g(y) cdot a , d y
- int_0^1 y , b , d y \
& = a cdot b - a cdot b
= 0 , .
end{align*}
answered Nov 25 '18 at 18:17
PhoemueX
27.3k22456
edited Nov 28 '18 at 10:44
answered Nov 25 '18 at 18:17
PhoemueX
27.3k22456
answered Nov 25 '18 at 18:17
PhoemueX
27.3k22456
answered Nov 25 '18 at 18:17
PhoemueX
27.3k22456
27.3k22456
add a comment |
add a comment |
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Hint: How does $f$ change if you change $g$ to $widetilde{g}(x) = ax + g(x)$?
– PhoemueX
Nov 23 '18 at 19:03
@PhoemueX, there is no change, but I still can't see for where it is going
– Eduardo Silva
Nov 23 '18 at 19:20
I now also added an answer for part b)
– PhoemueX
Nov 28 '18 at 10:48