Show $frac{partial}{partial t} int_{f(x) < t} varphi(x) text{d}x in C_0^infty(mathbb{R}_t )$ if $varphi in...
$begingroup$
Let $X$ be an open subset of $mathbb{R}^n$. Suppose that $f in C^infty(X)$ is real-valued, with the property that $nabla f(x) neq 0$ for all $x in X$. For any $varphi in C_0^infty(X)$, and $t in mathbb{R}$, define
$$varphi_f(t) = frac{partial}{partial t} int_{f(x) < t} varphi(x) text{d}x.$$
I would like to show that $varphi_f in C_0^infty(mathbb{R})$.
This is a claim made in Chapter 7 of Friedlander's book on distributions. I am having trouble following his justification.
My attempt at a solution:
My idea is to first show that $frac{partial}{partial t} int_{f(x) < t} varphi(x) text{d}x$ is well-defined (i.e., that the derivative exists) and find it's value, with the hope this will give a clue about how to calculate the higher-order derivatives. For $h > 0$:
$$frac{1}{h} left(int_{f(x) < t} varphi(x) text{d}x - int_{f(x) < t + h} varphi(x) text{d}x right) = frac{1}{h} int mathbf{1}_{t le f < t + h }(x) varphi(x) text{d}x.$$
But it's not clear to me that the right side converges to anything as $h to 0^+$. I think maybe a change of variables (made possible by the implicit function Theorem?) is in order to make the dependence on $f$ more explicit.
Hints or solutions are greatly appreciated!
real-analysis pde partial-derivative
edited Dec 20 '18 at 12:11
Falrach
1,664223
asked Aug 7 '17 at 18:11
JZShapiroJZShapiro
1,99511125
$endgroup$
add a comment |
$begingroup$
Let $X$ be an open subset of $mathbb{R}^n$. Suppose that $f in C^infty(X)$ is real-valued, with the property that $nabla f(x) neq 0$ for all $x in X$. For any $varphi in C_0^infty(X)$, and $t in mathbb{R}$, define
$$varphi_f(t) = frac{partial}{partial t} int_{f(x) < t} varphi(x) text{d}x.$$
I would like to show that $varphi_f in C_0^infty(mathbb{R})$.
This is a claim made in Chapter 7 of Friedlander's book on distributions. I am having trouble following his justification.
My attempt at a solution:
My idea is to first show that $frac{partial}{partial t} int_{f(x) < t} varphi(x) text{d}x$ is well-defined (i.e., that the derivative exists) and find it's value, with the hope this will give a clue about how to calculate the higher-order derivatives. For $h > 0$:
$$frac{1}{h} left(int_{f(x) < t} varphi(x) text{d}x - int_{f(x) < t + h} varphi(x) text{d}x right) = frac{1}{h} int mathbf{1}_{t le f < t + h }(x) varphi(x) text{d}x.$$
But it's not clear to me that the right side converges to anything as $h to 0^+$. I think maybe a change of variables (made possible by the implicit function Theorem?) is in order to make the dependence on $f$ more explicit.
Hints or solutions are greatly appreciated!
real-analysis pde partial-derivative
edited Dec 20 '18 at 12:11
Falrach
1,664223
asked Aug 7 '17 at 18:11
JZShapiroJZShapiro
1,99511125
$endgroup$
1
$begingroup$
If you apply the coarea formula with $g = varphi / |nabla f|$, you can at least find a nice formula for $varphi_f$.
$endgroup$
– Michał Miśkiewicz
Aug 7 '17 at 19:57
add a comment |
$begingroup$
Let $X$ be an open subset of $mathbb{R}^n$. Suppose that $f in C^infty(X)$ is real-valued, with the property that $nabla f(x) neq 0$ for all $x in X$. For any $varphi in C_0^infty(X)$, and $t in mathbb{R}$, define
$$varphi_f(t) = frac{partial}{partial t} int_{f(x) < t} varphi(x) text{d}x.$$
I would like to show that $varphi_f in C_0^infty(mathbb{R})$.
This is a claim made in Chapter 7 of Friedlander's book on distributions. I am having trouble following his justification.
My attempt at a solution:
My idea is to first show that $frac{partial}{partial t} int_{f(x) < t} varphi(x) text{d}x$ is well-defined (i.e., that the derivative exists) and find it's value, with the hope this will give a clue about how to calculate the higher-order derivatives. For $h > 0$:
$$frac{1}{h} left(int_{f(x) < t} varphi(x) text{d}x - int_{f(x) < t + h} varphi(x) text{d}x right) = frac{1}{h} int mathbf{1}_{t le f < t + h }(x) varphi(x) text{d}x.$$
But it's not clear to me that the right side converges to anything as $h to 0^+$. I think maybe a change of variables (made possible by the implicit function Theorem?) is in order to make the dependence on $f$ more explicit.
Hints or solutions are greatly appreciated!
real-analysis pde partial-derivative
edited Dec 20 '18 at 12:11
Falrach
1,664223
asked Aug 7 '17 at 18:11
JZShapiroJZShapiro
1,99511125
$endgroup$
Let $X$ be an open subset of $mathbb{R}^n$. Suppose that $f in C^infty(X)$ is real-valued, with the property that $nabla f(x) neq 0$ for all $x in X$. For any $varphi in C_0^infty(X)$, and $t in mathbb{R}$, define
$$varphi_f(t) = frac{partial}{partial t} int_{f(x) < t} varphi(x) text{d}x.$$
I would like to show that $varphi_f in C_0^infty(mathbb{R})$.
This is a claim made in Chapter 7 of Friedlander's book on distributions. I am having trouble following his justification.
My attempt at a solution:
My idea is to first show that $frac{partial}{partial t} int_{f(x) < t} varphi(x) text{d}x$ is well-defined (i.e., that the derivative exists) and find it's value, with the hope this will give a clue about how to calculate the higher-order derivatives. For $h > 0$:
$$frac{1}{h} left(int_{f(x) < t} varphi(x) text{d}x - int_{f(x) < t + h} varphi(x) text{d}x right) = frac{1}{h} int mathbf{1}_{t le f < t + h }(x) varphi(x) text{d}x.$$
But it's not clear to me that the right side converges to anything as $h to 0^+$. I think maybe a change of variables (made possible by the implicit function Theorem?) is in order to make the dependence on $f$ more explicit.
Hints or solutions are greatly appreciated!
real-analysis pde partial-derivative
real-analysis pde partial-derivative
edited Dec 20 '18 at 12:11
Falrach
1,664223
asked Aug 7 '17 at 18:11
JZShapiroJZShapiro
1,99511125
edited Dec 20 '18 at 12:11
Falrach
1,664223
asked Aug 7 '17 at 18:11
JZShapiroJZShapiro
1,99511125
edited Dec 20 '18 at 12:11
Falrach
1,664223
edited Dec 20 '18 at 12:11
Falrach
1,664223
edited Dec 20 '18 at 12:11
Falrach
1,664223
1,664223
asked Aug 7 '17 at 18:11
JZShapiroJZShapiro
1,99511125
asked Aug 7 '17 at 18:11
JZShapiroJZShapiro
1,99511125
asked Aug 7 '17 at 18:11
JZShapiroJZShapiro
1,99511125
1,99511125
1
$begingroup$
If you apply the coarea formula with $g = varphi / |nabla f|$, you can at least find a nice formula for $varphi_f$.
$endgroup$
– Michał Miśkiewicz
Aug 7 '17 at 19:57
add a comment |
1
$begingroup$
If you apply the coarea formula with $g = varphi / |nabla f|$, you can at least find a nice formula for $varphi_f$.
$endgroup$
– Michał Miśkiewicz
Aug 7 '17 at 19:57
1
1
$begingroup$
If you apply the coarea formula with $g = varphi / |nabla f|$, you can at least find a nice formula for $varphi_f$.
$endgroup$
– Michał Miśkiewicz
Aug 7 '17 at 19:57
$begingroup$
If you apply the coarea formula with $g = varphi / |nabla f|$, you can at least find a nice formula for $varphi_f$.
$endgroup$
– Michał Miśkiewicz
Aug 7 '17 at 19:57
add a comment |
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$begingroup$
If you apply the coarea formula with $g = varphi / |nabla f|$, you can at least find a nice formula for $varphi_f$.
$endgroup$
– Michał Miśkiewicz
Aug 7 '17 at 19:57