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Convolution of locally integrable and compactly supported infinitely differentiable function
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A function $f$ on $mathbbR$ is called locally integrable if $f$ is integrable on
every bounded interval $[a, b]$ for $a < b$ in $mathbbR$. Show that if $g in C_c^infty(mathbbR)$
and $f$ is locally integrable, then $$f∗g(y)=int_-infty^inftyf(t)g(y-t)dt$$ exists and is infinitely
differentiable on $mathbbR$.
I feel pretty much stuck from the beginning. After taking modulus using triangle inequality I was trying to bound things individually but it didn't work out.
Please help.
real-analysis convolution
edited Jul 22 at 7:21
Travis
55.7k764137
asked Jul 22 at 7:04
user101
564213
add a comment |Â
up vote
2
down vote
favorite
A function $f$ on $mathbbR$ is called locally integrable if $f$ is integrable on
every bounded interval $[a, b]$ for $a < b$ in $mathbbR$. Show that if $g in C_c^infty(mathbbR)$
and $f$ is locally integrable, then $$f∗g(y)=int_-infty^inftyf(t)g(y-t)dt$$ exists and is infinitely
differentiable on $mathbbR$.
I feel pretty much stuck from the beginning. After taking modulus using triangle inequality I was trying to bound things individually but it didn't work out.
Please help.
real-analysis convolution
edited Jul 22 at 7:21
Travis
55.7k764137
asked Jul 22 at 7:04
user101
564213
2
g has compact support (and so it is bounded). this allows you to easily see that the convolution is well defined since for is locally integrable. to show that this is infinitely differentiable look at the difference quotient and see if you can pass the limit through the integral (pass the derivatives onto get instead of f!)
– yousuf soliman
Jul 22 at 7:12
I missed the subscript $c$ in $C_c^infty(Bbb R)$.
– Kenny Lau
Jul 22 at 7:13
@yousufsoliman The subscript means it's having a compact support?
– user101
Jul 22 at 7:24
@mathjack51 that is correct
– yousuf soliman
Jul 22 at 7:25
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
A function $f$ on $mathbbR$ is called locally integrable if $f$ is integrable on
every bounded interval $[a, b]$ for $a < b$ in $mathbbR$. Show that if $g in C_c^infty(mathbbR)$
and $f$ is locally integrable, then $$f∗g(y)=int_-infty^inftyf(t)g(y-t)dt$$ exists and is infinitely
differentiable on $mathbbR$.
I feel pretty much stuck from the beginning. After taking modulus using triangle inequality I was trying to bound things individually but it didn't work out.
Please help.
real-analysis convolution
edited Jul 22 at 7:21
Travis
55.7k764137
asked Jul 22 at 7:04
user101
564213
A function $f$ on $mathbbR$ is called locally integrable if $f$ is integrable on
every bounded interval $[a, b]$ for $a < b$ in $mathbbR$. Show that if $g in C_c^infty(mathbbR)$
and $f$ is locally integrable, then $$f∗g(y)=int_-infty^inftyf(t)g(y-t)dt$$ exists and is infinitely
differentiable on $mathbbR$.
I feel pretty much stuck from the beginning. After taking modulus using triangle inequality I was trying to bound things individually but it didn't work out.
Please help.
real-analysis convolution
edited Jul 22 at 7:21
Travis
55.7k764137
asked Jul 22 at 7:04
user101
564213
edited Jul 22 at 7:21
Travis
55.7k764137
edited Jul 22 at 7:21
Travis
55.7k764137
edited Jul 22 at 7:21
Travis
55.7k764137
55.7k764137
asked Jul 22 at 7:04
user101
564213
asked Jul 22 at 7:04
user101
564213
asked Jul 22 at 7:04
user101
564213
564213
2
g has compact support (and so it is bounded). this allows you to easily see that the convolution is well defined since for is locally integrable. to show that this is infinitely differentiable look at the difference quotient and see if you can pass the limit through the integral (pass the derivatives onto get instead of f!)
– yousuf soliman
Jul 22 at 7:12
I missed the subscript $c$ in $C_c^infty(Bbb R)$.
– Kenny Lau
Jul 22 at 7:13
@yousufsoliman The subscript means it's having a compact support?
– user101
Jul 22 at 7:24
@mathjack51 that is correct
– yousuf soliman
Jul 22 at 7:25
add a comment |Â
2
g has compact support (and so it is bounded). this allows you to easily see that the convolution is well defined since for is locally integrable. to show that this is infinitely differentiable look at the difference quotient and see if you can pass the limit through the integral (pass the derivatives onto get instead of f!)
– yousuf soliman
Jul 22 at 7:12
I missed the subscript $c$ in $C_c^infty(Bbb R)$.
– Kenny Lau
Jul 22 at 7:13
@yousufsoliman The subscript means it's having a compact support?
– user101
Jul 22 at 7:24
@mathjack51 that is correct
– yousuf soliman
Jul 22 at 7:25
2
2
g has compact support (and so it is bounded). this allows you to easily see that the convolution is well defined since for is locally integrable. to show that this is infinitely differentiable look at the difference quotient and see if you can pass the limit through the integral (pass the derivatives onto get instead of f!)
– yousuf soliman
Jul 22 at 7:12
g has compact support (and so it is bounded). this allows you to easily see that the convolution is well defined since for is locally integrable. to show that this is infinitely differentiable look at the difference quotient and see if you can pass the limit through the integral (pass the derivatives onto get instead of f!)
– yousuf soliman
Jul 22 at 7:12
I missed the subscript $c$ in $C_c^infty(Bbb R)$.
– Kenny Lau
Jul 22 at 7:13
I missed the subscript $c$ in $C_c^infty(Bbb R)$.
– Kenny Lau
Jul 22 at 7:13
@yousufsoliman The subscript means it's having a compact support?
– user101
Jul 22 at 7:24
@yousufsoliman The subscript means it's having a compact support?
– user101
Jul 22 at 7:24
@mathjack51 that is correct
– yousuf soliman
Jul 22 at 7:25
@mathjack51 that is correct
– yousuf soliman
Jul 22 at 7:25
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
The comment of yousuf soliman is exactly right. Since I've been wanting to write the details of this computation for myself, I'll give them here. Just for some context, some typical examples of unbounded, locally (but not globally) integrable $f$'s are $f(y) = |y|^-p$ for $0< p< 1$ and $f(y) = |log |y||$; the singularities of these functions at the origin are mild enough that they can be integrated over.
Here are the details as outlined by yousef. Fix $gin C_c^infty(mathbb R)$ and choose $R>0$ such that the support of $g$ is contained in $[-R, R]$. For each $yin mathbb R$ the support of the function $tmapsto g(y - t)$ is contained in $[y - R, y+ R]$. First, I'll verify that the convolution is pointwise well-defined on $mathbb R$. For $yin mathbb R$ fixed, we have
begineqnarray*
left|f*g(y)right|
& = &
left|int_mathbb R f(t) g(y - t); rm d tright|\
& = &
left|int_y - R^y + R f(t) g(y - t); rm d tright|\
& leq &
int_y - R^y + R |f(t)|; |g(y - t)|; rm d t\
& leq &
max_xin mathbb R|g(x)|int_y - R^y + R |f(t)|; rm dt\
& < &
infty.
endeqnarray*
Next, I'll use the Dominated Convergence Theorem to show that $f*g$ is differentiable and that the derivative of $f*g$ is equal to the convolution of $f$ with $g'$. With $R$ as above, let $0< |h|< fracR100$ so that for each $yin mathbb R$, the supports of both of the functions $tmapsto g(y - t)$ and $tmapsto g(y + h- t)$ are contained in the interval $[y - 2R, y + 2R]$. Then
begineqnarray*
fracf*g(y + h) - f*g(y)h
& = &
int_mathbb Rf(t)fracg(y + h- t) - g(y- t)h; rm dt
\
& = &
int_y - 2R^y + 2Rf(t)fracg(y + h- t) - g(y- t)h; rm dt.
endeqnarray*
Since $f$ is integrable on the interval $[y - 2R, y+ 2R]$, we have $|f(t)|< infty$ for a.e. $tin [y - 2R, y + 2R]$. Therefore, for a.e. $tin [y - 2R, y + 2R]$ we have
beginequation*
lim_hto 0f(t)fracg(y + h- t) - g(y- t)h = f(t) g'(y - t).
endequation*
It remains to find a function $phi(t)$, integrable over $[y- 2R, y + 2R]$ for which
beginequation*
left|f(t)fracg(y + h - t) - g(y- t)hright| leq phi(t)
endequation*
for a.e. $tin [y - 2R, y+ 2R]$. By the Mean Value Theorem we have
begineqnarray*
left|g(y + h - t) - g(y - t)right|
& = &
left|
int_0^1fracrm drm ds g(x - t + sh); rm d s
right|
\
&leq &
|h|max_xin mathbb R |g'(x)|.
endeqnarray*
As a consequence, for all $0< |h| < fracR100$,
beginequation*
left|f(t) fracg(y + h -t) - g(y - t)hright|
leq
(max_xin mathbb R |g'(x)|)|f(t)|
endequation*
with the function on the right-hand side integrable over $[y - 2R, y+ 2R]$. Finally, the Dominated Convergence Theorem gives
begineqnarray*
lim_hto 0fracf*g(y + h) - f*g(y)h
& = &
int_y - 2R^y + 2R lim_hto 0 f(t) fracg(y + h - t) - g(y - t)h ; rm dt
\
& = &
int_y - 2R^y + 2R f(t)g'(y - t); rm dt
\
& = &
int_mathbb R f(t) g'(y - t); rm dt
\
& = &
f*g'(y).
endeqnarray*
Finally, to see that $f*g$ is infinitely differentiable, one may proceed by induction on the order of differentiability of $f*g$. If $f*g$ is $k$-times differentiable with $(f*g)^(k) = f*(g^(k))$ then apply the above argument with $g$ replaced by $g^(k)$ to find that $f*g$ is $(k + 1)$-times differentiable with $(f*g)^(k+1) = f*(g^(k+1))$.
answered Jul 22 at 14:48
BindersFull
498110
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The comment of yousuf soliman is exactly right. Since I've been wanting to write the details of this computation for myself, I'll give them here. Just for some context, some typical examples of unbounded, locally (but not globally) integrable $f$'s are $f(y) = |y|^-p$ for $0< p< 1$ and $f(y) = |log |y||$; the singularities of these functions at the origin are mild enough that they can be integrated over.
Here are the details as outlined by yousef. Fix $gin C_c^infty(mathbb R)$ and choose $R>0$ such that the support of $g$ is contained in $[-R, R]$. For each $yin mathbb R$ the support of the function $tmapsto g(y - t)$ is contained in $[y - R, y+ R]$. First, I'll verify that the convolution is pointwise well-defined on $mathbb R$. For $yin mathbb R$ fixed, we have
begineqnarray*
left|f*g(y)right|
& = &
left|int_mathbb R f(t) g(y - t); rm d tright|\
& = &
left|int_y - R^y + R f(t) g(y - t); rm d tright|\
& leq &
int_y - R^y + R |f(t)|; |g(y - t)|; rm d t\
& leq &
max_xin mathbb R|g(x)|int_y - R^y + R |f(t)|; rm dt\
& < &
infty.
endeqnarray*
Next, I'll use the Dominated Convergence Theorem to show that $f*g$ is differentiable and that the derivative of $f*g$ is equal to the convolution of $f$ with $g'$. With $R$ as above, let $0< |h|< fracR100$ so that for each $yin mathbb R$, the supports of both of the functions $tmapsto g(y - t)$ and $tmapsto g(y + h- t)$ are contained in the interval $[y - 2R, y + 2R]$. Then
begineqnarray*
fracf*g(y + h) - f*g(y)h
& = &
int_mathbb Rf(t)fracg(y + h- t) - g(y- t)h; rm dt
\
& = &
int_y - 2R^y + 2Rf(t)fracg(y + h- t) - g(y- t)h; rm dt.
endeqnarray*
Since $f$ is integrable on the interval $[y - 2R, y+ 2R]$, we have $|f(t)|< infty$ for a.e. $tin [y - 2R, y + 2R]$. Therefore, for a.e. $tin [y - 2R, y + 2R]$ we have
beginequation*
lim_hto 0f(t)fracg(y + h- t) - g(y- t)h = f(t) g'(y - t).
endequation*
It remains to find a function $phi(t)$, integrable over $[y- 2R, y + 2R]$ for which
beginequation*
left|f(t)fracg(y + h - t) - g(y- t)hright| leq phi(t)
endequation*
for a.e. $tin [y - 2R, y+ 2R]$. By the Mean Value Theorem we have
begineqnarray*
left|g(y + h - t) - g(y - t)right|
& = &
left|
int_0^1fracrm drm ds g(x - t + sh); rm d s
right|
\
&leq &
|h|max_xin mathbb R |g'(x)|.
endeqnarray*
As a consequence, for all $0< |h| < fracR100$,
beginequation*
left|f(t) fracg(y + h -t) - g(y - t)hright|
leq
(max_xin mathbb R |g'(x)|)|f(t)|
endequation*
with the function on the right-hand side integrable over $[y - 2R, y+ 2R]$. Finally, the Dominated Convergence Theorem gives
begineqnarray*
lim_hto 0fracf*g(y + h) - f*g(y)h
& = &
int_y - 2R^y + 2R lim_hto 0 f(t) fracg(y + h - t) - g(y - t)h ; rm dt
\
& = &
int_y - 2R^y + 2R f(t)g'(y - t); rm dt
\
& = &
int_mathbb R f(t) g'(y - t); rm dt
\
& = &
f*g'(y).
endeqnarray*
Finally, to see that $f*g$ is infinitely differentiable, one may proceed by induction on the order of differentiability of $f*g$. If $f*g$ is $k$-times differentiable with $(f*g)^(k) = f*(g^(k))$ then apply the above argument with $g$ replaced by $g^(k)$ to find that $f*g$ is $(k + 1)$-times differentiable with $(f*g)^(k+1) = f*(g^(k+1))$.
answered Jul 22 at 14:48
BindersFull
498110
add a comment |Â
up vote
1
down vote
accepted
The comment of yousuf soliman is exactly right. Since I've been wanting to write the details of this computation for myself, I'll give them here. Just for some context, some typical examples of unbounded, locally (but not globally) integrable $f$'s are $f(y) = |y|^-p$ for $0< p< 1$ and $f(y) = |log |y||$; the singularities of these functions at the origin are mild enough that they can be integrated over.
Here are the details as outlined by yousef. Fix $gin C_c^infty(mathbb R)$ and choose $R>0$ such that the support of $g$ is contained in $[-R, R]$. For each $yin mathbb R$ the support of the function $tmapsto g(y - t)$ is contained in $[y - R, y+ R]$. First, I'll verify that the convolution is pointwise well-defined on $mathbb R$. For $yin mathbb R$ fixed, we have
begineqnarray*
left|f*g(y)right|
& = &
left|int_mathbb R f(t) g(y - t); rm d tright|\
& = &
left|int_y - R^y + R f(t) g(y - t); rm d tright|\
& leq &
int_y - R^y + R |f(t)|; |g(y - t)|; rm d t\
& leq &
max_xin mathbb R|g(x)|int_y - R^y + R |f(t)|; rm dt\
& < &
infty.
endeqnarray*
Next, I'll use the Dominated Convergence Theorem to show that $f*g$ is differentiable and that the derivative of $f*g$ is equal to the convolution of $f$ with $g'$. With $R$ as above, let $0< |h|< fracR100$ so that for each $yin mathbb R$, the supports of both of the functions $tmapsto g(y - t)$ and $tmapsto g(y + h- t)$ are contained in the interval $[y - 2R, y + 2R]$. Then
begineqnarray*
fracf*g(y + h) - f*g(y)h
& = &
int_mathbb Rf(t)fracg(y + h- t) - g(y- t)h; rm dt
\
& = &
int_y - 2R^y + 2Rf(t)fracg(y + h- t) - g(y- t)h; rm dt.
endeqnarray*
Since $f$ is integrable on the interval $[y - 2R, y+ 2R]$, we have $|f(t)|< infty$ for a.e. $tin [y - 2R, y + 2R]$. Therefore, for a.e. $tin [y - 2R, y + 2R]$ we have
beginequation*
lim_hto 0f(t)fracg(y + h- t) - g(y- t)h = f(t) g'(y - t).
endequation*
It remains to find a function $phi(t)$, integrable over $[y- 2R, y + 2R]$ for which
beginequation*
left|f(t)fracg(y + h - t) - g(y- t)hright| leq phi(t)
endequation*
for a.e. $tin [y - 2R, y+ 2R]$. By the Mean Value Theorem we have
begineqnarray*
left|g(y + h - t) - g(y - t)right|
& = &
left|
int_0^1fracrm drm ds g(x - t + sh); rm d s
right|
\
&leq &
|h|max_xin mathbb R |g'(x)|.
endeqnarray*
As a consequence, for all $0< |h| < fracR100$,
beginequation*
left|f(t) fracg(y + h -t) - g(y - t)hright|
leq
(max_xin mathbb R |g'(x)|)|f(t)|
endequation*
with the function on the right-hand side integrable over $[y - 2R, y+ 2R]$. Finally, the Dominated Convergence Theorem gives
begineqnarray*
lim_hto 0fracf*g(y + h) - f*g(y)h
& = &
int_y - 2R^y + 2R lim_hto 0 f(t) fracg(y + h - t) - g(y - t)h ; rm dt
\
& = &
int_y - 2R^y + 2R f(t)g'(y - t); rm dt
\
& = &
int_mathbb R f(t) g'(y - t); rm dt
\
& = &
f*g'(y).
endeqnarray*
Finally, to see that $f*g$ is infinitely differentiable, one may proceed by induction on the order of differentiability of $f*g$. If $f*g$ is $k$-times differentiable with $(f*g)^(k) = f*(g^(k))$ then apply the above argument with $g$ replaced by $g^(k)$ to find that $f*g$ is $(k + 1)$-times differentiable with $(f*g)^(k+1) = f*(g^(k+1))$.
answered Jul 22 at 14:48
BindersFull
498110
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The comment of yousuf soliman is exactly right. Since I've been wanting to write the details of this computation for myself, I'll give them here. Just for some context, some typical examples of unbounded, locally (but not globally) integrable $f$'s are $f(y) = |y|^-p$ for $0< p< 1$ and $f(y) = |log |y||$; the singularities of these functions at the origin are mild enough that they can be integrated over.
Here are the details as outlined by yousef. Fix $gin C_c^infty(mathbb R)$ and choose $R>0$ such that the support of $g$ is contained in $[-R, R]$. For each $yin mathbb R$ the support of the function $tmapsto g(y - t)$ is contained in $[y - R, y+ R]$. First, I'll verify that the convolution is pointwise well-defined on $mathbb R$. For $yin mathbb R$ fixed, we have
begineqnarray*
left|f*g(y)right|
& = &
left|int_mathbb R f(t) g(y - t); rm d tright|\
& = &
left|int_y - R^y + R f(t) g(y - t); rm d tright|\
& leq &
int_y - R^y + R |f(t)|; |g(y - t)|; rm d t\
& leq &
max_xin mathbb R|g(x)|int_y - R^y + R |f(t)|; rm dt\
& < &
infty.
endeqnarray*
Next, I'll use the Dominated Convergence Theorem to show that $f*g$ is differentiable and that the derivative of $f*g$ is equal to the convolution of $f$ with $g'$. With $R$ as above, let $0< |h|< fracR100$ so that for each $yin mathbb R$, the supports of both of the functions $tmapsto g(y - t)$ and $tmapsto g(y + h- t)$ are contained in the interval $[y - 2R, y + 2R]$. Then
begineqnarray*
fracf*g(y + h) - f*g(y)h
& = &
int_mathbb Rf(t)fracg(y + h- t) - g(y- t)h; rm dt
\
& = &
int_y - 2R^y + 2Rf(t)fracg(y + h- t) - g(y- t)h; rm dt.
endeqnarray*
Since $f$ is integrable on the interval $[y - 2R, y+ 2R]$, we have $|f(t)|< infty$ for a.e. $tin [y - 2R, y + 2R]$. Therefore, for a.e. $tin [y - 2R, y + 2R]$ we have
beginequation*
lim_hto 0f(t)fracg(y + h- t) - g(y- t)h = f(t) g'(y - t).
endequation*
It remains to find a function $phi(t)$, integrable over $[y- 2R, y + 2R]$ for which
beginequation*
left|f(t)fracg(y + h - t) - g(y- t)hright| leq phi(t)
endequation*
for a.e. $tin [y - 2R, y+ 2R]$. By the Mean Value Theorem we have
begineqnarray*
left|g(y + h - t) - g(y - t)right|
& = &
left|
int_0^1fracrm drm ds g(x - t + sh); rm d s
right|
\
&leq &
|h|max_xin mathbb R |g'(x)|.
endeqnarray*
As a consequence, for all $0< |h| < fracR100$,
beginequation*
left|f(t) fracg(y + h -t) - g(y - t)hright|
leq
(max_xin mathbb R |g'(x)|)|f(t)|
endequation*
with the function on the right-hand side integrable over $[y - 2R, y+ 2R]$. Finally, the Dominated Convergence Theorem gives
begineqnarray*
lim_hto 0fracf*g(y + h) - f*g(y)h
& = &
int_y - 2R^y + 2R lim_hto 0 f(t) fracg(y + h - t) - g(y - t)h ; rm dt
\
& = &
int_y - 2R^y + 2R f(t)g'(y - t); rm dt
\
& = &
int_mathbb R f(t) g'(y - t); rm dt
\
& = &
f*g'(y).
endeqnarray*
Finally, to see that $f*g$ is infinitely differentiable, one may proceed by induction on the order of differentiability of $f*g$. If $f*g$ is $k$-times differentiable with $(f*g)^(k) = f*(g^(k))$ then apply the above argument with $g$ replaced by $g^(k)$ to find that $f*g$ is $(k + 1)$-times differentiable with $(f*g)^(k+1) = f*(g^(k+1))$.
answered Jul 22 at 14:48
BindersFull
498110
The comment of yousuf soliman is exactly right. Since I've been wanting to write the details of this computation for myself, I'll give them here. Just for some context, some typical examples of unbounded, locally (but not globally) integrable $f$'s are $f(y) = |y|^-p$ for $0< p< 1$ and $f(y) = |log |y||$; the singularities of these functions at the origin are mild enough that they can be integrated over.
Here are the details as outlined by yousef. Fix $gin C_c^infty(mathbb R)$ and choose $R>0$ such that the support of $g$ is contained in $[-R, R]$. For each $yin mathbb R$ the support of the function $tmapsto g(y - t)$ is contained in $[y - R, y+ R]$. First, I'll verify that the convolution is pointwise well-defined on $mathbb R$. For $yin mathbb R$ fixed, we have
begineqnarray*
left|f*g(y)right|
& = &
left|int_mathbb R f(t) g(y - t); rm d tright|\
& = &
left|int_y - R^y + R f(t) g(y - t); rm d tright|\
& leq &
int_y - R^y + R |f(t)|; |g(y - t)|; rm d t\
& leq &
max_xin mathbb R|g(x)|int_y - R^y + R |f(t)|; rm dt\
& < &
infty.
endeqnarray*
Next, I'll use the Dominated Convergence Theorem to show that $f*g$ is differentiable and that the derivative of $f*g$ is equal to the convolution of $f$ with $g'$. With $R$ as above, let $0< |h|< fracR100$ so that for each $yin mathbb R$, the supports of both of the functions $tmapsto g(y - t)$ and $tmapsto g(y + h- t)$ are contained in the interval $[y - 2R, y + 2R]$. Then
begineqnarray*
fracf*g(y + h) - f*g(y)h
& = &
int_mathbb Rf(t)fracg(y + h- t) - g(y- t)h; rm dt
\
& = &
int_y - 2R^y + 2Rf(t)fracg(y + h- t) - g(y- t)h; rm dt.
endeqnarray*
Since $f$ is integrable on the interval $[y - 2R, y+ 2R]$, we have $|f(t)|< infty$ for a.e. $tin [y - 2R, y + 2R]$. Therefore, for a.e. $tin [y - 2R, y + 2R]$ we have
beginequation*
lim_hto 0f(t)fracg(y + h- t) - g(y- t)h = f(t) g'(y - t).
endequation*
It remains to find a function $phi(t)$, integrable over $[y- 2R, y + 2R]$ for which
beginequation*
left|f(t)fracg(y + h - t) - g(y- t)hright| leq phi(t)
endequation*
for a.e. $tin [y - 2R, y+ 2R]$. By the Mean Value Theorem we have
begineqnarray*
left|g(y + h - t) - g(y - t)right|
& = &
left|
int_0^1fracrm drm ds g(x - t + sh); rm d s
right|
\
&leq &
|h|max_xin mathbb R |g'(x)|.
endeqnarray*
As a consequence, for all $0< |h| < fracR100$,
beginequation*
left|f(t) fracg(y + h -t) - g(y - t)hright|
leq
(max_xin mathbb R |g'(x)|)|f(t)|
endequation*
with the function on the right-hand side integrable over $[y - 2R, y+ 2R]$. Finally, the Dominated Convergence Theorem gives
begineqnarray*
lim_hto 0fracf*g(y + h) - f*g(y)h
& = &
int_y - 2R^y + 2R lim_hto 0 f(t) fracg(y + h - t) - g(y - t)h ; rm dt
\
& = &
int_y - 2R^y + 2R f(t)g'(y - t); rm dt
\
& = &
int_mathbb R f(t) g'(y - t); rm dt
\
& = &
f*g'(y).
endeqnarray*
Finally, to see that $f*g$ is infinitely differentiable, one may proceed by induction on the order of differentiability of $f*g$. If $f*g$ is $k$-times differentiable with $(f*g)^(k) = f*(g^(k))$ then apply the above argument with $g$ replaced by $g^(k)$ to find that $f*g$ is $(k + 1)$-times differentiable with $(f*g)^(k+1) = f*(g^(k+1))$.
answered Jul 22 at 14:48
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answered Jul 22 at 14:48
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2
g has compact support (and so it is bounded). this allows you to easily see that the convolution is well defined since for is locally integrable. to show that this is infinitely differentiable look at the difference quotient and see if you can pass the limit through the integral (pass the derivatives onto get instead of f!)
– yousuf soliman
Jul 22 at 7:12
I missed the subscript $c$ in $C_c^infty(Bbb R)$.
– Kenny Lau
Jul 22 at 7:13
@yousufsoliman The subscript means it's having a compact support?
– user101
Jul 22 at 7:24
@mathjack51 that is correct
– yousuf soliman
Jul 22 at 7:25