Show that the function $f$ is a tempered distribution
$begingroup$
Let $f$ be a locally integrable function on $mathbb{R}^n$ such that $f$ is of polynomial growth at infinity. Prove that $f$ is a tempered distribution.
Here the distribution is defined as $f(phi)=int_{mathbb{R}^n}fphi.$
I think that I have no reason to be confused of this problem, but I am suddenly stuck.
The linearity is immediate, but I'm figuring out to prove the continuity.
The continuity is equivalent to proving that $int fphi_kto 0$ whenever $phi_kto 0$ in the Schwarz class $mathcal{S}(mathbb{R}^n)$, i.e. the sequence of functions $phi_k$ itself and all of the sequences of partial derivatives converge uniformly to zero.
Of course I can insert the limit into the integral by uniform convergence if the integral is done in a bounded set, but $mathbb{R}^n$ is unbounded, so I'm finding another way.
Another way I'm considering is to apply DCT, but how should I find an integrable majorant of ${f_n}$?
Thanks in advance!
functional-analysis distribution-theory schwartz-space
edited Dec 31 '18 at 21:14
Davide Giraudo
128k17154268
asked Dec 31 '18 at 12:45
bellcirclebellcircle
1,373411
$endgroup$
add a comment |
$begingroup$
Let $f$ be a locally integrable function on $mathbb{R}^n$ such that $f$ is of polynomial growth at infinity. Prove that $f$ is a tempered distribution.
Here the distribution is defined as $f(phi)=int_{mathbb{R}^n}fphi.$
I think that I have no reason to be confused of this problem, but I am suddenly stuck.
The linearity is immediate, but I'm figuring out to prove the continuity.
The continuity is equivalent to proving that $int fphi_kto 0$ whenever $phi_kto 0$ in the Schwarz class $mathcal{S}(mathbb{R}^n)$, i.e. the sequence of functions $phi_k$ itself and all of the sequences of partial derivatives converge uniformly to zero.
Of course I can insert the limit into the integral by uniform convergence if the integral is done in a bounded set, but $mathbb{R}^n$ is unbounded, so I'm finding another way.
Another way I'm considering is to apply DCT, but how should I find an integrable majorant of ${f_n}$?
Thanks in advance!
functional-analysis distribution-theory schwartz-space
edited Dec 31 '18 at 21:14
Davide Giraudo
128k17154268
asked Dec 31 '18 at 12:45
bellcirclebellcircle
1,373411
$endgroup$
add a comment |
$begingroup$
Let $f$ be a locally integrable function on $mathbb{R}^n$ such that $f$ is of polynomial growth at infinity. Prove that $f$ is a tempered distribution.
Here the distribution is defined as $f(phi)=int_{mathbb{R}^n}fphi.$
I think that I have no reason to be confused of this problem, but I am suddenly stuck.
The linearity is immediate, but I'm figuring out to prove the continuity.
The continuity is equivalent to proving that $int fphi_kto 0$ whenever $phi_kto 0$ in the Schwarz class $mathcal{S}(mathbb{R}^n)$, i.e. the sequence of functions $phi_k$ itself and all of the sequences of partial derivatives converge uniformly to zero.
Of course I can insert the limit into the integral by uniform convergence if the integral is done in a bounded set, but $mathbb{R}^n$ is unbounded, so I'm finding another way.
Another way I'm considering is to apply DCT, but how should I find an integrable majorant of ${f_n}$?
Thanks in advance!
functional-analysis distribution-theory schwartz-space
edited Dec 31 '18 at 21:14
Davide Giraudo
128k17154268
asked Dec 31 '18 at 12:45
bellcirclebellcircle
1,373411
$endgroup$
Let $f$ be a locally integrable function on $mathbb{R}^n$ such that $f$ is of polynomial growth at infinity. Prove that $f$ is a tempered distribution.
Here the distribution is defined as $f(phi)=int_{mathbb{R}^n}fphi.$
I think that I have no reason to be confused of this problem, but I am suddenly stuck.
The linearity is immediate, but I'm figuring out to prove the continuity.
The continuity is equivalent to proving that $int fphi_kto 0$ whenever $phi_kto 0$ in the Schwarz class $mathcal{S}(mathbb{R}^n)$, i.e. the sequence of functions $phi_k$ itself and all of the sequences of partial derivatives converge uniformly to zero.
Of course I can insert the limit into the integral by uniform convergence if the integral is done in a bounded set, but $mathbb{R}^n$ is unbounded, so I'm finding another way.
Another way I'm considering is to apply DCT, but how should I find an integrable majorant of ${f_n}$?
Thanks in advance!
functional-analysis distribution-theory schwartz-space
functional-analysis distribution-theory schwartz-space
edited Dec 31 '18 at 21:14
Davide Giraudo
128k17154268
asked Dec 31 '18 at 12:45
bellcirclebellcircle
1,373411
edited Dec 31 '18 at 21:14
Davide Giraudo
128k17154268
asked Dec 31 '18 at 12:45
bellcirclebellcircle
1,373411
edited Dec 31 '18 at 21:14
Davide Giraudo
128k17154268
edited Dec 31 '18 at 21:14
Davide Giraudo
128k17154268
edited Dec 31 '18 at 21:14
Davide Giraudo
128k17154268
128k17154268
asked Dec 31 '18 at 12:45
bellcirclebellcircle
1,373411
asked Dec 31 '18 at 12:45
bellcirclebellcircle
1,373411
asked Dec 31 '18 at 12:45
bellcirclebellcircle
1,373411
1,373411
add a comment |
add a comment |
1 Answer
1
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oldest
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$begingroup$
Actually, convergence to zero in the Schwartz space refers to something stronger, namely, that for all $alpha, betainmathbb N^n$,
$$
lim_{kto +infty}sup_{xinmathbb R^n}leftlvert x^alpha left(D^{beta}phi_kright)(x)rightrvert=0.
$$
All we need is that for all $pinmathbb N$,
$$
lim_{kto +infty}sup_{xinmathbb R^n}leftlVert xrightrVert^p leftlvert phi_k (x)rightrvert=0.
$$
Indeed, from the polynomial growth assumption, we know that there exists $qin mathbb N $ and $cgt 0$ such that for all $xinmathbb R^n$, $leftlvert f(x)rightrvertleqslant cleft(1+leftlVert x rightrVert^qright)$. Therefore,
$$
leftlvert int_{mathbb R^n}
fphi_k rightrvertleqslant cint_{mathbb R^n} left(1+leftlVert x rightrVert^qright)leftlVert xrightrVert^{-p} leftlVert xrightrVert^p leftlvert phi_k (x)rightrvertmathrm dx.
$$
It thus suffices to choose $p$ such that the integral $int_{mathbb R^n} left(1+leftlVert x rightrVert^qright)leftlVert xrightrVert^{-p}mathrm dx$ is finite.
answered Dec 31 '18 at 14:35
Davide GiraudoDavide Giraudo
128k17154268
$endgroup$
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Actually, convergence to zero in the Schwartz space refers to something stronger, namely, that for all $alpha, betainmathbb N^n$,
$$
lim_{kto +infty}sup_{xinmathbb R^n}leftlvert x^alpha left(D^{beta}phi_kright)(x)rightrvert=0.
$$
All we need is that for all $pinmathbb N$,
$$
lim_{kto +infty}sup_{xinmathbb R^n}leftlVert xrightrVert^p leftlvert phi_k (x)rightrvert=0.
$$
Indeed, from the polynomial growth assumption, we know that there exists $qin mathbb N $ and $cgt 0$ such that for all $xinmathbb R^n$, $leftlvert f(x)rightrvertleqslant cleft(1+leftlVert x rightrVert^qright)$. Therefore,
$$
leftlvert int_{mathbb R^n}
fphi_k rightrvertleqslant cint_{mathbb R^n} left(1+leftlVert x rightrVert^qright)leftlVert xrightrVert^{-p} leftlVert xrightrVert^p leftlvert phi_k (x)rightrvertmathrm dx.
$$
It thus suffices to choose $p$ such that the integral $int_{mathbb R^n} left(1+leftlVert x rightrVert^qright)leftlVert xrightrVert^{-p}mathrm dx$ is finite.
answered Dec 31 '18 at 14:35
Davide GiraudoDavide Giraudo
128k17154268
$endgroup$
add a comment |
$begingroup$
Actually, convergence to zero in the Schwartz space refers to something stronger, namely, that for all $alpha, betainmathbb N^n$,
$$
lim_{kto +infty}sup_{xinmathbb R^n}leftlvert x^alpha left(D^{beta}phi_kright)(x)rightrvert=0.
$$
All we need is that for all $pinmathbb N$,
$$
lim_{kto +infty}sup_{xinmathbb R^n}leftlVert xrightrVert^p leftlvert phi_k (x)rightrvert=0.
$$
Indeed, from the polynomial growth assumption, we know that there exists $qin mathbb N $ and $cgt 0$ such that for all $xinmathbb R^n$, $leftlvert f(x)rightrvertleqslant cleft(1+leftlVert x rightrVert^qright)$. Therefore,
$$
leftlvert int_{mathbb R^n}
fphi_k rightrvertleqslant cint_{mathbb R^n} left(1+leftlVert x rightrVert^qright)leftlVert xrightrVert^{-p} leftlVert xrightrVert^p leftlvert phi_k (x)rightrvertmathrm dx.
$$
It thus suffices to choose $p$ such that the integral $int_{mathbb R^n} left(1+leftlVert x rightrVert^qright)leftlVert xrightrVert^{-p}mathrm dx$ is finite.
answered Dec 31 '18 at 14:35
Davide GiraudoDavide Giraudo
128k17154268
$endgroup$
add a comment |
$begingroup$
Actually, convergence to zero in the Schwartz space refers to something stronger, namely, that for all $alpha, betainmathbb N^n$,
$$
lim_{kto +infty}sup_{xinmathbb R^n}leftlvert x^alpha left(D^{beta}phi_kright)(x)rightrvert=0.
$$
All we need is that for all $pinmathbb N$,
$$
lim_{kto +infty}sup_{xinmathbb R^n}leftlVert xrightrVert^p leftlvert phi_k (x)rightrvert=0.
$$
Indeed, from the polynomial growth assumption, we know that there exists $qin mathbb N $ and $cgt 0$ such that for all $xinmathbb R^n$, $leftlvert f(x)rightrvertleqslant cleft(1+leftlVert x rightrVert^qright)$. Therefore,
$$
leftlvert int_{mathbb R^n}
fphi_k rightrvertleqslant cint_{mathbb R^n} left(1+leftlVert x rightrVert^qright)leftlVert xrightrVert^{-p} leftlVert xrightrVert^p leftlvert phi_k (x)rightrvertmathrm dx.
$$
It thus suffices to choose $p$ such that the integral $int_{mathbb R^n} left(1+leftlVert x rightrVert^qright)leftlVert xrightrVert^{-p}mathrm dx$ is finite.
answered Dec 31 '18 at 14:35
Davide GiraudoDavide Giraudo
128k17154268
$endgroup$
Actually, convergence to zero in the Schwartz space refers to something stronger, namely, that for all $alpha, betainmathbb N^n$,
$$
lim_{kto +infty}sup_{xinmathbb R^n}leftlvert x^alpha left(D^{beta}phi_kright)(x)rightrvert=0.
$$
All we need is that for all $pinmathbb N$,
$$
lim_{kto +infty}sup_{xinmathbb R^n}leftlVert xrightrVert^p leftlvert phi_k (x)rightrvert=0.
$$
Indeed, from the polynomial growth assumption, we know that there exists $qin mathbb N $ and $cgt 0$ such that for all $xinmathbb R^n$, $leftlvert f(x)rightrvertleqslant cleft(1+leftlVert x rightrVert^qright)$. Therefore,
$$
leftlvert int_{mathbb R^n}
fphi_k rightrvertleqslant cint_{mathbb R^n} left(1+leftlVert x rightrVert^qright)leftlVert xrightrVert^{-p} leftlVert xrightrVert^p leftlvert phi_k (x)rightrvertmathrm dx.
$$
It thus suffices to choose $p$ such that the integral $int_{mathbb R^n} left(1+leftlVert x rightrVert^qright)leftlVert xrightrVert^{-p}mathrm dx$ is finite.
answered Dec 31 '18 at 14:35
Davide GiraudoDavide Giraudo
128k17154268
answered Dec 31 '18 at 14:35
Davide GiraudoDavide Giraudo
128k17154268
answered Dec 31 '18 at 14:35
Davide GiraudoDavide Giraudo
128k17154268
answered Dec 31 '18 at 14:35
Davide GiraudoDavide Giraudo
128k17154268
128k17154268
add a comment |
add a comment |
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