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Is this estimate true for functionals on Frechet spaces?
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In class the other day my professor made the following claim about the Schwartz class $mathcal S$:
Let $u: mathcal S to mathbb C$ be linear. $u$ is continuous iff there exist $C,N>0$ such that for all $varphi in mathcal S$,
$$|u(varphi)| leq C sup_{|alpha|,|beta| leq N} ||x^alpha D^beta varphi||_{L^infty}$$
where $alpha,beta$ denote multiindices.
(so $||x^alpha D^beta varphi||_{L^infty}$ is just the $(alpha,beta)$th seminorm of $mathcal S$).
He claims this follows immediately from uniform boundedness for Frechet spaces. I'm largely unfamiliar with this area of functional analysis, so it isn't at all clear to me. In particular, I don't see why we'd only need to use finitely many seminorms.
So, is the claim
Let $F$ be a Frechet space with seminorms $||cdot||_j$. A linear map $u: F to mathbb C$ is continuous iff there exist $C,N>0$ such that for all $f in F$,
$$|u(f)| leq Csup_{j leq N} ||f||_j$$
valid, or is this just something true for the Schwartz class? Does it follow from Banach-Steinhaus?
functional-analysis distribution-theory topological-vector-spaces
asked Jan 24 at 0:09
Aidan BackusAidan Backus
12916
$endgroup$
add a comment |
$begingroup$
In class the other day my professor made the following claim about the Schwartz class $mathcal S$:
Let $u: mathcal S to mathbb C$ be linear. $u$ is continuous iff there exist $C,N>0$ such that for all $varphi in mathcal S$,
$$|u(varphi)| leq C sup_{|alpha|,|beta| leq N} ||x^alpha D^beta varphi||_{L^infty}$$
where $alpha,beta$ denote multiindices.
(so $||x^alpha D^beta varphi||_{L^infty}$ is just the $(alpha,beta)$th seminorm of $mathcal S$).
He claims this follows immediately from uniform boundedness for Frechet spaces. I'm largely unfamiliar with this area of functional analysis, so it isn't at all clear to me. In particular, I don't see why we'd only need to use finitely many seminorms.
So, is the claim
Let $F$ be a Frechet space with seminorms $||cdot||_j$. A linear map $u: F to mathbb C$ is continuous iff there exist $C,N>0$ such that for all $f in F$,
$$|u(f)| leq Csup_{j leq N} ||f||_j$$
valid, or is this just something true for the Schwartz class? Does it follow from Banach-Steinhaus?
functional-analysis distribution-theory topological-vector-spaces
asked Jan 24 at 0:09
Aidan BackusAidan Backus
12916
$endgroup$
add a comment |
$begingroup$
In class the other day my professor made the following claim about the Schwartz class $mathcal S$:
Let $u: mathcal S to mathbb C$ be linear. $u$ is continuous iff there exist $C,N>0$ such that for all $varphi in mathcal S$,
$$|u(varphi)| leq C sup_{|alpha|,|beta| leq N} ||x^alpha D^beta varphi||_{L^infty}$$
where $alpha,beta$ denote multiindices.
(so $||x^alpha D^beta varphi||_{L^infty}$ is just the $(alpha,beta)$th seminorm of $mathcal S$).
He claims this follows immediately from uniform boundedness for Frechet spaces. I'm largely unfamiliar with this area of functional analysis, so it isn't at all clear to me. In particular, I don't see why we'd only need to use finitely many seminorms.
So, is the claim
Let $F$ be a Frechet space with seminorms $||cdot||_j$. A linear map $u: F to mathbb C$ is continuous iff there exist $C,N>0$ such that for all $f in F$,
$$|u(f)| leq Csup_{j leq N} ||f||_j$$
valid, or is this just something true for the Schwartz class? Does it follow from Banach-Steinhaus?
functional-analysis distribution-theory topological-vector-spaces
asked Jan 24 at 0:09
Aidan BackusAidan Backus
12916
$endgroup$
In class the other day my professor made the following claim about the Schwartz class $mathcal S$:
Let $u: mathcal S to mathbb C$ be linear. $u$ is continuous iff there exist $C,N>0$ such that for all $varphi in mathcal S$,
$$|u(varphi)| leq C sup_{|alpha|,|beta| leq N} ||x^alpha D^beta varphi||_{L^infty}$$
where $alpha,beta$ denote multiindices.
(so $||x^alpha D^beta varphi||_{L^infty}$ is just the $(alpha,beta)$th seminorm of $mathcal S$).
He claims this follows immediately from uniform boundedness for Frechet spaces. I'm largely unfamiliar with this area of functional analysis, so it isn't at all clear to me. In particular, I don't see why we'd only need to use finitely many seminorms.
So, is the claim
Let $F$ be a Frechet space with seminorms $||cdot||_j$. A linear map $u: F to mathbb C$ is continuous iff there exist $C,N>0$ such that for all $f in F$,
$$|u(f)| leq Csup_{j leq N} ||f||_j$$
valid, or is this just something true for the Schwartz class? Does it follow from Banach-Steinhaus?
functional-analysis distribution-theory topological-vector-spaces
functional-analysis distribution-theory topological-vector-spaces
asked Jan 24 at 0:09
Aidan BackusAidan Backus
12916
asked Jan 24 at 0:09
Aidan BackusAidan Backus
12916
asked Jan 24 at 0:09
Aidan BackusAidan Backus
12916
asked Jan 24 at 0:09
Aidan BackusAidan Backus
12916
asked Jan 24 at 0:09
Aidan BackusAidan Backus
12916
12916
add a comment |
add a comment |
1 Answer
1
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oldest
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The result is true in any locally convex space $F$. It is a consequence of the fact that the topology is given by a family of seminorms.
Since $u$ is continuous at $0$, the set $u^{-1}(B_1(0))$ is open (preimage of the unit disk). So there exists a neighbourhood $V$ of zero, of the form $V={f: |f(x)|_j<varepsilon, j=1,ldots,N}$, with $Vsubset u^{-1}(B_1(0))$. Given any $fin F$, we have
$$
frac{varepsilon f}{2sum_{j=1}^N|f|_j}in V,
$$
so
$$
left|uleft(frac{varepsilon f}{2sum_{j=1}^N|f|_j}right)right|leq1,
$$
which we may write as
$$
|u(f)|leq sum_{j=1}^Nfrac2varepsilon,|f|_jleqfrac{2N}varepsilon,sup_{jleq N}|f|_j.
$$
There is a small exception to the above, which is the case in which $|f|_j=0$ for $j=1,ldots,N$. But in that case $mfin V$ for all $minmathbb N$, so $|u(mf)|leq1$ and so $|u(f)|leq1/m$, forcing $u(f)=0$.
answered Jan 24 at 0:32
Martin ArgeramiMartin Argerami
128k1184184
$endgroup$
$begingroup$
Thank you, this is exactly what I needed!
$endgroup$
– Aidan Backus
Jan 24 at 0:36
add a comment |
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1 Answer
1
active
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votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
The result is true in any locally convex space $F$. It is a consequence of the fact that the topology is given by a family of seminorms.
Since $u$ is continuous at $0$, the set $u^{-1}(B_1(0))$ is open (preimage of the unit disk). So there exists a neighbourhood $V$ of zero, of the form $V={f: |f(x)|_j<varepsilon, j=1,ldots,N}$, with $Vsubset u^{-1}(B_1(0))$. Given any $fin F$, we have
$$
frac{varepsilon f}{2sum_{j=1}^N|f|_j}in V,
$$
so
$$
left|uleft(frac{varepsilon f}{2sum_{j=1}^N|f|_j}right)right|leq1,
$$
which we may write as
$$
|u(f)|leq sum_{j=1}^Nfrac2varepsilon,|f|_jleqfrac{2N}varepsilon,sup_{jleq N}|f|_j.
$$
There is a small exception to the above, which is the case in which $|f|_j=0$ for $j=1,ldots,N$. But in that case $mfin V$ for all $minmathbb N$, so $|u(mf)|leq1$ and so $|u(f)|leq1/m$, forcing $u(f)=0$.
answered Jan 24 at 0:32
Martin ArgeramiMartin Argerami
128k1184184
$endgroup$
$begingroup$
Thank you, this is exactly what I needed!
$endgroup$
– Aidan Backus
Jan 24 at 0:36
add a comment |
$begingroup$
The result is true in any locally convex space $F$. It is a consequence of the fact that the topology is given by a family of seminorms.
Since $u$ is continuous at $0$, the set $u^{-1}(B_1(0))$ is open (preimage of the unit disk). So there exists a neighbourhood $V$ of zero, of the form $V={f: |f(x)|_j<varepsilon, j=1,ldots,N}$, with $Vsubset u^{-1}(B_1(0))$. Given any $fin F$, we have
$$
frac{varepsilon f}{2sum_{j=1}^N|f|_j}in V,
$$
so
$$
left|uleft(frac{varepsilon f}{2sum_{j=1}^N|f|_j}right)right|leq1,
$$
which we may write as
$$
|u(f)|leq sum_{j=1}^Nfrac2varepsilon,|f|_jleqfrac{2N}varepsilon,sup_{jleq N}|f|_j.
$$
There is a small exception to the above, which is the case in which $|f|_j=0$ for $j=1,ldots,N$. But in that case $mfin V$ for all $minmathbb N$, so $|u(mf)|leq1$ and so $|u(f)|leq1/m$, forcing $u(f)=0$.
answered Jan 24 at 0:32
Martin ArgeramiMartin Argerami
128k1184184
$endgroup$
$begingroup$
Thank you, this is exactly what I needed!
$endgroup$
– Aidan Backus
Jan 24 at 0:36
add a comment |
$begingroup$
The result is true in any locally convex space $F$. It is a consequence of the fact that the topology is given by a family of seminorms.
Since $u$ is continuous at $0$, the set $u^{-1}(B_1(0))$ is open (preimage of the unit disk). So there exists a neighbourhood $V$ of zero, of the form $V={f: |f(x)|_j<varepsilon, j=1,ldots,N}$, with $Vsubset u^{-1}(B_1(0))$. Given any $fin F$, we have
$$
frac{varepsilon f}{2sum_{j=1}^N|f|_j}in V,
$$
so
$$
left|uleft(frac{varepsilon f}{2sum_{j=1}^N|f|_j}right)right|leq1,
$$
which we may write as
$$
|u(f)|leq sum_{j=1}^Nfrac2varepsilon,|f|_jleqfrac{2N}varepsilon,sup_{jleq N}|f|_j.
$$
There is a small exception to the above, which is the case in which $|f|_j=0$ for $j=1,ldots,N$. But in that case $mfin V$ for all $minmathbb N$, so $|u(mf)|leq1$ and so $|u(f)|leq1/m$, forcing $u(f)=0$.
answered Jan 24 at 0:32
Martin ArgeramiMartin Argerami
128k1184184
$endgroup$
The result is true in any locally convex space $F$. It is a consequence of the fact that the topology is given by a family of seminorms.
Since $u$ is continuous at $0$, the set $u^{-1}(B_1(0))$ is open (preimage of the unit disk). So there exists a neighbourhood $V$ of zero, of the form $V={f: |f(x)|_j<varepsilon, j=1,ldots,N}$, with $Vsubset u^{-1}(B_1(0))$. Given any $fin F$, we have
$$
frac{varepsilon f}{2sum_{j=1}^N|f|_j}in V,
$$
so
$$
left|uleft(frac{varepsilon f}{2sum_{j=1}^N|f|_j}right)right|leq1,
$$
which we may write as
$$
|u(f)|leq sum_{j=1}^Nfrac2varepsilon,|f|_jleqfrac{2N}varepsilon,sup_{jleq N}|f|_j.
$$
There is a small exception to the above, which is the case in which $|f|_j=0$ for $j=1,ldots,N$. But in that case $mfin V$ for all $minmathbb N$, so $|u(mf)|leq1$ and so $|u(f)|leq1/m$, forcing $u(f)=0$.
answered Jan 24 at 0:32
Martin ArgeramiMartin Argerami
128k1184184
edited Jan 24 at 1:33
answered Jan 24 at 0:32
Martin ArgeramiMartin Argerami
128k1184184
answered Jan 24 at 0:32
Martin ArgeramiMartin Argerami
128k1184184
answered Jan 24 at 0:32
Martin ArgeramiMartin Argerami
128k1184184
128k1184184
$begingroup$
Thank you, this is exactly what I needed!
$endgroup$
– Aidan Backus
Jan 24 at 0:36
add a comment |
$begingroup$
Thank you, this is exactly what I needed!
$endgroup$
– Aidan Backus
Jan 24 at 0:36
$begingroup$
Thank you, this is exactly what I needed!
$endgroup$
– Aidan Backus
Jan 24 at 0:36
$begingroup$
Thank you, this is exactly what I needed!
$endgroup$
– Aidan Backus
Jan 24 at 0:36
add a comment |
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