Uniform Boundedness principle for bounded linear maps from Frechet Space into a Banach Space
$begingroup$
I am looking for a proof of the uniform boundedness principle where the domain is a Frechet space, instead of the usual setting of a Banach Space.
This is used in proving the space of tempered distributions is complete but I can't find a proof of it anywhere.
When I try to prove it myself I get stuck on the final part(which uses the scaling property of linear maps).
Does anyone have a proof that they could share?
real-analysis functional-analysis distribution-theory
edited Jan 17 '17 at 0:17
spaceisdarkgreen
34k21754
asked Jan 16 '17 at 23:44
FourierFluxFourierFlux
987
$endgroup$
add a comment |
$begingroup$
I am looking for a proof of the uniform boundedness principle where the domain is a Frechet space, instead of the usual setting of a Banach Space.
This is used in proving the space of tempered distributions is complete but I can't find a proof of it anywhere.
When I try to prove it myself I get stuck on the final part(which uses the scaling property of linear maps).
Does anyone have a proof that they could share?
real-analysis functional-analysis distribution-theory
edited Jan 17 '17 at 0:17
spaceisdarkgreen
34k21754
asked Jan 16 '17 at 23:44
FourierFluxFourierFlux
987
$endgroup$
$begingroup$
Doesn't the standard proof via the Baire category theorem just work?
$endgroup$
– tomasz
Jan 17 '17 at 0:46
$begingroup$
The problem the last part relies upon the scaling property of a linear map(pulling out a scaler). I admit I'm not well versed in Frechet spaces though - if I'm wrong please correct me.
$endgroup$
– FourierFlux
Jan 17 '17 at 0:49
$begingroup$
HAHAHA, Ok I see. Being stupid I was thinking the norm was of the domain and not the range.
$endgroup$
– FourierFlux
Jan 17 '17 at 1:05
add a comment |
$begingroup$
I am looking for a proof of the uniform boundedness principle where the domain is a Frechet space, instead of the usual setting of a Banach Space.
This is used in proving the space of tempered distributions is complete but I can't find a proof of it anywhere.
When I try to prove it myself I get stuck on the final part(which uses the scaling property of linear maps).
Does anyone have a proof that they could share?
real-analysis functional-analysis distribution-theory
edited Jan 17 '17 at 0:17
spaceisdarkgreen
34k21754
asked Jan 16 '17 at 23:44
FourierFluxFourierFlux
987
$endgroup$
I am looking for a proof of the uniform boundedness principle where the domain is a Frechet space, instead of the usual setting of a Banach Space.
This is used in proving the space of tempered distributions is complete but I can't find a proof of it anywhere.
When I try to prove it myself I get stuck on the final part(which uses the scaling property of linear maps).
Does anyone have a proof that they could share?
real-analysis functional-analysis distribution-theory
real-analysis functional-analysis distribution-theory
edited Jan 17 '17 at 0:17
spaceisdarkgreen
34k21754
asked Jan 16 '17 at 23:44
FourierFluxFourierFlux
987
edited Jan 17 '17 at 0:17
spaceisdarkgreen
34k21754
asked Jan 16 '17 at 23:44
FourierFluxFourierFlux
987
edited Jan 17 '17 at 0:17
spaceisdarkgreen
34k21754
edited Jan 17 '17 at 0:17
spaceisdarkgreen
34k21754
edited Jan 17 '17 at 0:17
spaceisdarkgreen
34k21754
34k21754
asked Jan 16 '17 at 23:44
FourierFluxFourierFlux
987
asked Jan 16 '17 at 23:44
FourierFluxFourierFlux
987
asked Jan 16 '17 at 23:44
FourierFluxFourierFlux
987
987
$begingroup$
Doesn't the standard proof via the Baire category theorem just work?
$endgroup$
– tomasz
Jan 17 '17 at 0:46
$begingroup$
The problem the last part relies upon the scaling property of a linear map(pulling out a scaler). I admit I'm not well versed in Frechet spaces though - if I'm wrong please correct me.
$endgroup$
– FourierFlux
Jan 17 '17 at 0:49
$begingroup$
HAHAHA, Ok I see. Being stupid I was thinking the norm was of the domain and not the range.
$endgroup$
– FourierFlux
Jan 17 '17 at 1:05
add a comment |
$begingroup$
Doesn't the standard proof via the Baire category theorem just work?
$endgroup$
– tomasz
Jan 17 '17 at 0:46
$begingroup$
The problem the last part relies upon the scaling property of a linear map(pulling out a scaler). I admit I'm not well versed in Frechet spaces though - if I'm wrong please correct me.
$endgroup$
– FourierFlux
Jan 17 '17 at 0:49
$begingroup$
HAHAHA, Ok I see. Being stupid I was thinking the norm was of the domain and not the range.
$endgroup$
– FourierFlux
Jan 17 '17 at 1:05
$begingroup$
Doesn't the standard proof via the Baire category theorem just work?
$endgroup$
– tomasz
Jan 17 '17 at 0:46
$begingroup$
Doesn't the standard proof via the Baire category theorem just work?
$endgroup$
– tomasz
Jan 17 '17 at 0:46
$begingroup$
The problem the last part relies upon the scaling property of a linear map(pulling out a scaler). I admit I'm not well versed in Frechet spaces though - if I'm wrong please correct me.
$endgroup$
– FourierFlux
Jan 17 '17 at 0:49
$begingroup$
The problem the last part relies upon the scaling property of a linear map(pulling out a scaler). I admit I'm not well versed in Frechet spaces though - if I'm wrong please correct me.
$endgroup$
– FourierFlux
Jan 17 '17 at 0:49
$begingroup$
HAHAHA, Ok I see. Being stupid I was thinking the norm was of the domain and not the range.
$endgroup$
– FourierFlux
Jan 17 '17 at 1:05
$begingroup$
HAHAHA, Ok I see. Being stupid I was thinking the norm was of the domain and not the range.
$endgroup$
– FourierFlux
Jan 17 '17 at 1:05
add a comment |
1 Answer
1
active
oldest
votes
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Let $X$ be Frechet and $Y$ be locally convex. Let $T_a: X rightarrow Y$ be continuous linear and $q: Y rightarrow [0,infty)$ be a continuous semi-norm. If $sup {q(T_a x):ain A} <infty$ (pointwise bounded), then $sup {q(T_a x):ain A}$ is a continuous semi-norm (uniformly bounded/equicontinuous).
To prove this, let
$$E_n = {xin X: q(T_a x) le n, forall a}$$
By pointwise boundedness, we see that $E_n nearrow X$. Since $E_n$ are closed and $X$ is a Baire space, we see that there exists $E_n$ with an interior point $x$. Since $X$ is Frechet, its topology is generated by countable semi-norms $p_n$ and thus there exists $N,r$ such that
$$
x + bigcap_{k=1}^N {p_k < r} subseteq E_n
$$
Hence,
$$
bigcap_{k=1}^N {p_k < 2r} subseteq E_n -E_n subseteq E_{2n}
$$
Let
$$
p(x) = frac{1}{r} sum_{k=1}^N p_k (x)
$$
Hence,
$$
0in {p < 1} subseteq {sup {q(T_a x):ain A} <3n}
$$
Since $p$ is continuous, we see that $0$ is an interior point of ${sup {q(T_a x):ain A} <3n} $ and thus $sup {q(T_a x):ain A}$ is a continuous semi-norm.
answered Jan 6 at 10:28
Andrew YuanAndrew Yuan
530210
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Let $X$ be Frechet and $Y$ be locally convex. Let $T_a: X rightarrow Y$ be continuous linear and $q: Y rightarrow [0,infty)$ be a continuous semi-norm. If $sup {q(T_a x):ain A} <infty$ (pointwise bounded), then $sup {q(T_a x):ain A}$ is a continuous semi-norm (uniformly bounded/equicontinuous).
To prove this, let
$$E_n = {xin X: q(T_a x) le n, forall a}$$
By pointwise boundedness, we see that $E_n nearrow X$. Since $E_n$ are closed and $X$ is a Baire space, we see that there exists $E_n$ with an interior point $x$. Since $X$ is Frechet, its topology is generated by countable semi-norms $p_n$ and thus there exists $N,r$ such that
$$
x + bigcap_{k=1}^N {p_k < r} subseteq E_n
$$
Hence,
$$
bigcap_{k=1}^N {p_k < 2r} subseteq E_n -E_n subseteq E_{2n}
$$
Let
$$
p(x) = frac{1}{r} sum_{k=1}^N p_k (x)
$$
Hence,
$$
0in {p < 1} subseteq {sup {q(T_a x):ain A} <3n}
$$
Since $p$ is continuous, we see that $0$ is an interior point of ${sup {q(T_a x):ain A} <3n} $ and thus $sup {q(T_a x):ain A}$ is a continuous semi-norm.
answered Jan 6 at 10:28
Andrew YuanAndrew Yuan
530210
$endgroup$
add a comment |
$begingroup$
Let $X$ be Frechet and $Y$ be locally convex. Let $T_a: X rightarrow Y$ be continuous linear and $q: Y rightarrow [0,infty)$ be a continuous semi-norm. If $sup {q(T_a x):ain A} <infty$ (pointwise bounded), then $sup {q(T_a x):ain A}$ is a continuous semi-norm (uniformly bounded/equicontinuous).
To prove this, let
$$E_n = {xin X: q(T_a x) le n, forall a}$$
By pointwise boundedness, we see that $E_n nearrow X$. Since $E_n$ are closed and $X$ is a Baire space, we see that there exists $E_n$ with an interior point $x$. Since $X$ is Frechet, its topology is generated by countable semi-norms $p_n$ and thus there exists $N,r$ such that
$$
x + bigcap_{k=1}^N {p_k < r} subseteq E_n
$$
Hence,
$$
bigcap_{k=1}^N {p_k < 2r} subseteq E_n -E_n subseteq E_{2n}
$$
Let
$$
p(x) = frac{1}{r} sum_{k=1}^N p_k (x)
$$
Hence,
$$
0in {p < 1} subseteq {sup {q(T_a x):ain A} <3n}
$$
Since $p$ is continuous, we see that $0$ is an interior point of ${sup {q(T_a x):ain A} <3n} $ and thus $sup {q(T_a x):ain A}$ is a continuous semi-norm.
answered Jan 6 at 10:28
Andrew YuanAndrew Yuan
530210
$endgroup$
add a comment |
$begingroup$
Let $X$ be Frechet and $Y$ be locally convex. Let $T_a: X rightarrow Y$ be continuous linear and $q: Y rightarrow [0,infty)$ be a continuous semi-norm. If $sup {q(T_a x):ain A} <infty$ (pointwise bounded), then $sup {q(T_a x):ain A}$ is a continuous semi-norm (uniformly bounded/equicontinuous).
To prove this, let
$$E_n = {xin X: q(T_a x) le n, forall a}$$
By pointwise boundedness, we see that $E_n nearrow X$. Since $E_n$ are closed and $X$ is a Baire space, we see that there exists $E_n$ with an interior point $x$. Since $X$ is Frechet, its topology is generated by countable semi-norms $p_n$ and thus there exists $N,r$ such that
$$
x + bigcap_{k=1}^N {p_k < r} subseteq E_n
$$
Hence,
$$
bigcap_{k=1}^N {p_k < 2r} subseteq E_n -E_n subseteq E_{2n}
$$
Let
$$
p(x) = frac{1}{r} sum_{k=1}^N p_k (x)
$$
Hence,
$$
0in {p < 1} subseteq {sup {q(T_a x):ain A} <3n}
$$
Since $p$ is continuous, we see that $0$ is an interior point of ${sup {q(T_a x):ain A} <3n} $ and thus $sup {q(T_a x):ain A}$ is a continuous semi-norm.
answered Jan 6 at 10:28
Andrew YuanAndrew Yuan
530210
$endgroup$
Let $X$ be Frechet and $Y$ be locally convex. Let $T_a: X rightarrow Y$ be continuous linear and $q: Y rightarrow [0,infty)$ be a continuous semi-norm. If $sup {q(T_a x):ain A} <infty$ (pointwise bounded), then $sup {q(T_a x):ain A}$ is a continuous semi-norm (uniformly bounded/equicontinuous).
To prove this, let
$$E_n = {xin X: q(T_a x) le n, forall a}$$
By pointwise boundedness, we see that $E_n nearrow X$. Since $E_n$ are closed and $X$ is a Baire space, we see that there exists $E_n$ with an interior point $x$. Since $X$ is Frechet, its topology is generated by countable semi-norms $p_n$ and thus there exists $N,r$ such that
$$
x + bigcap_{k=1}^N {p_k < r} subseteq E_n
$$
Hence,
$$
bigcap_{k=1}^N {p_k < 2r} subseteq E_n -E_n subseteq E_{2n}
$$
Let
$$
p(x) = frac{1}{r} sum_{k=1}^N p_k (x)
$$
Hence,
$$
0in {p < 1} subseteq {sup {q(T_a x):ain A} <3n}
$$
Since $p$ is continuous, we see that $0$ is an interior point of ${sup {q(T_a x):ain A} <3n} $ and thus $sup {q(T_a x):ain A}$ is a continuous semi-norm.
answered Jan 6 at 10:28
Andrew YuanAndrew Yuan
530210
edited Jan 6 at 20:48
answered Jan 6 at 10:28
Andrew YuanAndrew Yuan
530210
answered Jan 6 at 10:28
Andrew YuanAndrew Yuan
530210
answered Jan 6 at 10:28
Andrew YuanAndrew Yuan
530210
530210
add a comment |
add a comment |
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$begingroup$
Doesn't the standard proof via the Baire category theorem just work?
$endgroup$
– tomasz
Jan 17 '17 at 0:46
$begingroup$
The problem the last part relies upon the scaling property of a linear map(pulling out a scaler). I admit I'm not well versed in Frechet spaces though - if I'm wrong please correct me.
$endgroup$
– FourierFlux
Jan 17 '17 at 0:49
$begingroup$
HAHAHA, Ok I see. Being stupid I was thinking the norm was of the domain and not the range.
$endgroup$
– FourierFlux
Jan 17 '17 at 1:05