consequences of the uniform boundedness theorem
$begingroup$
Let $X$ be the function space defined in the following way. A function $x$ belongs to $X$ iff the two conditions are satisfied:
i) $x : mathbb{R} rightarrow mathbb{R}$
ii) there exists a compact interval $I_x$ of $mathbb{R}$ such that $x(t) = 0$ for every $t in mathbb{R} | I_x$.
$T_nx = int^n_0 x(s) ds$ for every $x in X$.
where the norm $||x|| = max_{t in mathbb{R}} |x(t)|$ and $n in mathbb{N}$.
I have already shown $X$ is a vector space over $mathbb{R}$ and proved $T_n$ is linear and bounded.
I now need to show that for every $x in X$, there exists $c_x > 0$ such that $sup_{n in mathbb{N}}|T_nx| leq c_x$.
I'm not sure how to begin.
functional-analysis vector-spaces normed-spaces
edited Dec 6 '18 at 19:13
Aweygan
14.1k21441
asked Dec 6 '18 at 16:32
Zombiegit123Zombiegit123
306113
$endgroup$
add a comment |
$begingroup$
Let $X$ be the function space defined in the following way. A function $x$ belongs to $X$ iff the two conditions are satisfied:
i) $x : mathbb{R} rightarrow mathbb{R}$
ii) there exists a compact interval $I_x$ of $mathbb{R}$ such that $x(t) = 0$ for every $t in mathbb{R} | I_x$.
$T_nx = int^n_0 x(s) ds$ for every $x in X$.
where the norm $||x|| = max_{t in mathbb{R}} |x(t)|$ and $n in mathbb{N}$.
I have already shown $X$ is a vector space over $mathbb{R}$ and proved $T_n$ is linear and bounded.
I now need to show that for every $x in X$, there exists $c_x > 0$ such that $sup_{n in mathbb{N}}|T_nx| leq c_x$.
I'm not sure how to begin.
functional-analysis vector-spaces normed-spaces
edited Dec 6 '18 at 19:13
Aweygan
14.1k21441
asked Dec 6 '18 at 16:32
Zombiegit123Zombiegit123
306113
$endgroup$
add a comment |
$begingroup$
Let $X$ be the function space defined in the following way. A function $x$ belongs to $X$ iff the two conditions are satisfied:
i) $x : mathbb{R} rightarrow mathbb{R}$
ii) there exists a compact interval $I_x$ of $mathbb{R}$ such that $x(t) = 0$ for every $t in mathbb{R} | I_x$.
$T_nx = int^n_0 x(s) ds$ for every $x in X$.
where the norm $||x|| = max_{t in mathbb{R}} |x(t)|$ and $n in mathbb{N}$.
I have already shown $X$ is a vector space over $mathbb{R}$ and proved $T_n$ is linear and bounded.
I now need to show that for every $x in X$, there exists $c_x > 0$ such that $sup_{n in mathbb{N}}|T_nx| leq c_x$.
I'm not sure how to begin.
functional-analysis vector-spaces normed-spaces
edited Dec 6 '18 at 19:13
Aweygan
14.1k21441
asked Dec 6 '18 at 16:32
Zombiegit123Zombiegit123
306113
$endgroup$
Let $X$ be the function space defined in the following way. A function $x$ belongs to $X$ iff the two conditions are satisfied:
i) $x : mathbb{R} rightarrow mathbb{R}$
ii) there exists a compact interval $I_x$ of $mathbb{R}$ such that $x(t) = 0$ for every $t in mathbb{R} | I_x$.
$T_nx = int^n_0 x(s) ds$ for every $x in X$.
where the norm $||x|| = max_{t in mathbb{R}} |x(t)|$ and $n in mathbb{N}$.
I have already shown $X$ is a vector space over $mathbb{R}$ and proved $T_n$ is linear and bounded.
I now need to show that for every $x in X$, there exists $c_x > 0$ such that $sup_{n in mathbb{N}}|T_nx| leq c_x$.
I'm not sure how to begin.
functional-analysis vector-spaces normed-spaces
functional-analysis vector-spaces normed-spaces
edited Dec 6 '18 at 19:13
Aweygan
14.1k21441
asked Dec 6 '18 at 16:32
Zombiegit123Zombiegit123
306113
edited Dec 6 '18 at 19:13
Aweygan
14.1k21441
asked Dec 6 '18 at 16:32
Zombiegit123Zombiegit123
306113
edited Dec 6 '18 at 19:13
Aweygan
14.1k21441
edited Dec 6 '18 at 19:13
Aweygan
14.1k21441
edited Dec 6 '18 at 19:13
Aweygan
14.1k21441
14.1k21441
asked Dec 6 '18 at 16:32
Zombiegit123Zombiegit123
306113
asked Dec 6 '18 at 16:32
Zombiegit123Zombiegit123
306113
asked Dec 6 '18 at 16:32
Zombiegit123Zombiegit123
306113
306113
add a comment |
add a comment |
1 Answer
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$begingroup$
I assume you also want your functions in $X$ to be continuous, or at least bounded and integrable, so that $|x|$ is well-defined for $xin X$. If this is the case, here's one way to obtain a bound:
begin{align*}
|T_nx|&=left|int_0^nx(s) dsright|\
&leqint_0^n|x(s)| ds\
&=int_{[0,n]capoperatorname{supp}(x)}|x(s)| ds\
&leq|x|m([0,n]capoperatorname{supp}(x))\
&leq|x|m(operatorname{supp}(x))
end{align*}
Where $m$ denotes Lebesgue measure on $mathbb R$ and $operatorname{supp}(x)$ is the support of $x$.
You mention the uniform boundedness principle, but I should warn you that you can't use it here, since $X$ is not complete
answered Dec 6 '18 at 19:12
AweyganAweygan
14.1k21441
$endgroup$
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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votes
$begingroup$
I assume you also want your functions in $X$ to be continuous, or at least bounded and integrable, so that $|x|$ is well-defined for $xin X$. If this is the case, here's one way to obtain a bound:
begin{align*}
|T_nx|&=left|int_0^nx(s) dsright|\
&leqint_0^n|x(s)| ds\
&=int_{[0,n]capoperatorname{supp}(x)}|x(s)| ds\
&leq|x|m([0,n]capoperatorname{supp}(x))\
&leq|x|m(operatorname{supp}(x))
end{align*}
Where $m$ denotes Lebesgue measure on $mathbb R$ and $operatorname{supp}(x)$ is the support of $x$.
You mention the uniform boundedness principle, but I should warn you that you can't use it here, since $X$ is not complete
answered Dec 6 '18 at 19:12
AweyganAweygan
14.1k21441
$endgroup$
add a comment |
$begingroup$
I assume you also want your functions in $X$ to be continuous, or at least bounded and integrable, so that $|x|$ is well-defined for $xin X$. If this is the case, here's one way to obtain a bound:
begin{align*}
|T_nx|&=left|int_0^nx(s) dsright|\
&leqint_0^n|x(s)| ds\
&=int_{[0,n]capoperatorname{supp}(x)}|x(s)| ds\
&leq|x|m([0,n]capoperatorname{supp}(x))\
&leq|x|m(operatorname{supp}(x))
end{align*}
Where $m$ denotes Lebesgue measure on $mathbb R$ and $operatorname{supp}(x)$ is the support of $x$.
You mention the uniform boundedness principle, but I should warn you that you can't use it here, since $X$ is not complete
answered Dec 6 '18 at 19:12
AweyganAweygan
14.1k21441
$endgroup$
add a comment |
$begingroup$
I assume you also want your functions in $X$ to be continuous, or at least bounded and integrable, so that $|x|$ is well-defined for $xin X$. If this is the case, here's one way to obtain a bound:
begin{align*}
|T_nx|&=left|int_0^nx(s) dsright|\
&leqint_0^n|x(s)| ds\
&=int_{[0,n]capoperatorname{supp}(x)}|x(s)| ds\
&leq|x|m([0,n]capoperatorname{supp}(x))\
&leq|x|m(operatorname{supp}(x))
end{align*}
Where $m$ denotes Lebesgue measure on $mathbb R$ and $operatorname{supp}(x)$ is the support of $x$.
You mention the uniform boundedness principle, but I should warn you that you can't use it here, since $X$ is not complete
answered Dec 6 '18 at 19:12
AweyganAweygan
14.1k21441
$endgroup$
I assume you also want your functions in $X$ to be continuous, or at least bounded and integrable, so that $|x|$ is well-defined for $xin X$. If this is the case, here's one way to obtain a bound:
begin{align*}
|T_nx|&=left|int_0^nx(s) dsright|\
&leqint_0^n|x(s)| ds\
&=int_{[0,n]capoperatorname{supp}(x)}|x(s)| ds\
&leq|x|m([0,n]capoperatorname{supp}(x))\
&leq|x|m(operatorname{supp}(x))
end{align*}
Where $m$ denotes Lebesgue measure on $mathbb R$ and $operatorname{supp}(x)$ is the support of $x$.
You mention the uniform boundedness principle, but I should warn you that you can't use it here, since $X$ is not complete
answered Dec 6 '18 at 19:12
AweyganAweygan
14.1k21441
answered Dec 6 '18 at 19:12
AweyganAweygan
14.1k21441
answered Dec 6 '18 at 19:12
AweyganAweygan
14.1k21441
answered Dec 6 '18 at 19:12
AweyganAweygan
14.1k21441
14.1k21441
add a comment |
add a comment |
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