Proving uniform boundedness
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Let $-infty < a<b<infty$ and $A subseteq C([a,b]) cap C^1((a,b))$ satisfying $forall fin A$ $int_{b}^{a} |f(x)| dx + int_{b}^{a} |f'(x)|^p dx < C$, where $C>0$ and $p>1$. I want to show uniform boundedness of functions from $A$.
It is easy to show using Hölder inequality that $int_{b}^{a} |f(x)+f'(x) | dx < C'$ where $C'$ is some positive constant but I can't figure out uniform boundedness.
real-analysis functional-analysis sobolev-spaces
edited Nov 13 at 12:54
Giuseppe Negro
17k328121
asked Nov 13 at 11:14
MadChemist
12119
add a comment |
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Let $-infty < a<b<infty$ and $A subseteq C([a,b]) cap C^1((a,b))$ satisfying $forall fin A$ $int_{b}^{a} |f(x)| dx + int_{b}^{a} |f'(x)|^p dx < C$, where $C>0$ and $p>1$. I want to show uniform boundedness of functions from $A$.
It is easy to show using Hölder inequality that $int_{b}^{a} |f(x)+f'(x) | dx < C'$ where $C'$ is some positive constant but I can't figure out uniform boundedness.
real-analysis functional-analysis sobolev-spaces
edited Nov 13 at 12:54
Giuseppe Negro
17k328121
asked Nov 13 at 11:14
MadChemist
12119
This is a kind of Sobolev embedding, that's why I retagged.
– Giuseppe Negro
Nov 13 at 12:55
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $-infty < a<b<infty$ and $A subseteq C([a,b]) cap C^1((a,b))$ satisfying $forall fin A$ $int_{b}^{a} |f(x)| dx + int_{b}^{a} |f'(x)|^p dx < C$, where $C>0$ and $p>1$. I want to show uniform boundedness of functions from $A$.
It is easy to show using Hölder inequality that $int_{b}^{a} |f(x)+f'(x) | dx < C'$ where $C'$ is some positive constant but I can't figure out uniform boundedness.
real-analysis functional-analysis sobolev-spaces
edited Nov 13 at 12:54
Giuseppe Negro
17k328121
asked Nov 13 at 11:14
MadChemist
12119
Let $-infty < a<b<infty$ and $A subseteq C([a,b]) cap C^1((a,b))$ satisfying $forall fin A$ $int_{b}^{a} |f(x)| dx + int_{b}^{a} |f'(x)|^p dx < C$, where $C>0$ and $p>1$. I want to show uniform boundedness of functions from $A$.
It is easy to show using Hölder inequality that $int_{b}^{a} |f(x)+f'(x) | dx < C'$ where $C'$ is some positive constant but I can't figure out uniform boundedness.
real-analysis functional-analysis sobolev-spaces
real-analysis functional-analysis sobolev-spaces
edited Nov 13 at 12:54
Giuseppe Negro
17k328121
asked Nov 13 at 11:14
MadChemist
12119
edited Nov 13 at 12:54
Giuseppe Negro
17k328121
asked Nov 13 at 11:14
MadChemist
12119
edited Nov 13 at 12:54
Giuseppe Negro
17k328121
edited Nov 13 at 12:54
Giuseppe Negro
17k328121
edited Nov 13 at 12:54
Giuseppe Negro
17k328121
17k328121
asked Nov 13 at 11:14
MadChemist
12119
asked Nov 13 at 11:14
MadChemist
12119
asked Nov 13 at 11:14
MadChemist
12119
12119
This is a kind of Sobolev embedding, that's why I retagged.
– Giuseppe Negro
Nov 13 at 12:55
add a comment |
This is a kind of Sobolev embedding, that's why I retagged.
– Giuseppe Negro
Nov 13 at 12:55
This is a kind of Sobolev embedding, that's why I retagged.
– Giuseppe Negro
Nov 13 at 12:55
This is a kind of Sobolev embedding, that's why I retagged.
– Giuseppe Negro
Nov 13 at 12:55
add a comment |
1 Answer
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Use $f(x)-f(0) =int_0^{x}f'(t) , dt$ and Holder's inequality to show that $|f(x)-f(0)|$ is bounded, say $|f(x)-f(0)| leq M$ for all $x$ for all $fin A$. It is enough to show that ${f(0):f in A}$ is bounded. Suppose, if possible, there exist $f_1,f_2,... in A$ such that $f_n(0) to +infty$. Then $f_n(0) >C/(b-a)+M$ for $n$ sufficiently large. Hence $f_n(x) >C/(b-a)$ for all $x$ for $n$ sufficiently large. But then $int_a^{b} |f_n(x)|, dx >C$ for $n$ sufficiently large which is a contradiction. Similar argument holds if $f_n(0) to -infty$.
answered Nov 13 at 12:07
Kavi Rama Murthy
39.8k31749
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Use $f(x)-f(0) =int_0^{x}f'(t) , dt$ and Holder's inequality to show that $|f(x)-f(0)|$ is bounded, say $|f(x)-f(0)| leq M$ for all $x$ for all $fin A$. It is enough to show that ${f(0):f in A}$ is bounded. Suppose, if possible, there exist $f_1,f_2,... in A$ such that $f_n(0) to +infty$. Then $f_n(0) >C/(b-a)+M$ for $n$ sufficiently large. Hence $f_n(x) >C/(b-a)$ for all $x$ for $n$ sufficiently large. But then $int_a^{b} |f_n(x)|, dx >C$ for $n$ sufficiently large which is a contradiction. Similar argument holds if $f_n(0) to -infty$.
answered Nov 13 at 12:07
Kavi Rama Murthy
39.8k31749
add a comment |
up vote
2
down vote
accepted
Use $f(x)-f(0) =int_0^{x}f'(t) , dt$ and Holder's inequality to show that $|f(x)-f(0)|$ is bounded, say $|f(x)-f(0)| leq M$ for all $x$ for all $fin A$. It is enough to show that ${f(0):f in A}$ is bounded. Suppose, if possible, there exist $f_1,f_2,... in A$ such that $f_n(0) to +infty$. Then $f_n(0) >C/(b-a)+M$ for $n$ sufficiently large. Hence $f_n(x) >C/(b-a)$ for all $x$ for $n$ sufficiently large. But then $int_a^{b} |f_n(x)|, dx >C$ for $n$ sufficiently large which is a contradiction. Similar argument holds if $f_n(0) to -infty$.
answered Nov 13 at 12:07
Kavi Rama Murthy
39.8k31749
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Use $f(x)-f(0) =int_0^{x}f'(t) , dt$ and Holder's inequality to show that $|f(x)-f(0)|$ is bounded, say $|f(x)-f(0)| leq M$ for all $x$ for all $fin A$. It is enough to show that ${f(0):f in A}$ is bounded. Suppose, if possible, there exist $f_1,f_2,... in A$ such that $f_n(0) to +infty$. Then $f_n(0) >C/(b-a)+M$ for $n$ sufficiently large. Hence $f_n(x) >C/(b-a)$ for all $x$ for $n$ sufficiently large. But then $int_a^{b} |f_n(x)|, dx >C$ for $n$ sufficiently large which is a contradiction. Similar argument holds if $f_n(0) to -infty$.
answered Nov 13 at 12:07
Kavi Rama Murthy
39.8k31749
Use $f(x)-f(0) =int_0^{x}f'(t) , dt$ and Holder's inequality to show that $|f(x)-f(0)|$ is bounded, say $|f(x)-f(0)| leq M$ for all $x$ for all $fin A$. It is enough to show that ${f(0):f in A}$ is bounded. Suppose, if possible, there exist $f_1,f_2,... in A$ such that $f_n(0) to +infty$. Then $f_n(0) >C/(b-a)+M$ for $n$ sufficiently large. Hence $f_n(x) >C/(b-a)$ for all $x$ for $n$ sufficiently large. But then $int_a^{b} |f_n(x)|, dx >C$ for $n$ sufficiently large which is a contradiction. Similar argument holds if $f_n(0) to -infty$.
answered Nov 13 at 12:07
Kavi Rama Murthy
39.8k31749
answered Nov 13 at 12:07
Kavi Rama Murthy
39.8k31749
answered Nov 13 at 12:07
Kavi Rama Murthy
39.8k31749
answered Nov 13 at 12:07
Kavi Rama Murthy
39.8k31749
39.8k31749
add a comment |
add a comment |
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This is a kind of Sobolev embedding, that's why I retagged.
– Giuseppe Negro
Nov 13 at 12:55