Normed linear space
$begingroup$
In Walter Rudin's Complex Analysis, it states that by definition$$|Lambda|=text{sup}{|Lambda x|: xin X, |x|leq1}$$
and then later he shows that
$|Lambda x|leq |Lambda||x|.$
But, the only thing I know is $|Lambda|geq|Lambda x|$.
How to show that indeed $|Lambda x|leq |Lambda||x|.$
real-analysis inequality normed-spaces
edited Dec 9 '18 at 22:25
José Carlos Santos
162k22129233
asked Dec 9 '18 at 22:04
beginnerbeginner
227
$endgroup$
add a comment |
$begingroup$
In Walter Rudin's Complex Analysis, it states that by definition$$|Lambda|=text{sup}{|Lambda x|: xin X, |x|leq1}$$
and then later he shows that
$|Lambda x|leq |Lambda||x|.$
But, the only thing I know is $|Lambda|geq|Lambda x|$.
How to show that indeed $|Lambda x|leq |Lambda||x|.$
real-analysis inequality normed-spaces
edited Dec 9 '18 at 22:25
José Carlos Santos
162k22129233
asked Dec 9 '18 at 22:04
beginnerbeginner
227
$endgroup$
add a comment |
$begingroup$
In Walter Rudin's Complex Analysis, it states that by definition$$|Lambda|=text{sup}{|Lambda x|: xin X, |x|leq1}$$
and then later he shows that
$|Lambda x|leq |Lambda||x|.$
But, the only thing I know is $|Lambda|geq|Lambda x|$.
How to show that indeed $|Lambda x|leq |Lambda||x|.$
real-analysis inequality normed-spaces
edited Dec 9 '18 at 22:25
José Carlos Santos
162k22129233
asked Dec 9 '18 at 22:04
beginnerbeginner
227
$endgroup$
In Walter Rudin's Complex Analysis, it states that by definition$$|Lambda|=text{sup}{|Lambda x|: xin X, |x|leq1}$$
and then later he shows that
$|Lambda x|leq |Lambda||x|.$
But, the only thing I know is $|Lambda|geq|Lambda x|$.
How to show that indeed $|Lambda x|leq |Lambda||x|.$
real-analysis inequality normed-spaces
real-analysis inequality normed-spaces
edited Dec 9 '18 at 22:25
José Carlos Santos
162k22129233
asked Dec 9 '18 at 22:04
beginnerbeginner
227
edited Dec 9 '18 at 22:25
José Carlos Santos
162k22129233
asked Dec 9 '18 at 22:04
beginnerbeginner
227
edited Dec 9 '18 at 22:25
José Carlos Santos
162k22129233
edited Dec 9 '18 at 22:25
José Carlos Santos
162k22129233
edited Dec 9 '18 at 22:25
José Carlos Santos
162k22129233
162k22129233
asked Dec 9 '18 at 22:04
beginnerbeginner
227
asked Dec 9 '18 at 22:04
beginnerbeginner
227
asked Dec 9 '18 at 22:04
beginnerbeginner
227
227
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
If $x=0$, that inequality is trivial.
In the other cases, $x=lVert xrVerttimesfrac x{lVert xrVert}$. Since, $leftlVertfrac x{lVert xrVert}rightrVert=1$ and $Lambda$ is linear,$$lvertLambda xrvert=lVert xrVertleftlvertLambdafrac x{lVert xrVert}rightrvertleqslantlVert xrVertlVertLambdarVert.$$
answered Dec 9 '18 at 22:15
José Carlos SantosJosé Carlos Santos
162k22129233
$endgroup$
$begingroup$
Since, $lVertfrac x{lVert xrVert}rVert=1$ then $lvertLambdafrac x{lVert xrVert}rvertleqlvertLambdarVert$?
$endgroup$
– beginner
Dec 10 '18 at 0:05
$begingroup$
Sure, by the definition of $lVertLambdarVert$.
$endgroup$
– José Carlos Santos
Dec 10 '18 at 0:09
add a comment |
$begingroup$
You can prove that
$$
|Lambda|=sup_{substack{xin X\xne0}}frac{|Lambda x|}{|x|}
$$
by noticing that
$$
left|frac{1}{|x|}x,right|=1
$$
answered Dec 9 '18 at 22:15
egregegreg
182k1485204
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If $x=0$, that inequality is trivial.
In the other cases, $x=lVert xrVerttimesfrac x{lVert xrVert}$. Since, $leftlVertfrac x{lVert xrVert}rightrVert=1$ and $Lambda$ is linear,$$lvertLambda xrvert=lVert xrVertleftlvertLambdafrac x{lVert xrVert}rightrvertleqslantlVert xrVertlVertLambdarVert.$$
answered Dec 9 '18 at 22:15
José Carlos SantosJosé Carlos Santos
162k22129233
$endgroup$
$begingroup$
Since, $lVertfrac x{lVert xrVert}rVert=1$ then $lvertLambdafrac x{lVert xrVert}rvertleqlvertLambdarVert$?
$endgroup$
– beginner
Dec 10 '18 at 0:05
$begingroup$
Sure, by the definition of $lVertLambdarVert$.
$endgroup$
– José Carlos Santos
Dec 10 '18 at 0:09
add a comment |
$begingroup$
If $x=0$, that inequality is trivial.
In the other cases, $x=lVert xrVerttimesfrac x{lVert xrVert}$. Since, $leftlVertfrac x{lVert xrVert}rightrVert=1$ and $Lambda$ is linear,$$lvertLambda xrvert=lVert xrVertleftlvertLambdafrac x{lVert xrVert}rightrvertleqslantlVert xrVertlVertLambdarVert.$$
answered Dec 9 '18 at 22:15
José Carlos SantosJosé Carlos Santos
162k22129233
$endgroup$
$begingroup$
Since, $lVertfrac x{lVert xrVert}rVert=1$ then $lvertLambdafrac x{lVert xrVert}rvertleqlvertLambdarVert$?
$endgroup$
– beginner
Dec 10 '18 at 0:05
$begingroup$
Sure, by the definition of $lVertLambdarVert$.
$endgroup$
– José Carlos Santos
Dec 10 '18 at 0:09
add a comment |
$begingroup$
If $x=0$, that inequality is trivial.
In the other cases, $x=lVert xrVerttimesfrac x{lVert xrVert}$. Since, $leftlVertfrac x{lVert xrVert}rightrVert=1$ and $Lambda$ is linear,$$lvertLambda xrvert=lVert xrVertleftlvertLambdafrac x{lVert xrVert}rightrvertleqslantlVert xrVertlVertLambdarVert.$$
answered Dec 9 '18 at 22:15
José Carlos SantosJosé Carlos Santos
162k22129233
$endgroup$
If $x=0$, that inequality is trivial.
In the other cases, $x=lVert xrVerttimesfrac x{lVert xrVert}$. Since, $leftlVertfrac x{lVert xrVert}rightrVert=1$ and $Lambda$ is linear,$$lvertLambda xrvert=lVert xrVertleftlvertLambdafrac x{lVert xrVert}rightrvertleqslantlVert xrVertlVertLambdarVert.$$
answered Dec 9 '18 at 22:15
José Carlos SantosJosé Carlos Santos
162k22129233
answered Dec 9 '18 at 22:15
José Carlos SantosJosé Carlos Santos
162k22129233
answered Dec 9 '18 at 22:15
José Carlos SantosJosé Carlos Santos
162k22129233
answered Dec 9 '18 at 22:15
José Carlos SantosJosé Carlos Santos
162k22129233
162k22129233
$begingroup$
Since, $lVertfrac x{lVert xrVert}rVert=1$ then $lvertLambdafrac x{lVert xrVert}rvertleqlvertLambdarVert$?
$endgroup$
– beginner
Dec 10 '18 at 0:05
$begingroup$
Sure, by the definition of $lVertLambdarVert$.
$endgroup$
– José Carlos Santos
Dec 10 '18 at 0:09
add a comment |
$begingroup$
Since, $lVertfrac x{lVert xrVert}rVert=1$ then $lvertLambdafrac x{lVert xrVert}rvertleqlvertLambdarVert$?
$endgroup$
– beginner
Dec 10 '18 at 0:05
$begingroup$
Sure, by the definition of $lVertLambdarVert$.
$endgroup$
– José Carlos Santos
Dec 10 '18 at 0:09
$begingroup$
Since, $lVertfrac x{lVert xrVert}rVert=1$ then $lvertLambdafrac x{lVert xrVert}rvertleqlvertLambdarVert$?
$endgroup$
– beginner
Dec 10 '18 at 0:05
$begingroup$
Since, $lVertfrac x{lVert xrVert}rVert=1$ then $lvertLambdafrac x{lVert xrVert}rvertleqlvertLambdarVert$?
$endgroup$
– beginner
Dec 10 '18 at 0:05
$begingroup$
Sure, by the definition of $lVertLambdarVert$.
$endgroup$
– José Carlos Santos
Dec 10 '18 at 0:09
$begingroup$
Sure, by the definition of $lVertLambdarVert$.
$endgroup$
– José Carlos Santos
Dec 10 '18 at 0:09
add a comment |
$begingroup$
You can prove that
$$
|Lambda|=sup_{substack{xin X\xne0}}frac{|Lambda x|}{|x|}
$$
by noticing that
$$
left|frac{1}{|x|}x,right|=1
$$
answered Dec 9 '18 at 22:15
egregegreg
182k1485204
$endgroup$
add a comment |
$begingroup$
You can prove that
$$
|Lambda|=sup_{substack{xin X\xne0}}frac{|Lambda x|}{|x|}
$$
by noticing that
$$
left|frac{1}{|x|}x,right|=1
$$
answered Dec 9 '18 at 22:15
egregegreg
182k1485204
$endgroup$
add a comment |
$begingroup$
You can prove that
$$
|Lambda|=sup_{substack{xin X\xne0}}frac{|Lambda x|}{|x|}
$$
by noticing that
$$
left|frac{1}{|x|}x,right|=1
$$
answered Dec 9 '18 at 22:15
egregegreg
182k1485204
$endgroup$
You can prove that
$$
|Lambda|=sup_{substack{xin X\xne0}}frac{|Lambda x|}{|x|}
$$
by noticing that
$$
left|frac{1}{|x|}x,right|=1
$$
answered Dec 9 '18 at 22:15
egregegreg
182k1485204
answered Dec 9 '18 at 22:15
egregegreg
182k1485204
answered Dec 9 '18 at 22:15
egregegreg
182k1485204
answered Dec 9 '18 at 22:15
egregegreg
182k1485204
182k1485204
add a comment |
add a comment |
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