If $T:C[0,1] rightarrow mathbb{R}$ is defined by $T_{x_0}(f)=f(x_0)$ then $||T||=1$.
$begingroup$
Let $x_0 in [0,1]$. Define $T_{x_0}:C[0,1] rightarrow mathbb{R}$ by $T_{x_0}(f)=f(x_0)$.
$||T||:= sup{|T_{x_0}(f)|: ||f||_{infty} leq 1}$ where $||f||_{infty}:=max{|f(x)|:x in [0,1]}$.
Prove that $||T||=1$.
For every $f in C[0,1]$, if $||f||_infty leq 1$ then $|f(x)| leq 1$ for all $x in [0,1]$. In particular, $|f(x_0)| leq 1$. So the set ${|T_{x_0}(f)|: ||f||_{infty} leq 1}$ is bounded above by $1$.
Assume some $a$ such that $0 leq a <1$ is the least upper bound. How can I derive a contradiction?
real-analysis functional-analysis analysis normed-spaces
edited Nov 29 '18 at 23:10
José Carlos Santos
160k22126232
asked Nov 29 '18 at 23:05
bbwbbw
50038
$endgroup$
add a comment |
$begingroup$
Let $x_0 in [0,1]$. Define $T_{x_0}:C[0,1] rightarrow mathbb{R}$ by $T_{x_0}(f)=f(x_0)$.
$||T||:= sup{|T_{x_0}(f)|: ||f||_{infty} leq 1}$ where $||f||_{infty}:=max{|f(x)|:x in [0,1]}$.
Prove that $||T||=1$.
For every $f in C[0,1]$, if $||f||_infty leq 1$ then $|f(x)| leq 1$ for all $x in [0,1]$. In particular, $|f(x_0)| leq 1$. So the set ${|T_{x_0}(f)|: ||f||_{infty} leq 1}$ is bounded above by $1$.
Assume some $a$ such that $0 leq a <1$ is the least upper bound. How can I derive a contradiction?
real-analysis functional-analysis analysis normed-spaces
edited Nov 29 '18 at 23:10
José Carlos Santos
160k22126232
asked Nov 29 '18 at 23:05
bbwbbw
50038
$endgroup$
1
$begingroup$
Explicitly find $x_0$ and $f$ so that $|T_{x_0}(f)| = 1$ i.e. $f(x_0) = 1$ for some $x_0 in [0,1]$ and $f in C[0,1]$ Then, $1$ belongs to the set you are trying to bound, hence will definitely be the supremum(maximum).
$endgroup$
– астон вілла олоф мэллбэрг
Nov 29 '18 at 23:09
add a comment |
$begingroup$
Let $x_0 in [0,1]$. Define $T_{x_0}:C[0,1] rightarrow mathbb{R}$ by $T_{x_0}(f)=f(x_0)$.
$||T||:= sup{|T_{x_0}(f)|: ||f||_{infty} leq 1}$ where $||f||_{infty}:=max{|f(x)|:x in [0,1]}$.
Prove that $||T||=1$.
For every $f in C[0,1]$, if $||f||_infty leq 1$ then $|f(x)| leq 1$ for all $x in [0,1]$. In particular, $|f(x_0)| leq 1$. So the set ${|T_{x_0}(f)|: ||f||_{infty} leq 1}$ is bounded above by $1$.
Assume some $a$ such that $0 leq a <1$ is the least upper bound. How can I derive a contradiction?
real-analysis functional-analysis analysis normed-spaces
edited Nov 29 '18 at 23:10
José Carlos Santos
160k22126232
asked Nov 29 '18 at 23:05
bbwbbw
50038
$endgroup$
Let $x_0 in [0,1]$. Define $T_{x_0}:C[0,1] rightarrow mathbb{R}$ by $T_{x_0}(f)=f(x_0)$.
$||T||:= sup{|T_{x_0}(f)|: ||f||_{infty} leq 1}$ where $||f||_{infty}:=max{|f(x)|:x in [0,1]}$.
Prove that $||T||=1$.
For every $f in C[0,1]$, if $||f||_infty leq 1$ then $|f(x)| leq 1$ for all $x in [0,1]$. In particular, $|f(x_0)| leq 1$. So the set ${|T_{x_0}(f)|: ||f||_{infty} leq 1}$ is bounded above by $1$.
Assume some $a$ such that $0 leq a <1$ is the least upper bound. How can I derive a contradiction?
real-analysis functional-analysis analysis normed-spaces
real-analysis functional-analysis analysis normed-spaces
edited Nov 29 '18 at 23:10
José Carlos Santos
160k22126232
asked Nov 29 '18 at 23:05
bbwbbw
50038
edited Nov 29 '18 at 23:10
José Carlos Santos
160k22126232
asked Nov 29 '18 at 23:05
bbwbbw
50038
edited Nov 29 '18 at 23:10
José Carlos Santos
160k22126232
edited Nov 29 '18 at 23:10
José Carlos Santos
160k22126232
edited Nov 29 '18 at 23:10
José Carlos Santos
160k22126232
160k22126232
asked Nov 29 '18 at 23:05
bbwbbw
50038
asked Nov 29 '18 at 23:05
bbwbbw
50038
asked Nov 29 '18 at 23:05
bbwbbw
50038
50038
1
$begingroup$
Explicitly find $x_0$ and $f$ so that $|T_{x_0}(f)| = 1$ i.e. $f(x_0) = 1$ for some $x_0 in [0,1]$ and $f in C[0,1]$ Then, $1$ belongs to the set you are trying to bound, hence will definitely be the supremum(maximum).
$endgroup$
– астон вілла олоф мэллбэрг
Nov 29 '18 at 23:09
add a comment |
1
$begingroup$
Explicitly find $x_0$ and $f$ so that $|T_{x_0}(f)| = 1$ i.e. $f(x_0) = 1$ for some $x_0 in [0,1]$ and $f in C[0,1]$ Then, $1$ belongs to the set you are trying to bound, hence will definitely be the supremum(maximum).
$endgroup$
– астон вілла олоф мэллбэрг
Nov 29 '18 at 23:09
1
1
$begingroup$
Explicitly find $x_0$ and $f$ so that $|T_{x_0}(f)| = 1$ i.e. $f(x_0) = 1$ for some $x_0 in [0,1]$ and $f in C[0,1]$ Then, $1$ belongs to the set you are trying to bound, hence will definitely be the supremum(maximum).
$endgroup$
– астон вілла олоф мэллбэрг
Nov 29 '18 at 23:09
$begingroup$
Explicitly find $x_0$ and $f$ so that $|T_{x_0}(f)| = 1$ i.e. $f(x_0) = 1$ for some $x_0 in [0,1]$ and $f in C[0,1]$ Then, $1$ belongs to the set you are trying to bound, hence will definitely be the supremum(maximum).
$endgroup$
– астон вілла олоф мэллбэрг
Nov 29 '18 at 23:09
add a comment |
1 Answer
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$begingroup$
Suppose that $f$ is the constant function $1$. Then $lVert frVert=1$ and $bigllvert f(x_0)bigrrvert=1$. So, $lVert T_{x_0}rVert=1$.
answered Nov 29 '18 at 23:08
José Carlos SantosJosé Carlos Santos
160k22126232
$endgroup$
add a comment |
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$begingroup$
Suppose that $f$ is the constant function $1$. Then $lVert frVert=1$ and $bigllvert f(x_0)bigrrvert=1$. So, $lVert T_{x_0}rVert=1$.
answered Nov 29 '18 at 23:08
José Carlos SantosJosé Carlos Santos
160k22126232
$endgroup$
add a comment |
$begingroup$
Suppose that $f$ is the constant function $1$. Then $lVert frVert=1$ and $bigllvert f(x_0)bigrrvert=1$. So, $lVert T_{x_0}rVert=1$.
answered Nov 29 '18 at 23:08
José Carlos SantosJosé Carlos Santos
160k22126232
$endgroup$
add a comment |
$begingroup$
Suppose that $f$ is the constant function $1$. Then $lVert frVert=1$ and $bigllvert f(x_0)bigrrvert=1$. So, $lVert T_{x_0}rVert=1$.
answered Nov 29 '18 at 23:08
José Carlos SantosJosé Carlos Santos
160k22126232
$endgroup$
Suppose that $f$ is the constant function $1$. Then $lVert frVert=1$ and $bigllvert f(x_0)bigrrvert=1$. So, $lVert T_{x_0}rVert=1$.
answered Nov 29 '18 at 23:08
José Carlos SantosJosé Carlos Santos
160k22126232
answered Nov 29 '18 at 23:08
José Carlos SantosJosé Carlos Santos
160k22126232
answered Nov 29 '18 at 23:08
José Carlos SantosJosé Carlos Santos
160k22126232
answered Nov 29 '18 at 23:08
José Carlos SantosJosé Carlos Santos
160k22126232
160k22126232
add a comment |
add a comment |
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1
$begingroup$
Explicitly find $x_0$ and $f$ so that $|T_{x_0}(f)| = 1$ i.e. $f(x_0) = 1$ for some $x_0 in [0,1]$ and $f in C[0,1]$ Then, $1$ belongs to the set you are trying to bound, hence will definitely be the supremum(maximum).
$endgroup$
– астон вілла олоф мэллбэрг
Nov 29 '18 at 23:09