Bounded operator but not Compact
Multi tool use
$begingroup$
Let $T : (C([-1,1]),||.||_{infty}) rightarrow (C([-1,1]),||.||_{infty}) $
Such as : $(Tf)(x)=frac{1}{2}(f(x)+f(-x))$ . For all $fin C([-1,1])$
Why $T$ isn't Compact ?
I tried to use the sequence $f_n(x)=x^n$. For $xin [-1,1]$.
But I couldn't prove that $(T(f_n))_n$ has no convergent subsequence.
functional-analysis operator-theory compact-operators unbounded-operators
asked Dec 2 '18 at 17:10
Anas BOUALIIAnas BOUALII
1337
$endgroup$
add a comment |
$begingroup$
Let $T : (C([-1,1]),||.||_{infty}) rightarrow (C([-1,1]),||.||_{infty}) $
Such as : $(Tf)(x)=frac{1}{2}(f(x)+f(-x))$ . For all $fin C([-1,1])$
Why $T$ isn't Compact ?
I tried to use the sequence $f_n(x)=x^n$. For $xin [-1,1]$.
But I couldn't prove that $(T(f_n))_n$ has no convergent subsequence.
functional-analysis operator-theory compact-operators unbounded-operators
asked Dec 2 '18 at 17:10
Anas BOUALIIAnas BOUALII
1337
$endgroup$
1
$begingroup$
It has a convergent subsequence: $lim_{ntoinfty}T(f_{2n-1})=0$.
$endgroup$
– José Carlos Santos
Dec 2 '18 at 17:15
add a comment |
$begingroup$
Let $T : (C([-1,1]),||.||_{infty}) rightarrow (C([-1,1]),||.||_{infty}) $
Such as : $(Tf)(x)=frac{1}{2}(f(x)+f(-x))$ . For all $fin C([-1,1])$
Why $T$ isn't Compact ?
I tried to use the sequence $f_n(x)=x^n$. For $xin [-1,1]$.
But I couldn't prove that $(T(f_n))_n$ has no convergent subsequence.
functional-analysis operator-theory compact-operators unbounded-operators
asked Dec 2 '18 at 17:10
Anas BOUALIIAnas BOUALII
1337
$endgroup$
Let $T : (C([-1,1]),||.||_{infty}) rightarrow (C([-1,1]),||.||_{infty}) $
Such as : $(Tf)(x)=frac{1}{2}(f(x)+f(-x))$ . For all $fin C([-1,1])$
Why $T$ isn't Compact ?
I tried to use the sequence $f_n(x)=x^n$. For $xin [-1,1]$.
But I couldn't prove that $(T(f_n))_n$ has no convergent subsequence.
functional-analysis operator-theory compact-operators unbounded-operators
functional-analysis operator-theory compact-operators unbounded-operators
asked Dec 2 '18 at 17:10
Anas BOUALIIAnas BOUALII
1337
asked Dec 2 '18 at 17:10
Anas BOUALIIAnas BOUALII
1337
asked Dec 2 '18 at 17:10
Anas BOUALIIAnas BOUALII
1337
asked Dec 2 '18 at 17:10
Anas BOUALIIAnas BOUALII
1337
asked Dec 2 '18 at 17:10
Anas BOUALIIAnas BOUALII
1337
1337
1
$begingroup$
It has a convergent subsequence: $lim_{ntoinfty}T(f_{2n-1})=0$.
$endgroup$
– José Carlos Santos
Dec 2 '18 at 17:15
add a comment |
1
$begingroup$
It has a convergent subsequence: $lim_{ntoinfty}T(f_{2n-1})=0$.
$endgroup$
– José Carlos Santos
Dec 2 '18 at 17:15
1
1
$begingroup$
It has a convergent subsequence: $lim_{ntoinfty}T(f_{2n-1})=0$.
$endgroup$
– José Carlos Santos
Dec 2 '18 at 17:15
$begingroup$
It has a convergent subsequence: $lim_{ntoinfty}T(f_{2n-1})=0$.
$endgroup$
– José Carlos Santos
Dec 2 '18 at 17:15
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hint: Instead, try
$$ f_n(x) = x^{2n}, $$
and observe that any subsequence of $Tf_n=f_n$ converges pointwise to a discontinuous function.
answered Dec 2 '18 at 17:13
MisterRiemannMisterRiemann
5,8451624
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
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$begingroup$
Hint: Instead, try
$$ f_n(x) = x^{2n}, $$
and observe that any subsequence of $Tf_n=f_n$ converges pointwise to a discontinuous function.
answered Dec 2 '18 at 17:13
MisterRiemannMisterRiemann
5,8451624
$endgroup$
add a comment |
$begingroup$
Hint: Instead, try
$$ f_n(x) = x^{2n}, $$
and observe that any subsequence of $Tf_n=f_n$ converges pointwise to a discontinuous function.
answered Dec 2 '18 at 17:13
MisterRiemannMisterRiemann
5,8451624
$endgroup$
add a comment |
$begingroup$
Hint: Instead, try
$$ f_n(x) = x^{2n}, $$
and observe that any subsequence of $Tf_n=f_n$ converges pointwise to a discontinuous function.
answered Dec 2 '18 at 17:13
MisterRiemannMisterRiemann
5,8451624
$endgroup$
Hint: Instead, try
$$ f_n(x) = x^{2n}, $$
and observe that any subsequence of $Tf_n=f_n$ converges pointwise to a discontinuous function.
answered Dec 2 '18 at 17:13
MisterRiemannMisterRiemann
5,8451624
answered Dec 2 '18 at 17:13
MisterRiemannMisterRiemann
5,8451624
answered Dec 2 '18 at 17:13
MisterRiemannMisterRiemann
5,8451624
answered Dec 2 '18 at 17:13
MisterRiemannMisterRiemann
5,8451624
5,8451624
add a comment |
add a comment |
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1
$begingroup$
It has a convergent subsequence: $lim_{ntoinfty}T(f_{2n-1})=0$.
$endgroup$
– José Carlos Santos
Dec 2 '18 at 17:15