How to show the continuity of the following operator?
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Let $T:mathcal{l}^2tomathcal{l}^2$ be an operator with
$$(Tu)_n:= frac{1}{2}u_n + frac{1}{n^2}u_n.$$
I've found that the operator is not compact, since
$$(Tu)_n = left(frac{1}{2}+frac{1}{n^2}right)u_n$$
consists a sequence which converges not to zero. But how can I show whether the operator is continuous?
functional-analysis continuity compact-operators
asked Nov 21 at 14:43
MathCracky
445212
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up vote
0
down vote
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Let $T:mathcal{l}^2tomathcal{l}^2$ be an operator with
$$(Tu)_n:= frac{1}{2}u_n + frac{1}{n^2}u_n.$$
I've found that the operator is not compact, since
$$(Tu)_n = left(frac{1}{2}+frac{1}{n^2}right)u_n$$
consists a sequence which converges not to zero. But how can I show whether the operator is continuous?
functional-analysis continuity compact-operators
asked Nov 21 at 14:43
MathCracky
445212
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $T:mathcal{l}^2tomathcal{l}^2$ be an operator with
$$(Tu)_n:= frac{1}{2}u_n + frac{1}{n^2}u_n.$$
I've found that the operator is not compact, since
$$(Tu)_n = left(frac{1}{2}+frac{1}{n^2}right)u_n$$
consists a sequence which converges not to zero. But how can I show whether the operator is continuous?
functional-analysis continuity compact-operators
asked Nov 21 at 14:43
MathCracky
445212
Let $T:mathcal{l}^2tomathcal{l}^2$ be an operator with
$$(Tu)_n:= frac{1}{2}u_n + frac{1}{n^2}u_n.$$
I've found that the operator is not compact, since
$$(Tu)_n = left(frac{1}{2}+frac{1}{n^2}right)u_n$$
consists a sequence which converges not to zero. But how can I show whether the operator is continuous?
functional-analysis continuity compact-operators
functional-analysis continuity compact-operators
asked Nov 21 at 14:43
MathCracky
445212
asked Nov 21 at 14:43
MathCracky
445212
asked Nov 21 at 14:43
MathCracky
445212
asked Nov 21 at 14:43
MathCracky
445212
asked Nov 21 at 14:43
MathCracky
445212
445212
add a comment |
add a comment |
1 Answer
1
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Suppose that $u^{(k)} to u$ and $Tu^{(k)} to v$ in $ell^2$. Then, for all $ngeq 1$,
begin{align}
|(Tu)_n-v_n|&leq |(Tu)_n - (Tu^{(k)})_n|+|(Tu^{(k)})_n-v_n|\
&= left(frac{1}{2}+frac{1}{n^2}right)|u_n-u^{(k)}_n|+|(Tu^{(k)})_n-v_n|\
&leq frac{3}{2}Vert u-u^{(k)} Vert_2 + Vert Tu^{(k)}-v Vert_2 to 0,
end{align}
as $ktoinfty$,
so $(Tu)_n = v_n$ for each $ngeq 1$, and thus $Tu=v$.
The Closed graph theorem does the rest.
Alternatively, observe that
$$ Vert Tu Vert_2^2 = sum_{n=1}^infty left(frac{1}{2}+frac{1}{n^2}right)^2 |u_n|^2 leq frac{9}{4} sum_{n=1}^infty |u_n|^2 = frac{9}{4} Vert uVert_2^2. $$
answered Nov 21 at 14:57
MisterRiemann
5,7041624
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Suppose that $u^{(k)} to u$ and $Tu^{(k)} to v$ in $ell^2$. Then, for all $ngeq 1$,
begin{align}
|(Tu)_n-v_n|&leq |(Tu)_n - (Tu^{(k)})_n|+|(Tu^{(k)})_n-v_n|\
&= left(frac{1}{2}+frac{1}{n^2}right)|u_n-u^{(k)}_n|+|(Tu^{(k)})_n-v_n|\
&leq frac{3}{2}Vert u-u^{(k)} Vert_2 + Vert Tu^{(k)}-v Vert_2 to 0,
end{align}
as $ktoinfty$,
so $(Tu)_n = v_n$ for each $ngeq 1$, and thus $Tu=v$.
The Closed graph theorem does the rest.
Alternatively, observe that
$$ Vert Tu Vert_2^2 = sum_{n=1}^infty left(frac{1}{2}+frac{1}{n^2}right)^2 |u_n|^2 leq frac{9}{4} sum_{n=1}^infty |u_n|^2 = frac{9}{4} Vert uVert_2^2. $$
answered Nov 21 at 14:57
MisterRiemann
5,7041624
add a comment |
up vote
1
down vote
accepted
Suppose that $u^{(k)} to u$ and $Tu^{(k)} to v$ in $ell^2$. Then, for all $ngeq 1$,
begin{align}
|(Tu)_n-v_n|&leq |(Tu)_n - (Tu^{(k)})_n|+|(Tu^{(k)})_n-v_n|\
&= left(frac{1}{2}+frac{1}{n^2}right)|u_n-u^{(k)}_n|+|(Tu^{(k)})_n-v_n|\
&leq frac{3}{2}Vert u-u^{(k)} Vert_2 + Vert Tu^{(k)}-v Vert_2 to 0,
end{align}
as $ktoinfty$,
so $(Tu)_n = v_n$ for each $ngeq 1$, and thus $Tu=v$.
The Closed graph theorem does the rest.
Alternatively, observe that
$$ Vert Tu Vert_2^2 = sum_{n=1}^infty left(frac{1}{2}+frac{1}{n^2}right)^2 |u_n|^2 leq frac{9}{4} sum_{n=1}^infty |u_n|^2 = frac{9}{4} Vert uVert_2^2. $$
answered Nov 21 at 14:57
MisterRiemann
5,7041624
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Suppose that $u^{(k)} to u$ and $Tu^{(k)} to v$ in $ell^2$. Then, for all $ngeq 1$,
begin{align}
|(Tu)_n-v_n|&leq |(Tu)_n - (Tu^{(k)})_n|+|(Tu^{(k)})_n-v_n|\
&= left(frac{1}{2}+frac{1}{n^2}right)|u_n-u^{(k)}_n|+|(Tu^{(k)})_n-v_n|\
&leq frac{3}{2}Vert u-u^{(k)} Vert_2 + Vert Tu^{(k)}-v Vert_2 to 0,
end{align}
as $ktoinfty$,
so $(Tu)_n = v_n$ for each $ngeq 1$, and thus $Tu=v$.
The Closed graph theorem does the rest.
Alternatively, observe that
$$ Vert Tu Vert_2^2 = sum_{n=1}^infty left(frac{1}{2}+frac{1}{n^2}right)^2 |u_n|^2 leq frac{9}{4} sum_{n=1}^infty |u_n|^2 = frac{9}{4} Vert uVert_2^2. $$
answered Nov 21 at 14:57
MisterRiemann
5,7041624
Suppose that $u^{(k)} to u$ and $Tu^{(k)} to v$ in $ell^2$. Then, for all $ngeq 1$,
begin{align}
|(Tu)_n-v_n|&leq |(Tu)_n - (Tu^{(k)})_n|+|(Tu^{(k)})_n-v_n|\
&= left(frac{1}{2}+frac{1}{n^2}right)|u_n-u^{(k)}_n|+|(Tu^{(k)})_n-v_n|\
&leq frac{3}{2}Vert u-u^{(k)} Vert_2 + Vert Tu^{(k)}-v Vert_2 to 0,
end{align}
as $ktoinfty$,
so $(Tu)_n = v_n$ for each $ngeq 1$, and thus $Tu=v$.
The Closed graph theorem does the rest.
Alternatively, observe that
$$ Vert Tu Vert_2^2 = sum_{n=1}^infty left(frac{1}{2}+frac{1}{n^2}right)^2 |u_n|^2 leq frac{9}{4} sum_{n=1}^infty |u_n|^2 = frac{9}{4} Vert uVert_2^2. $$
answered Nov 21 at 14:57
MisterRiemann
5,7041624
edited Nov 21 at 15:05
answered Nov 21 at 14:57
MisterRiemann
5,7041624
answered Nov 21 at 14:57
MisterRiemann
5,7041624
answered Nov 21 at 14:57
MisterRiemann
5,7041624
5,7041624
add a comment |
add a comment |
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