A sequence of orthogonal projection in Hilbert spaceFinite dimensional approximations of operators on Hilbert...



A sequence of orthogonal projection in Hilbert space


Finite dimensional approximations of operators on Hilbert spacesDoes the Laplace-Beltrami/surface gradient commute with orthogonal projection? (related to Galerkin method)Existence of a projection operator onto subspace of Hilbert spaceBoundedness of a Hilbert space projection mapA double sequence in a Banach spaceA homeomorphism between the unit interval $[0,1]$ and a linearly independent subset of a Hilbert spaceThe closure of span of a linearly independent and convergent sequence in $ell^2$A nested sequence of closed subspaces of $ell^2$Limit of sequence of vectors in $ell^2$ with coefficients approaching $0$Almost orthonormal projection and orthonormal projection in Hilbert space













1












$begingroup$


Let $H$ be an infinite dimensional Hilbert space over $mathbb{C}$



Let ${v_n}_{n in mathbb{N}} subset H$ be a sequence of linearly independent vectors in $H$ such that $v_n to u$



Let $forall m in mathbb{N}: V_m = operatorname{span} {v_n}_{n geq m}$ and $P_m$ be the orthogonal projection on $V_m$



My question is if it is true that:
$$
forall v in V_1:
lim_{m to infty}
P_m(v)= a cdot u
$$
in $H$-norm and with $a in mathbb{C}$



Thanks.






fa.functional-analysis hilbert-spaces







share|cite|improve this question









$endgroup$












  • $begingroup$
    I suppose that $V_m$ is supposed to be the closed span of ${v_n: , n ge m}$? Since otherwise there is no orthogonal projection onto $V_m$, in general.
    $endgroup$
    – Jochen Glueck
    2 hours ago
















1












$begingroup$


Let $H$ be an infinite dimensional Hilbert space over $mathbb{C}$



Let ${v_n}_{n in mathbb{N}} subset H$ be a sequence of linearly independent vectors in $H$ such that $v_n to u$



Let $forall m in mathbb{N}: V_m = operatorname{span} {v_n}_{n geq m}$ and $P_m$ be the orthogonal projection on $V_m$



My question is if it is true that:
$$
forall v in V_1:
lim_{m to infty}
P_m(v)= a cdot u
$$
in $H$-norm and with $a in mathbb{C}$



Thanks.






fa.functional-analysis hilbert-spaces







share|cite|improve this question









$endgroup$












  • $begingroup$
    I suppose that $V_m$ is supposed to be the closed span of ${v_n: , n ge m}$? Since otherwise there is no orthogonal projection onto $V_m$, in general.
    $endgroup$
    – Jochen Glueck
    2 hours ago














1












1








1





$begingroup$


Let $H$ be an infinite dimensional Hilbert space over $mathbb{C}$



Let ${v_n}_{n in mathbb{N}} subset H$ be a sequence of linearly independent vectors in $H$ such that $v_n to u$



Let $forall m in mathbb{N}: V_m = operatorname{span} {v_n}_{n geq m}$ and $P_m$ be the orthogonal projection on $V_m$



My question is if it is true that:
$$
forall v in V_1:
lim_{m to infty}
P_m(v)= a cdot u
$$
in $H$-norm and with $a in mathbb{C}$



Thanks.






fa.functional-analysis hilbert-spaces







share|cite|improve this question









$endgroup$




Let $H$ be an infinite dimensional Hilbert space over $mathbb{C}$



Let ${v_n}_{n in mathbb{N}} subset H$ be a sequence of linearly independent vectors in $H$ such that $v_n to u$



Let $forall m in mathbb{N}: V_m = operatorname{span} {v_n}_{n geq m}$ and $P_m$ be the orthogonal projection on $V_m$



My question is if it is true that:
$$
forall v in V_1:
lim_{m to infty}
P_m(v)= a cdot u
$$
in $H$-norm and with $a in mathbb{C}$



Thanks.






fa.functional-analysis hilbert-spaces




fa.functional-analysis hilbert-spaces






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 4 hours ago









Matey MathMatey Math

21817




21817












  • $begingroup$
    I suppose that $V_m$ is supposed to be the closed span of ${v_n: , n ge m}$? Since otherwise there is no orthogonal projection onto $V_m$, in general.
    $endgroup$
    – Jochen Glueck
    2 hours ago


















  • $begingroup$
    I suppose that $V_m$ is supposed to be the closed span of ${v_n: , n ge m}$? Since otherwise there is no orthogonal projection onto $V_m$, in general.
    $endgroup$
    – Jochen Glueck
    2 hours ago
















$begingroup$
I suppose that $V_m$ is supposed to be the closed span of ${v_n: , n ge m}$? Since otherwise there is no orthogonal projection onto $V_m$, in general.
$endgroup$
– Jochen Glueck
2 hours ago




$begingroup$
I suppose that $V_m$ is supposed to be the closed span of ${v_n: , n ge m}$? Since otherwise there is no orthogonal projection onto $V_m$, in general.
$endgroup$
– Jochen Glueck
2 hours ago










1 Answer
1






active

oldest

votes


















2












$begingroup$

The answer is no, in general.



As a counterexample, let $H = L^2([0,1])$, let $(q_n)_{n in mathbb{N}}$ be your favourite enumeration of $[0,1] cap mathbb{Q}$ and define
begin{align*}
v_n := 1 + frac{1}{n} 1_{[0,q_n]}
end{align*}
for each $n in mathbb{N}$.



Then the span of ${v_n: , n ge m}$ is dense in $L^2([0,1])$ for each $m$ (since ${q_n: , n ge m}$ is dense in $[0,1]$) and hence $P_m$ is the identity operator. However, $v_n$ converges to the constant function with value $1$ as $n to infty$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    ok @JochenGlueck thanks for your answer
    $endgroup$
    – Matey Math
    2 hours ago











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

The answer is no, in general.



As a counterexample, let $H = L^2([0,1])$, let $(q_n)_{n in mathbb{N}}$ be your favourite enumeration of $[0,1] cap mathbb{Q}$ and define
begin{align*}
v_n := 1 + frac{1}{n} 1_{[0,q_n]}
end{align*}
for each $n in mathbb{N}$.



Then the span of ${v_n: , n ge m}$ is dense in $L^2([0,1])$ for each $m$ (since ${q_n: , n ge m}$ is dense in $[0,1]$) and hence $P_m$ is the identity operator. However, $v_n$ converges to the constant function with value $1$ as $n to infty$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    ok @JochenGlueck thanks for your answer
    $endgroup$
    – Matey Math
    2 hours ago
















2












$begingroup$

The answer is no, in general.



As a counterexample, let $H = L^2([0,1])$, let $(q_n)_{n in mathbb{N}}$ be your favourite enumeration of $[0,1] cap mathbb{Q}$ and define
begin{align*}
v_n := 1 + frac{1}{n} 1_{[0,q_n]}
end{align*}
for each $n in mathbb{N}$.



Then the span of ${v_n: , n ge m}$ is dense in $L^2([0,1])$ for each $m$ (since ${q_n: , n ge m}$ is dense in $[0,1]$) and hence $P_m$ is the identity operator. However, $v_n$ converges to the constant function with value $1$ as $n to infty$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    ok @JochenGlueck thanks for your answer
    $endgroup$
    – Matey Math
    2 hours ago














2












2








2





$begingroup$

The answer is no, in general.



As a counterexample, let $H = L^2([0,1])$, let $(q_n)_{n in mathbb{N}}$ be your favourite enumeration of $[0,1] cap mathbb{Q}$ and define
begin{align*}
v_n := 1 + frac{1}{n} 1_{[0,q_n]}
end{align*}
for each $n in mathbb{N}$.



Then the span of ${v_n: , n ge m}$ is dense in $L^2([0,1])$ for each $m$ (since ${q_n: , n ge m}$ is dense in $[0,1]$) and hence $P_m$ is the identity operator. However, $v_n$ converges to the constant function with value $1$ as $n to infty$.






share|cite|improve this answer









$endgroup$



The answer is no, in general.



As a counterexample, let $H = L^2([0,1])$, let $(q_n)_{n in mathbb{N}}$ be your favourite enumeration of $[0,1] cap mathbb{Q}$ and define
begin{align*}
v_n := 1 + frac{1}{n} 1_{[0,q_n]}
end{align*}
for each $n in mathbb{N}$.



Then the span of ${v_n: , n ge m}$ is dense in $L^2([0,1])$ for each $m$ (since ${q_n: , n ge m}$ is dense in $[0,1]$) and hence $P_m$ is the identity operator. However, $v_n$ converges to the constant function with value $1$ as $n to infty$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 2 hours ago









Jochen GlueckJochen Glueck

2,99611423




2,99611423












  • $begingroup$
    ok @JochenGlueck thanks for your answer
    $endgroup$
    – Matey Math
    2 hours ago


















  • $begingroup$
    ok @JochenGlueck thanks for your answer
    $endgroup$
    – Matey Math
    2 hours ago
















$begingroup$
ok @JochenGlueck thanks for your answer
$endgroup$
– Matey Math
2 hours ago




$begingroup$
ok @JochenGlueck thanks for your answer
$endgroup$
– Matey Math
2 hours ago


















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