Mixing
Inner product on a sequence and its limit
$begingroup$
I am stuck on a question, and it seems like I'm missing a really obvious Cauchy-Schwarz application or something, but I am left scratching my head.
Let $(x_n):n in mathbb{N}$ be a sequence in a Hilbert space $H$. Let $x$ satisfy $|x_n|to |x|$ and $langle x,x_nrangle to langle x,xrangle$. Show that $x_n to x$.
I have found so far that
$|x_n-x|^2=langle x_n,x_n-xrangle -langle x,x_n-xrangle $, and I know that the rightmost term tends to zero which helps, but I don't know about the first one.
Any solutions? Thanks in advance.
hilbert-spaces banach-spaces inner-product-space
asked Jan 23 at 18:06
MarmosetMarmoset
186
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add a comment |
$begingroup$
I am stuck on a question, and it seems like I'm missing a really obvious Cauchy-Schwarz application or something, but I am left scratching my head.
Let $(x_n):n in mathbb{N}$ be a sequence in a Hilbert space $H$. Let $x$ satisfy $|x_n|to |x|$ and $langle x,x_nrangle to langle x,xrangle$. Show that $x_n to x$.
I have found so far that
$|x_n-x|^2=langle x_n,x_n-xrangle -langle x,x_n-xrangle $, and I know that the rightmost term tends to zero which helps, but I don't know about the first one.
Any solutions? Thanks in advance.
hilbert-spaces banach-spaces inner-product-space
asked Jan 23 at 18:06
MarmosetMarmoset
186
$endgroup$
add a comment |
$begingroup$
I am stuck on a question, and it seems like I'm missing a really obvious Cauchy-Schwarz application or something, but I am left scratching my head.
Let $(x_n):n in mathbb{N}$ be a sequence in a Hilbert space $H$. Let $x$ satisfy $|x_n|to |x|$ and $langle x,x_nrangle to langle x,xrangle$. Show that $x_n to x$.
I have found so far that
$|x_n-x|^2=langle x_n,x_n-xrangle -langle x,x_n-xrangle $, and I know that the rightmost term tends to zero which helps, but I don't know about the first one.
Any solutions? Thanks in advance.
hilbert-spaces banach-spaces inner-product-space
asked Jan 23 at 18:06
MarmosetMarmoset
186
$endgroup$
I am stuck on a question, and it seems like I'm missing a really obvious Cauchy-Schwarz application or something, but I am left scratching my head.
Let $(x_n):n in mathbb{N}$ be a sequence in a Hilbert space $H$. Let $x$ satisfy $|x_n|to |x|$ and $langle x,x_nrangle to langle x,xrangle$. Show that $x_n to x$.
I have found so far that
$|x_n-x|^2=langle x_n,x_n-xrangle -langle x,x_n-xrangle $, and I know that the rightmost term tends to zero which helps, but I don't know about the first one.
Any solutions? Thanks in advance.
hilbert-spaces banach-spaces inner-product-space
hilbert-spaces banach-spaces inner-product-space
asked Jan 23 at 18:06
MarmosetMarmoset
186
asked Jan 23 at 18:06
MarmosetMarmoset
186
asked Jan 23 at 18:06
MarmosetMarmoset
186
asked Jan 23 at 18:06
MarmosetMarmoset
186
asked Jan 23 at 18:06
MarmosetMarmoset
186
186
add a comment |
add a comment |
1 Answer
1
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oldest
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Also the left term tends to zero as it is
$$
langle x_n, x_n - x rangle = ||x_n||^2 -langle x_n,x rangle to ||x||^2 - ||x||^2 = 0
$$
where I used both the hypotheses.
answered Jan 23 at 18:13
FedericoFederico
918313
$endgroup$
$begingroup$
So is it true that $langle x, x_n rangle to langle x,x rangle implies langle x_n,x rangle to langle x,xrangle$? Thanks, this seems like the thing I was missing.
$endgroup$
– Marmoset
Jan 23 at 18:17
$begingroup$
Yes, it is true because $langle x_n,x rangle = langle x, x_n rangle$ as the product is simmetric.
$endgroup$
– Federico
Jan 23 at 18:20
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
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oldest
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active
oldest
votes
$begingroup$
Also the left term tends to zero as it is
$$
langle x_n, x_n - x rangle = ||x_n||^2 -langle x_n,x rangle to ||x||^2 - ||x||^2 = 0
$$
where I used both the hypotheses.
answered Jan 23 at 18:13
FedericoFederico
918313
$endgroup$
$begingroup$
So is it true that $langle x, x_n rangle to langle x,x rangle implies langle x_n,x rangle to langle x,xrangle$? Thanks, this seems like the thing I was missing.
$endgroup$
– Marmoset
Jan 23 at 18:17
$begingroup$
Yes, it is true because $langle x_n,x rangle = langle x, x_n rangle$ as the product is simmetric.
$endgroup$
– Federico
Jan 23 at 18:20
add a comment |
$begingroup$
Also the left term tends to zero as it is
$$
langle x_n, x_n - x rangle = ||x_n||^2 -langle x_n,x rangle to ||x||^2 - ||x||^2 = 0
$$
where I used both the hypotheses.
answered Jan 23 at 18:13
FedericoFederico
918313
$endgroup$
$begingroup$
So is it true that $langle x, x_n rangle to langle x,x rangle implies langle x_n,x rangle to langle x,xrangle$? Thanks, this seems like the thing I was missing.
$endgroup$
– Marmoset
Jan 23 at 18:17
$begingroup$
Yes, it is true because $langle x_n,x rangle = langle x, x_n rangle$ as the product is simmetric.
$endgroup$
– Federico
Jan 23 at 18:20
add a comment |
$begingroup$
Also the left term tends to zero as it is
$$
langle x_n, x_n - x rangle = ||x_n||^2 -langle x_n,x rangle to ||x||^2 - ||x||^2 = 0
$$
where I used both the hypotheses.
answered Jan 23 at 18:13
FedericoFederico
918313
$endgroup$
Also the left term tends to zero as it is
$$
langle x_n, x_n - x rangle = ||x_n||^2 -langle x_n,x rangle to ||x||^2 - ||x||^2 = 0
$$
where I used both the hypotheses.
answered Jan 23 at 18:13
FedericoFederico
918313
answered Jan 23 at 18:13
FedericoFederico
918313
answered Jan 23 at 18:13
FedericoFederico
918313
answered Jan 23 at 18:13
FedericoFederico
918313
918313
$begingroup$
So is it true that $langle x, x_n rangle to langle x,x rangle implies langle x_n,x rangle to langle x,xrangle$? Thanks, this seems like the thing I was missing.
$endgroup$
– Marmoset
Jan 23 at 18:17
$begingroup$
Yes, it is true because $langle x_n,x rangle = langle x, x_n rangle$ as the product is simmetric.
$endgroup$
– Federico
Jan 23 at 18:20
add a comment |
$begingroup$
So is it true that $langle x, x_n rangle to langle x,x rangle implies langle x_n,x rangle to langle x,xrangle$? Thanks, this seems like the thing I was missing.
$endgroup$
– Marmoset
Jan 23 at 18:17
$begingroup$
Yes, it is true because $langle x_n,x rangle = langle x, x_n rangle$ as the product is simmetric.
$endgroup$
– Federico
Jan 23 at 18:20
$begingroup$
So is it true that $langle x, x_n rangle to langle x,x rangle implies langle x_n,x rangle to langle x,xrangle$? Thanks, this seems like the thing I was missing.
$endgroup$
– Marmoset
Jan 23 at 18:17
$begingroup$
So is it true that $langle x, x_n rangle to langle x,x rangle implies langle x_n,x rangle to langle x,xrangle$? Thanks, this seems like the thing I was missing.
$endgroup$
– Marmoset
Jan 23 at 18:17
$begingroup$
Yes, it is true because $langle x_n,x rangle = langle x, x_n rangle$ as the product is simmetric.
$endgroup$
– Federico
Jan 23 at 18:20
$begingroup$
Yes, it is true because $langle x_n,x rangle = langle x, x_n rangle$ as the product is simmetric.
$endgroup$
– Federico
Jan 23 at 18:20
add a comment |
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