Mixing
${u_{k}}$ is a sequence in $mathbb{R}^{n}$ that converges to $u$. Prove that $lim_{ktoinfty} langle u_{k}, v...
$begingroup$
Let ${u_{k}}$ be a sequence in $mathbb{R}^{n}$ that converges to
$u$. Prove that $lim_{ktoinfty} langle u_{k}, v rangle = langle
u, v rangle$
Here, $langle u, v rangle$ denotes the inner product of $u$ and $v$.
My attempt:
Okay, so I want to show $forall epsilon > 0$, there exists some index $K$ such that
$$left|langle u_{k}, vrangle - langle u, vrangleright| < epsilon $$
for all $k geq K$. Using the definition of an inner product, we can write
$$left|langle u_{k}, vrangle - langle u, vrangleright| = left|sum_{i=1}^{n} u_{k_{i}} v_{i} - sum_{i=1}^{n}u_{i}v_{i} right| = left|sum_{i=1}^{n}v_{i}(u_{k_{i}} - u_{i}) right| = |langle u_{k} - u, vrangle|$$
I don't know how to finish from here. I think we need to use the fact that $lim_{ktoinfty u_{k}} = u$. Since we know ${u_{k}}$ converges to $u$, for all $epsilon > 0$, there is some $K'$ so that for $k geq K'$, the quantity $|u_{k} - u_{i}|$ is less than $epsilon$.
real-analysis
asked Jan 17 at 23:43
stackofhay42stackofhay42
2267
$endgroup$
add a comment |
$begingroup$
Let ${u_{k}}$ be a sequence in $mathbb{R}^{n}$ that converges to
$u$. Prove that $lim_{ktoinfty} langle u_{k}, v rangle = langle
u, v rangle$
Here, $langle u, v rangle$ denotes the inner product of $u$ and $v$.
My attempt:
Okay, so I want to show $forall epsilon > 0$, there exists some index $K$ such that
$$left|langle u_{k}, vrangle - langle u, vrangleright| < epsilon $$
for all $k geq K$. Using the definition of an inner product, we can write
$$left|langle u_{k}, vrangle - langle u, vrangleright| = left|sum_{i=1}^{n} u_{k_{i}} v_{i} - sum_{i=1}^{n}u_{i}v_{i} right| = left|sum_{i=1}^{n}v_{i}(u_{k_{i}} - u_{i}) right| = |langle u_{k} - u, vrangle|$$
I don't know how to finish from here. I think we need to use the fact that $lim_{ktoinfty u_{k}} = u$. Since we know ${u_{k}}$ converges to $u$, for all $epsilon > 0$, there is some $K'$ so that for $k geq K'$, the quantity $|u_{k} - u_{i}|$ is less than $epsilon$.
real-analysis
asked Jan 17 at 23:43
stackofhay42stackofhay42
2267
$endgroup$
$begingroup$
Triangle inequality. Note each component goes to $0$.
$endgroup$
– Zachary Selk
Jan 17 at 23:45
add a comment |
$begingroup$
Let ${u_{k}}$ be a sequence in $mathbb{R}^{n}$ that converges to
$u$. Prove that $lim_{ktoinfty} langle u_{k}, v rangle = langle
u, v rangle$
Here, $langle u, v rangle$ denotes the inner product of $u$ and $v$.
My attempt:
Okay, so I want to show $forall epsilon > 0$, there exists some index $K$ such that
$$left|langle u_{k}, vrangle - langle u, vrangleright| < epsilon $$
for all $k geq K$. Using the definition of an inner product, we can write
$$left|langle u_{k}, vrangle - langle u, vrangleright| = left|sum_{i=1}^{n} u_{k_{i}} v_{i} - sum_{i=1}^{n}u_{i}v_{i} right| = left|sum_{i=1}^{n}v_{i}(u_{k_{i}} - u_{i}) right| = |langle u_{k} - u, vrangle|$$
I don't know how to finish from here. I think we need to use the fact that $lim_{ktoinfty u_{k}} = u$. Since we know ${u_{k}}$ converges to $u$, for all $epsilon > 0$, there is some $K'$ so that for $k geq K'$, the quantity $|u_{k} - u_{i}|$ is less than $epsilon$.
real-analysis
asked Jan 17 at 23:43
stackofhay42stackofhay42
2267
$endgroup$
Let ${u_{k}}$ be a sequence in $mathbb{R}^{n}$ that converges to
$u$. Prove that $lim_{ktoinfty} langle u_{k}, v rangle = langle
u, v rangle$
Here, $langle u, v rangle$ denotes the inner product of $u$ and $v$.
My attempt:
Okay, so I want to show $forall epsilon > 0$, there exists some index $K$ such that
$$left|langle u_{k}, vrangle - langle u, vrangleright| < epsilon $$
for all $k geq K$. Using the definition of an inner product, we can write
$$left|langle u_{k}, vrangle - langle u, vrangleright| = left|sum_{i=1}^{n} u_{k_{i}} v_{i} - sum_{i=1}^{n}u_{i}v_{i} right| = left|sum_{i=1}^{n}v_{i}(u_{k_{i}} - u_{i}) right| = |langle u_{k} - u, vrangle|$$
I don't know how to finish from here. I think we need to use the fact that $lim_{ktoinfty u_{k}} = u$. Since we know ${u_{k}}$ converges to $u$, for all $epsilon > 0$, there is some $K'$ so that for $k geq K'$, the quantity $|u_{k} - u_{i}|$ is less than $epsilon$.
real-analysis
real-analysis
asked Jan 17 at 23:43
stackofhay42stackofhay42
2267
asked Jan 17 at 23:43
stackofhay42stackofhay42
2267
asked Jan 17 at 23:43
stackofhay42stackofhay42
2267
asked Jan 17 at 23:43
stackofhay42stackofhay42
2267
asked Jan 17 at 23:43
stackofhay42stackofhay42
2267
2267
$begingroup$
Triangle inequality. Note each component goes to $0$.
$endgroup$
– Zachary Selk
Jan 17 at 23:45
add a comment |
$begingroup$
Triangle inequality. Note each component goes to $0$.
$endgroup$
– Zachary Selk
Jan 17 at 23:45
$begingroup$
Triangle inequality. Note each component goes to $0$.
$endgroup$
– Zachary Selk
Jan 17 at 23:45
$begingroup$
Triangle inequality. Note each component goes to $0$.
$endgroup$
– Zachary Selk
Jan 17 at 23:45
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Use the linearity of the scalar product and the Cauchy-Schwarz-inequality
answered Jan 17 at 23:56
NicolasNicolas
472
$endgroup$
$begingroup$
I am having trouble finishing
$endgroup$
– stackofhay42
Jan 18 at 0:04
$begingroup$
Where do we use linearity? I can apply C-S Inequality to get $|langle u_{k} - u, vrangle| leq ||u_{k} - u|| cdot ||v|| = text{dist}(u_{k}, u) cdot ||v||$. So, do I define my index to be $K = K' cdot ||v||$, where $K'$ is the index for which $text{dist}(u_{k}, u) < epsilon$ for all $k geq K'$?
$endgroup$
– stackofhay42
Jan 18 at 0:12
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
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votes
$begingroup$
Use the linearity of the scalar product and the Cauchy-Schwarz-inequality
answered Jan 17 at 23:56
NicolasNicolas
472
$endgroup$
$begingroup$
I am having trouble finishing
$endgroup$
– stackofhay42
Jan 18 at 0:04
$begingroup$
Where do we use linearity? I can apply C-S Inequality to get $|langle u_{k} - u, vrangle| leq ||u_{k} - u|| cdot ||v|| = text{dist}(u_{k}, u) cdot ||v||$. So, do I define my index to be $K = K' cdot ||v||$, where $K'$ is the index for which $text{dist}(u_{k}, u) < epsilon$ for all $k geq K'$?
$endgroup$
– stackofhay42
Jan 18 at 0:12
add a comment |
$begingroup$
Use the linearity of the scalar product and the Cauchy-Schwarz-inequality
answered Jan 17 at 23:56
NicolasNicolas
472
$endgroup$
$begingroup$
I am having trouble finishing
$endgroup$
– stackofhay42
Jan 18 at 0:04
$begingroup$
Where do we use linearity? I can apply C-S Inequality to get $|langle u_{k} - u, vrangle| leq ||u_{k} - u|| cdot ||v|| = text{dist}(u_{k}, u) cdot ||v||$. So, do I define my index to be $K = K' cdot ||v||$, where $K'$ is the index for which $text{dist}(u_{k}, u) < epsilon$ for all $k geq K'$?
$endgroup$
– stackofhay42
Jan 18 at 0:12
add a comment |
$begingroup$
Use the linearity of the scalar product and the Cauchy-Schwarz-inequality
answered Jan 17 at 23:56
NicolasNicolas
472
$endgroup$
Use the linearity of the scalar product and the Cauchy-Schwarz-inequality
answered Jan 17 at 23:56
NicolasNicolas
472
answered Jan 17 at 23:56
NicolasNicolas
472
answered Jan 17 at 23:56
NicolasNicolas
472
answered Jan 17 at 23:56
NicolasNicolas
472
472
$begingroup$
I am having trouble finishing
$endgroup$
– stackofhay42
Jan 18 at 0:04
$begingroup$
Where do we use linearity? I can apply C-S Inequality to get $|langle u_{k} - u, vrangle| leq ||u_{k} - u|| cdot ||v|| = text{dist}(u_{k}, u) cdot ||v||$. So, do I define my index to be $K = K' cdot ||v||$, where $K'$ is the index for which $text{dist}(u_{k}, u) < epsilon$ for all $k geq K'$?
$endgroup$
– stackofhay42
Jan 18 at 0:12
add a comment |
$begingroup$
I am having trouble finishing
$endgroup$
– stackofhay42
Jan 18 at 0:04
$begingroup$
Where do we use linearity? I can apply C-S Inequality to get $|langle u_{k} - u, vrangle| leq ||u_{k} - u|| cdot ||v|| = text{dist}(u_{k}, u) cdot ||v||$. So, do I define my index to be $K = K' cdot ||v||$, where $K'$ is the index for which $text{dist}(u_{k}, u) < epsilon$ for all $k geq K'$?
$endgroup$
– stackofhay42
Jan 18 at 0:12
$begingroup$
I am having trouble finishing
$endgroup$
– stackofhay42
Jan 18 at 0:04
$begingroup$
I am having trouble finishing
$endgroup$
– stackofhay42
Jan 18 at 0:04
$begingroup$
Where do we use linearity? I can apply C-S Inequality to get $|langle u_{k} - u, vrangle| leq ||u_{k} - u|| cdot ||v|| = text{dist}(u_{k}, u) cdot ||v||$. So, do I define my index to be $K = K' cdot ||v||$, where $K'$ is the index for which $text{dist}(u_{k}, u) < epsilon$ for all $k geq K'$?
$endgroup$
– stackofhay42
Jan 18 at 0:12
$begingroup$
Where do we use linearity? I can apply C-S Inequality to get $|langle u_{k} - u, vrangle| leq ||u_{k} - u|| cdot ||v|| = text{dist}(u_{k}, u) cdot ||v||$. So, do I define my index to be $K = K' cdot ||v||$, where $K'$ is the index for which $text{dist}(u_{k}, u) < epsilon$ for all $k geq K'$?
$endgroup$
– stackofhay42
Jan 18 at 0:12
add a comment |
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$begingroup$
Triangle inequality. Note each component goes to $0$.
$endgroup$
– Zachary Selk
Jan 17 at 23:45