Showing that $bigg| x - sum_{k=1}^n lambda_ke_kbigg| geq bigg|x-sum_{k=1}^nlangle x,e_krangle e_k bigg|$











up vote
3
down vote

favorite












Exercise :




Let ${e_1,e_2,dots, e_n}$ be an orthonormal set over the Hilbert space $H$. Show that :
$$bigg| x - sum_{k=1}^n lambda_ke_kbigg| geq bigg|x-sum_{k=1}^nlangle x,e_krangle e_k bigg|$$
for every $lambda_1,lambda_2,dots,lambda_n in mathbb R$.




Attempt :



Let $x in H$ and $varepsilon > 0$. Then, we can find $lambda_1,dots, lambda_n in mathbb R$ such that :



$$bigg| x - sum_{k=1}^n lambda_ke_kbigg| < varepsilon$$



But the element $w = x-sum_{k=1}^nlangle x,e_krangle e_k$ achieves the minimum distance of $x$ from the finite dimensional space $ F = langle e_1,dots, e_n rangle$ and then it will be



$$bigg| x-sum_{k=1}^nlangle x,e_krangle e_k bigg| leq bigg| x - sum_{k=1}^n lambda_ke_kbigg| < varepsilon$$



and thus the inequality which we were asked to show is proven.



Question : Is my approach rigorous and legit enough ? If not, any recommendations, hints or elaborations will be greatly appreciated !










share|cite|improve this question




















  • 2




    Just a remark: Since ${e_n}$ is a fixed orthonormal set, you might not be able to make the sum $sum_{k=1}^n lambda_k e_k$ arbitrarily close to $x$ in norm (which your solution seems to suggest since you're using $epsilon$).
    – MisterRiemann
    Dec 2 at 13:08












  • @MisterRiemann If that's the case, how would my proof be altered ? I assume that since ${e_n}$ is orthonormal then it would also be an orthonormal basis of the finite dimensional space $F$.
    – Rebellos
    Dec 2 at 13:11










  • I think that your idea is correct, i.e. you should use the fact that the projection of $x$ onto $F$ is indeed the closest point to $x$ in $F$. I'm just saying that you should remove $<epsilon$ in your argument (I don't even see the necessity for it). But let's wait for an answer from someone more knowledgeable than myself.
    – MisterRiemann
    Dec 2 at 13:13












  • @MisterRiemann I just elaborated using $varepsilon$ to show that the expression has some essence (meaning it wouldnt diverge).
    – Rebellos
    Dec 2 at 13:22






  • 2




    @MisterRiemann I get what you're saying, then it really is straightforward since the other distance is the smallest one anyway (or equal in case of being the same minimum distance).
    – Rebellos
    Dec 2 at 13:29















up vote
3
down vote

favorite












Exercise :




Let ${e_1,e_2,dots, e_n}$ be an orthonormal set over the Hilbert space $H$. Show that :
$$bigg| x - sum_{k=1}^n lambda_ke_kbigg| geq bigg|x-sum_{k=1}^nlangle x,e_krangle e_k bigg|$$
for every $lambda_1,lambda_2,dots,lambda_n in mathbb R$.




Attempt :



Let $x in H$ and $varepsilon > 0$. Then, we can find $lambda_1,dots, lambda_n in mathbb R$ such that :



$$bigg| x - sum_{k=1}^n lambda_ke_kbigg| < varepsilon$$



But the element $w = x-sum_{k=1}^nlangle x,e_krangle e_k$ achieves the minimum distance of $x$ from the finite dimensional space $ F = langle e_1,dots, e_n rangle$ and then it will be



$$bigg| x-sum_{k=1}^nlangle x,e_krangle e_k bigg| leq bigg| x - sum_{k=1}^n lambda_ke_kbigg| < varepsilon$$



and thus the inequality which we were asked to show is proven.



Question : Is my approach rigorous and legit enough ? If not, any recommendations, hints or elaborations will be greatly appreciated !










share|cite|improve this question




















  • 2




    Just a remark: Since ${e_n}$ is a fixed orthonormal set, you might not be able to make the sum $sum_{k=1}^n lambda_k e_k$ arbitrarily close to $x$ in norm (which your solution seems to suggest since you're using $epsilon$).
    – MisterRiemann
    Dec 2 at 13:08












  • @MisterRiemann If that's the case, how would my proof be altered ? I assume that since ${e_n}$ is orthonormal then it would also be an orthonormal basis of the finite dimensional space $F$.
    – Rebellos
    Dec 2 at 13:11










  • I think that your idea is correct, i.e. you should use the fact that the projection of $x$ onto $F$ is indeed the closest point to $x$ in $F$. I'm just saying that you should remove $<epsilon$ in your argument (I don't even see the necessity for it). But let's wait for an answer from someone more knowledgeable than myself.
    – MisterRiemann
    Dec 2 at 13:13












  • @MisterRiemann I just elaborated using $varepsilon$ to show that the expression has some essence (meaning it wouldnt diverge).
    – Rebellos
    Dec 2 at 13:22






  • 2




    @MisterRiemann I get what you're saying, then it really is straightforward since the other distance is the smallest one anyway (or equal in case of being the same minimum distance).
    – Rebellos
    Dec 2 at 13:29













up vote
3
down vote

favorite









up vote
3
down vote

favorite











Exercise :




Let ${e_1,e_2,dots, e_n}$ be an orthonormal set over the Hilbert space $H$. Show that :
$$bigg| x - sum_{k=1}^n lambda_ke_kbigg| geq bigg|x-sum_{k=1}^nlangle x,e_krangle e_k bigg|$$
for every $lambda_1,lambda_2,dots,lambda_n in mathbb R$.




Attempt :



Let $x in H$ and $varepsilon > 0$. Then, we can find $lambda_1,dots, lambda_n in mathbb R$ such that :



$$bigg| x - sum_{k=1}^n lambda_ke_kbigg| < varepsilon$$



But the element $w = x-sum_{k=1}^nlangle x,e_krangle e_k$ achieves the minimum distance of $x$ from the finite dimensional space $ F = langle e_1,dots, e_n rangle$ and then it will be



$$bigg| x-sum_{k=1}^nlangle x,e_krangle e_k bigg| leq bigg| x - sum_{k=1}^n lambda_ke_kbigg| < varepsilon$$



and thus the inequality which we were asked to show is proven.



Question : Is my approach rigorous and legit enough ? If not, any recommendations, hints or elaborations will be greatly appreciated !










share|cite|improve this question















Exercise :




Let ${e_1,e_2,dots, e_n}$ be an orthonormal set over the Hilbert space $H$. Show that :
$$bigg| x - sum_{k=1}^n lambda_ke_kbigg| geq bigg|x-sum_{k=1}^nlangle x,e_krangle e_k bigg|$$
for every $lambda_1,lambda_2,dots,lambda_n in mathbb R$.




Attempt :



Let $x in H$ and $varepsilon > 0$. Then, we can find $lambda_1,dots, lambda_n in mathbb R$ such that :



$$bigg| x - sum_{k=1}^n lambda_ke_kbigg| < varepsilon$$



But the element $w = x-sum_{k=1}^nlangle x,e_krangle e_k$ achieves the minimum distance of $x$ from the finite dimensional space $ F = langle e_1,dots, e_n rangle$ and then it will be



$$bigg| x-sum_{k=1}^nlangle x,e_krangle e_k bigg| leq bigg| x - sum_{k=1}^n lambda_ke_kbigg| < varepsilon$$



and thus the inequality which we were asked to show is proven.



Question : Is my approach rigorous and legit enough ? If not, any recommendations, hints or elaborations will be greatly appreciated !







real-analysis functional-analysis hilbert-spaces normed-spaces inner-product-space






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 2 at 13:08

























asked Dec 2 at 13:05









Rebellos

13k21042




13k21042








  • 2




    Just a remark: Since ${e_n}$ is a fixed orthonormal set, you might not be able to make the sum $sum_{k=1}^n lambda_k e_k$ arbitrarily close to $x$ in norm (which your solution seems to suggest since you're using $epsilon$).
    – MisterRiemann
    Dec 2 at 13:08












  • @MisterRiemann If that's the case, how would my proof be altered ? I assume that since ${e_n}$ is orthonormal then it would also be an orthonormal basis of the finite dimensional space $F$.
    – Rebellos
    Dec 2 at 13:11










  • I think that your idea is correct, i.e. you should use the fact that the projection of $x$ onto $F$ is indeed the closest point to $x$ in $F$. I'm just saying that you should remove $<epsilon$ in your argument (I don't even see the necessity for it). But let's wait for an answer from someone more knowledgeable than myself.
    – MisterRiemann
    Dec 2 at 13:13












  • @MisterRiemann I just elaborated using $varepsilon$ to show that the expression has some essence (meaning it wouldnt diverge).
    – Rebellos
    Dec 2 at 13:22






  • 2




    @MisterRiemann I get what you're saying, then it really is straightforward since the other distance is the smallest one anyway (or equal in case of being the same minimum distance).
    – Rebellos
    Dec 2 at 13:29














  • 2




    Just a remark: Since ${e_n}$ is a fixed orthonormal set, you might not be able to make the sum $sum_{k=1}^n lambda_k e_k$ arbitrarily close to $x$ in norm (which your solution seems to suggest since you're using $epsilon$).
    – MisterRiemann
    Dec 2 at 13:08












  • @MisterRiemann If that's the case, how would my proof be altered ? I assume that since ${e_n}$ is orthonormal then it would also be an orthonormal basis of the finite dimensional space $F$.
    – Rebellos
    Dec 2 at 13:11










  • I think that your idea is correct, i.e. you should use the fact that the projection of $x$ onto $F$ is indeed the closest point to $x$ in $F$. I'm just saying that you should remove $<epsilon$ in your argument (I don't even see the necessity for it). But let's wait for an answer from someone more knowledgeable than myself.
    – MisterRiemann
    Dec 2 at 13:13












  • @MisterRiemann I just elaborated using $varepsilon$ to show that the expression has some essence (meaning it wouldnt diverge).
    – Rebellos
    Dec 2 at 13:22






  • 2




    @MisterRiemann I get what you're saying, then it really is straightforward since the other distance is the smallest one anyway (or equal in case of being the same minimum distance).
    – Rebellos
    Dec 2 at 13:29








2




2




Just a remark: Since ${e_n}$ is a fixed orthonormal set, you might not be able to make the sum $sum_{k=1}^n lambda_k e_k$ arbitrarily close to $x$ in norm (which your solution seems to suggest since you're using $epsilon$).
– MisterRiemann
Dec 2 at 13:08






Just a remark: Since ${e_n}$ is a fixed orthonormal set, you might not be able to make the sum $sum_{k=1}^n lambda_k e_k$ arbitrarily close to $x$ in norm (which your solution seems to suggest since you're using $epsilon$).
– MisterRiemann
Dec 2 at 13:08














@MisterRiemann If that's the case, how would my proof be altered ? I assume that since ${e_n}$ is orthonormal then it would also be an orthonormal basis of the finite dimensional space $F$.
– Rebellos
Dec 2 at 13:11




@MisterRiemann If that's the case, how would my proof be altered ? I assume that since ${e_n}$ is orthonormal then it would also be an orthonormal basis of the finite dimensional space $F$.
– Rebellos
Dec 2 at 13:11












I think that your idea is correct, i.e. you should use the fact that the projection of $x$ onto $F$ is indeed the closest point to $x$ in $F$. I'm just saying that you should remove $<epsilon$ in your argument (I don't even see the necessity for it). But let's wait for an answer from someone more knowledgeable than myself.
– MisterRiemann
Dec 2 at 13:13






I think that your idea is correct, i.e. you should use the fact that the projection of $x$ onto $F$ is indeed the closest point to $x$ in $F$. I'm just saying that you should remove $<epsilon$ in your argument (I don't even see the necessity for it). But let's wait for an answer from someone more knowledgeable than myself.
– MisterRiemann
Dec 2 at 13:13














@MisterRiemann I just elaborated using $varepsilon$ to show that the expression has some essence (meaning it wouldnt diverge).
– Rebellos
Dec 2 at 13:22




@MisterRiemann I just elaborated using $varepsilon$ to show that the expression has some essence (meaning it wouldnt diverge).
– Rebellos
Dec 2 at 13:22




2




2




@MisterRiemann I get what you're saying, then it really is straightforward since the other distance is the smallest one anyway (or equal in case of being the same minimum distance).
– Rebellos
Dec 2 at 13:29




@MisterRiemann I get what you're saying, then it really is straightforward since the other distance is the smallest one anyway (or equal in case of being the same minimum distance).
– Rebellos
Dec 2 at 13:29










1 Answer
1






active

oldest

votes

















up vote
2
down vote



accepted










Notice that $r = x - sum_{i=1}^n langle x, e_i rangle e_i$ is orthogonal to each vector $e_i$. If $lambda_1,lambda_2,ldots, lambda_n in mathbb R$, then
begin{align}
| x - sum_i lambda_i e_i |^2 &=
|x - sum_i langle x, e_i rangle e_i + sum_i langle x, e_i rangle e_i - sum_i lambda_i e_i |^2 \
&= | r + sum_i (langle x, e_i rangle - lambda_i) e_i |^2 \
&= |r|^2 + sum_i | langle x, e_i rangle - lambda_i | |e_i|^2 qquad text{(by Pythagorean theorem)}\
& geq |r|^2.
end{align}






share|cite|improve this answer























  • Thanks a lot for the fast and swift algebraic proof.
    – Rebellos
    2 days ago










  • @Rebellos Sure. This proof is based in my mind on a picture. We know that the point $x^* = sum_i langle x, e_i rangle e_i$ has the property that $x - x^*$ is orthogonal to the subspace $S = text{span}(e_1,ldots,e_n)$. So, any other vector in $S$ must be further away from $x$, according to the Pythagorean theorem.
    – littleO
    2 days ago











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3022617%2fshowing-that-bigg-x-sum-k-1n-lambda-ke-k-bigg-geq-bigg-x-sum-k%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
2
down vote



accepted










Notice that $r = x - sum_{i=1}^n langle x, e_i rangle e_i$ is orthogonal to each vector $e_i$. If $lambda_1,lambda_2,ldots, lambda_n in mathbb R$, then
begin{align}
| x - sum_i lambda_i e_i |^2 &=
|x - sum_i langle x, e_i rangle e_i + sum_i langle x, e_i rangle e_i - sum_i lambda_i e_i |^2 \
&= | r + sum_i (langle x, e_i rangle - lambda_i) e_i |^2 \
&= |r|^2 + sum_i | langle x, e_i rangle - lambda_i | |e_i|^2 qquad text{(by Pythagorean theorem)}\
& geq |r|^2.
end{align}






share|cite|improve this answer























  • Thanks a lot for the fast and swift algebraic proof.
    – Rebellos
    2 days ago










  • @Rebellos Sure. This proof is based in my mind on a picture. We know that the point $x^* = sum_i langle x, e_i rangle e_i$ has the property that $x - x^*$ is orthogonal to the subspace $S = text{span}(e_1,ldots,e_n)$. So, any other vector in $S$ must be further away from $x$, according to the Pythagorean theorem.
    – littleO
    2 days ago















up vote
2
down vote



accepted










Notice that $r = x - sum_{i=1}^n langle x, e_i rangle e_i$ is orthogonal to each vector $e_i$. If $lambda_1,lambda_2,ldots, lambda_n in mathbb R$, then
begin{align}
| x - sum_i lambda_i e_i |^2 &=
|x - sum_i langle x, e_i rangle e_i + sum_i langle x, e_i rangle e_i - sum_i lambda_i e_i |^2 \
&= | r + sum_i (langle x, e_i rangle - lambda_i) e_i |^2 \
&= |r|^2 + sum_i | langle x, e_i rangle - lambda_i | |e_i|^2 qquad text{(by Pythagorean theorem)}\
& geq |r|^2.
end{align}






share|cite|improve this answer























  • Thanks a lot for the fast and swift algebraic proof.
    – Rebellos
    2 days ago










  • @Rebellos Sure. This proof is based in my mind on a picture. We know that the point $x^* = sum_i langle x, e_i rangle e_i$ has the property that $x - x^*$ is orthogonal to the subspace $S = text{span}(e_1,ldots,e_n)$. So, any other vector in $S$ must be further away from $x$, according to the Pythagorean theorem.
    – littleO
    2 days ago













up vote
2
down vote



accepted







up vote
2
down vote



accepted






Notice that $r = x - sum_{i=1}^n langle x, e_i rangle e_i$ is orthogonal to each vector $e_i$. If $lambda_1,lambda_2,ldots, lambda_n in mathbb R$, then
begin{align}
| x - sum_i lambda_i e_i |^2 &=
|x - sum_i langle x, e_i rangle e_i + sum_i langle x, e_i rangle e_i - sum_i lambda_i e_i |^2 \
&= | r + sum_i (langle x, e_i rangle - lambda_i) e_i |^2 \
&= |r|^2 + sum_i | langle x, e_i rangle - lambda_i | |e_i|^2 qquad text{(by Pythagorean theorem)}\
& geq |r|^2.
end{align}






share|cite|improve this answer














Notice that $r = x - sum_{i=1}^n langle x, e_i rangle e_i$ is orthogonal to each vector $e_i$. If $lambda_1,lambda_2,ldots, lambda_n in mathbb R$, then
begin{align}
| x - sum_i lambda_i e_i |^2 &=
|x - sum_i langle x, e_i rangle e_i + sum_i langle x, e_i rangle e_i - sum_i lambda_i e_i |^2 \
&= | r + sum_i (langle x, e_i rangle - lambda_i) e_i |^2 \
&= |r|^2 + sum_i | langle x, e_i rangle - lambda_i | |e_i|^2 qquad text{(by Pythagorean theorem)}\
& geq |r|^2.
end{align}







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 2 at 14:32

























answered Dec 2 at 14:09









littleO

27.6k642104




27.6k642104












  • Thanks a lot for the fast and swift algebraic proof.
    – Rebellos
    2 days ago










  • @Rebellos Sure. This proof is based in my mind on a picture. We know that the point $x^* = sum_i langle x, e_i rangle e_i$ has the property that $x - x^*$ is orthogonal to the subspace $S = text{span}(e_1,ldots,e_n)$. So, any other vector in $S$ must be further away from $x$, according to the Pythagorean theorem.
    – littleO
    2 days ago


















  • Thanks a lot for the fast and swift algebraic proof.
    – Rebellos
    2 days ago










  • @Rebellos Sure. This proof is based in my mind on a picture. We know that the point $x^* = sum_i langle x, e_i rangle e_i$ has the property that $x - x^*$ is orthogonal to the subspace $S = text{span}(e_1,ldots,e_n)$. So, any other vector in $S$ must be further away from $x$, according to the Pythagorean theorem.
    – littleO
    2 days ago
















Thanks a lot for the fast and swift algebraic proof.
– Rebellos
2 days ago




Thanks a lot for the fast and swift algebraic proof.
– Rebellos
2 days ago












@Rebellos Sure. This proof is based in my mind on a picture. We know that the point $x^* = sum_i langle x, e_i rangle e_i$ has the property that $x - x^*$ is orthogonal to the subspace $S = text{span}(e_1,ldots,e_n)$. So, any other vector in $S$ must be further away from $x$, according to the Pythagorean theorem.
– littleO
2 days ago




@Rebellos Sure. This proof is based in my mind on a picture. We know that the point $x^* = sum_i langle x, e_i rangle e_i$ has the property that $x - x^*$ is orthogonal to the subspace $S = text{span}(e_1,ldots,e_n)$. So, any other vector in $S$ must be further away from $x$, according to the Pythagorean theorem.
– littleO
2 days ago


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.





Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


Please pay close attention to the following guidance:


  • Please be sure to answer the question. Provide details and share your research!

But avoid



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3022617%2fshowing-that-bigg-x-sum-k-1n-lambda-ke-k-bigg-geq-bigg-x-sum-k%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







Popular posts from this blog

Måne

Storängen

VLT Carioca