Show that the projection from $C[0,1]$ onto $P_n[0,1]$ under $L^q$ norm is nonlinear when $q>1$ and $qneq 2$.
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Let $C[0,1]$ denote the space of real valued continuous functions defined on $[0,1]$. Let $P_n[0,1]$ the space of real polynomials with degree not greater than $n$ defined on $[0,1]$. Let $q>1$ denote a finite real number. For any $fin C[0,1]$, one can show by compactness argument and uniform convexity of $L^q$ norm that there exists an unique $P(f)in P_n[0,1]$ such that
beginequation
||f-P(f)||_L^q(0,1)=inflimits_pin P_n[0,1]||f-p||_L^q(0,1).
endequation
In this fashion, we define a projection operator: $fin C[0,1]mapsto P(f)in P_n[0,1]$. How to show that $P(cdot)$ is nonlinear when $qneq 2$? Could anyone help me show this? I really don't know how to start it.
functional-analysis polynomials continuity lp-spaces projection
asked Sep 6 at 7:22
Lin Xuelei
847
add a comment |Â
up vote
2
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Let $C[0,1]$ denote the space of real valued continuous functions defined on $[0,1]$. Let $P_n[0,1]$ the space of real polynomials with degree not greater than $n$ defined on $[0,1]$. Let $q>1$ denote a finite real number. For any $fin C[0,1]$, one can show by compactness argument and uniform convexity of $L^q$ norm that there exists an unique $P(f)in P_n[0,1]$ such that
beginequation
||f-P(f)||_L^q(0,1)=inflimits_pin P_n[0,1]||f-p||_L^q(0,1).
endequation
In this fashion, we define a projection operator: $fin C[0,1]mapsto P(f)in P_n[0,1]$. How to show that $P(cdot)$ is nonlinear when $qneq 2$? Could anyone help me show this? I really don't know how to start it.
functional-analysis polynomials continuity lp-spaces projection
asked Sep 6 at 7:22
Lin Xuelei
847
Since $P$ will be homogeneous for all $q$, the problem must be additivity. Have you tried examples where $f+g$ is a polynomial but $f$ and $g$ aren't?
– Christoph
Sep 6 at 7:31
What do you mean by homogeneous and additivity? Can you provided more details?
– Lin Xuelei
Sep 6 at 7:51
$P$ is homogeneous: $P(lambda f) = lambda P(f)$ for $lambdainmathbb R$. $P$ is additive: $P(f+g) = P(f) + P(g)$.
– Christoph
Sep 6 at 8:27
yes, when $q=2$, then $P(cdot)$ is linear, which is the result of projection theorem in Hilbert space. The question is how to show $q=2$ using assumption that $P(cdot)$ is linear.
– Lin Xuelei
Sep 6 at 8:33
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $C[0,1]$ denote the space of real valued continuous functions defined on $[0,1]$. Let $P_n[0,1]$ the space of real polynomials with degree not greater than $n$ defined on $[0,1]$. Let $q>1$ denote a finite real number. For any $fin C[0,1]$, one can show by compactness argument and uniform convexity of $L^q$ norm that there exists an unique $P(f)in P_n[0,1]$ such that
beginequation
||f-P(f)||_L^q(0,1)=inflimits_pin P_n[0,1]||f-p||_L^q(0,1).
endequation
In this fashion, we define a projection operator: $fin C[0,1]mapsto P(f)in P_n[0,1]$. How to show that $P(cdot)$ is nonlinear when $qneq 2$? Could anyone help me show this? I really don't know how to start it.
functional-analysis polynomials continuity lp-spaces projection
asked Sep 6 at 7:22
Lin Xuelei
847
Let $C[0,1]$ denote the space of real valued continuous functions defined on $[0,1]$. Let $P_n[0,1]$ the space of real polynomials with degree not greater than $n$ defined on $[0,1]$. Let $q>1$ denote a finite real number. For any $fin C[0,1]$, one can show by compactness argument and uniform convexity of $L^q$ norm that there exists an unique $P(f)in P_n[0,1]$ such that
beginequation
||f-P(f)||_L^q(0,1)=inflimits_pin P_n[0,1]||f-p||_L^q(0,1).
endequation
In this fashion, we define a projection operator: $fin C[0,1]mapsto P(f)in P_n[0,1]$. How to show that $P(cdot)$ is nonlinear when $qneq 2$? Could anyone help me show this? I really don't know how to start it.
functional-analysis polynomials continuity lp-spaces projection
functional-analysis polynomials continuity lp-spaces projection
asked Sep 6 at 7:22
Lin Xuelei
847
asked Sep 6 at 7:22
Lin Xuelei
847
edited Sep 6 at 14:27
asked Sep 6 at 7:22
Lin Xuelei
847
asked Sep 6 at 7:22
Lin Xuelei
847
asked Sep 6 at 7:22
Lin Xuelei
847
847
Since $P$ will be homogeneous for all $q$, the problem must be additivity. Have you tried examples where $f+g$ is a polynomial but $f$ and $g$ aren't?
– Christoph
Sep 6 at 7:31
What do you mean by homogeneous and additivity? Can you provided more details?
– Lin Xuelei
Sep 6 at 7:51
$P$ is homogeneous: $P(lambda f) = lambda P(f)$ for $lambdainmathbb R$. $P$ is additive: $P(f+g) = P(f) + P(g)$.
– Christoph
Sep 6 at 8:27
yes, when $q=2$, then $P(cdot)$ is linear, which is the result of projection theorem in Hilbert space. The question is how to show $q=2$ using assumption that $P(cdot)$ is linear.
– Lin Xuelei
Sep 6 at 8:33
add a comment |Â
Since $P$ will be homogeneous for all $q$, the problem must be additivity. Have you tried examples where $f+g$ is a polynomial but $f$ and $g$ aren't?
– Christoph
Sep 6 at 7:31
What do you mean by homogeneous and additivity? Can you provided more details?
– Lin Xuelei
Sep 6 at 7:51
$P$ is homogeneous: $P(lambda f) = lambda P(f)$ for $lambdainmathbb R$. $P$ is additive: $P(f+g) = P(f) + P(g)$.
– Christoph
Sep 6 at 8:27
yes, when $q=2$, then $P(cdot)$ is linear, which is the result of projection theorem in Hilbert space. The question is how to show $q=2$ using assumption that $P(cdot)$ is linear.
– Lin Xuelei
Sep 6 at 8:33
Since $P$ will be homogeneous for all $q$, the problem must be additivity. Have you tried examples where $f+g$ is a polynomial but $f$ and $g$ aren't?
– Christoph
Sep 6 at 7:31
Since $P$ will be homogeneous for all $q$, the problem must be additivity. Have you tried examples where $f+g$ is a polynomial but $f$ and $g$ aren't?
– Christoph
Sep 6 at 7:31
What do you mean by homogeneous and additivity? Can you provided more details?
– Lin Xuelei
Sep 6 at 7:51
What do you mean by homogeneous and additivity? Can you provided more details?
– Lin Xuelei
Sep 6 at 7:51
$P$ is homogeneous: $P(lambda f) = lambda P(f)$ for $lambdainmathbb R$. $P$ is additive: $P(f+g) = P(f) + P(g)$.
– Christoph
Sep 6 at 8:27
$P$ is homogeneous: $P(lambda f) = lambda P(f)$ for $lambdainmathbb R$. $P$ is additive: $P(f+g) = P(f) + P(g)$.
– Christoph
Sep 6 at 8:27
yes, when $q=2$, then $P(cdot)$ is linear, which is the result of projection theorem in Hilbert space. The question is how to show $q=2$ using assumption that $P(cdot)$ is linear.
– Lin Xuelei
Sep 6 at 8:33
yes, when $q=2$, then $P(cdot)$ is linear, which is the result of projection theorem in Hilbert space. The question is how to show $q=2$ using assumption that $P(cdot)$ is linear.
– Lin Xuelei
Sep 6 at 8:33
add a comment |Â
1 Answer
1
active
oldest
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up vote
1
down vote
accepted
I am just giving a (possibly a trivial) perception here.
Let $q=m$ an even positive integer.
beginequation
||f-sum_j=0^n a_i x^i||_m^m = int_0^1 (f-sum_j=0^n a_i x^i)^m dx\
= sum_i_1,...,i_n,i_m+1 fracm!i_1!...i_n+1! (-1)^i_1+...+i_m a_1^i_1...a_n^i_n int_0^1 f^i_n+1x^sum_k=0^n ki_k dx
endequation
for $m=2$, if we differentiate the above equation w.r.t $a_i$ and set to $0$, then its a system of linear equations in $f,a_1,...,a_n$ and for $m >2$, if we differentiate above and set to $0$, its a set of non-linear equations in $f,a_1,...,a_n$.
answered Sep 6 at 13:58
Balaji sb
40325
Yes, you provide a good idea to the proof. I think your answer can be regarded as a hint to the proof.
– Lin Xuelei
Sep 6 at 14:52
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
I am just giving a (possibly a trivial) perception here.
Let $q=m$ an even positive integer.
beginequation
||f-sum_j=0^n a_i x^i||_m^m = int_0^1 (f-sum_j=0^n a_i x^i)^m dx\
= sum_i_1,...,i_n,i_m+1 fracm!i_1!...i_n+1! (-1)^i_1+...+i_m a_1^i_1...a_n^i_n int_0^1 f^i_n+1x^sum_k=0^n ki_k dx
endequation
for $m=2$, if we differentiate the above equation w.r.t $a_i$ and set to $0$, then its a system of linear equations in $f,a_1,...,a_n$ and for $m >2$, if we differentiate above and set to $0$, its a set of non-linear equations in $f,a_1,...,a_n$.
answered Sep 6 at 13:58
Balaji sb
40325
Yes, you provide a good idea to the proof. I think your answer can be regarded as a hint to the proof.
– Lin Xuelei
Sep 6 at 14:52
add a comment |Â
up vote
1
down vote
accepted
I am just giving a (possibly a trivial) perception here.
Let $q=m$ an even positive integer.
beginequation
||f-sum_j=0^n a_i x^i||_m^m = int_0^1 (f-sum_j=0^n a_i x^i)^m dx\
= sum_i_1,...,i_n,i_m+1 fracm!i_1!...i_n+1! (-1)^i_1+...+i_m a_1^i_1...a_n^i_n int_0^1 f^i_n+1x^sum_k=0^n ki_k dx
endequation
for $m=2$, if we differentiate the above equation w.r.t $a_i$ and set to $0$, then its a system of linear equations in $f,a_1,...,a_n$ and for $m >2$, if we differentiate above and set to $0$, its a set of non-linear equations in $f,a_1,...,a_n$.
answered Sep 6 at 13:58
Balaji sb
40325
Yes, you provide a good idea to the proof. I think your answer can be regarded as a hint to the proof.
– Lin Xuelei
Sep 6 at 14:52
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
I am just giving a (possibly a trivial) perception here.
Let $q=m$ an even positive integer.
beginequation
||f-sum_j=0^n a_i x^i||_m^m = int_0^1 (f-sum_j=0^n a_i x^i)^m dx\
= sum_i_1,...,i_n,i_m+1 fracm!i_1!...i_n+1! (-1)^i_1+...+i_m a_1^i_1...a_n^i_n int_0^1 f^i_n+1x^sum_k=0^n ki_k dx
endequation
for $m=2$, if we differentiate the above equation w.r.t $a_i$ and set to $0$, then its a system of linear equations in $f,a_1,...,a_n$ and for $m >2$, if we differentiate above and set to $0$, its a set of non-linear equations in $f,a_1,...,a_n$.
answered Sep 6 at 13:58
Balaji sb
40325
I am just giving a (possibly a trivial) perception here.
Let $q=m$ an even positive integer.
beginequation
||f-sum_j=0^n a_i x^i||_m^m = int_0^1 (f-sum_j=0^n a_i x^i)^m dx\
= sum_i_1,...,i_n,i_m+1 fracm!i_1!...i_n+1! (-1)^i_1+...+i_m a_1^i_1...a_n^i_n int_0^1 f^i_n+1x^sum_k=0^n ki_k dx
endequation
for $m=2$, if we differentiate the above equation w.r.t $a_i$ and set to $0$, then its a system of linear equations in $f,a_1,...,a_n$ and for $m >2$, if we differentiate above and set to $0$, its a set of non-linear equations in $f,a_1,...,a_n$.
answered Sep 6 at 13:58
Balaji sb
40325
edited Sep 6 at 14:48
answered Sep 6 at 13:58
Balaji sb
40325
answered Sep 6 at 13:58
Balaji sb
40325
answered Sep 6 at 13:58
Balaji sb
40325
40325
Yes, you provide a good idea to the proof. I think your answer can be regarded as a hint to the proof.
– Lin Xuelei
Sep 6 at 14:52
add a comment |Â
Yes, you provide a good idea to the proof. I think your answer can be regarded as a hint to the proof.
– Lin Xuelei
Sep 6 at 14:52
Yes, you provide a good idea to the proof. I think your answer can be regarded as a hint to the proof.
– Lin Xuelei
Sep 6 at 14:52
Yes, you provide a good idea to the proof. I think your answer can be regarded as a hint to the proof.
– Lin Xuelei
Sep 6 at 14:52
add a comment |Â
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Since $P$ will be homogeneous for all $q$, the problem must be additivity. Have you tried examples where $f+g$ is a polynomial but $f$ and $g$ aren't?
– Christoph
Sep 6 at 7:31
What do you mean by homogeneous and additivity? Can you provided more details?
– Lin Xuelei
Sep 6 at 7:51
$P$ is homogeneous: $P(lambda f) = lambda P(f)$ for $lambdainmathbb R$. $P$ is additive: $P(f+g) = P(f) + P(g)$.
– Christoph
Sep 6 at 8:27
yes, when $q=2$, then $P(cdot)$ is linear, which is the result of projection theorem in Hilbert space. The question is how to show $q=2$ using assumption that $P(cdot)$ is linear.
– Lin Xuelei
Sep 6 at 8:33