Trouble calculating the operator norm
$begingroup$
I'm working with the following operator $R_{m}:Yrightarrow X$ defined as
$$R_{m}f:=sum_{mu_ngeqmu_m}frac{1}{mu_n}langle f,g_nrangle varphi_n$$
where $X$ and $Y$ are Hilbert spaces, $mu_n$ are singular values of an injective compact linear operator $A$, and ${g_n},{varphi_n}$ are orthonormal sequences of eigenelements. I'm supposed to show that
$$||(I-AR_{m})f||=1text{ for all }minmathbb{N}$$
and I'm having some trouble doing so. I've managed to establish an upper bound, since I have shown that
$$||(I-AR_{m})f||^2=sum_{mu_n<mu_m}|langle f,g_nrangle|^2leq ||f||^2$$
thus $||I-AR_{m}||leq 1$. My idea is to try and show that $||I-AR_{m}||geq 1$, since that would lead to the desired result, but I'm struggling with that, so if anyone can help that would be great.
Also, if you feel like more information is needed, please say so, and I will happily provide more information.
Thanks in advance!
functional-analysis
asked Jan 4 at 17:45
JamesJames
11910
$endgroup$
add a comment |
$begingroup$
I'm working with the following operator $R_{m}:Yrightarrow X$ defined as
$$R_{m}f:=sum_{mu_ngeqmu_m}frac{1}{mu_n}langle f,g_nrangle varphi_n$$
where $X$ and $Y$ are Hilbert spaces, $mu_n$ are singular values of an injective compact linear operator $A$, and ${g_n},{varphi_n}$ are orthonormal sequences of eigenelements. I'm supposed to show that
$$||(I-AR_{m})f||=1text{ for all }minmathbb{N}$$
and I'm having some trouble doing so. I've managed to establish an upper bound, since I have shown that
$$||(I-AR_{m})f||^2=sum_{mu_n<mu_m}|langle f,g_nrangle|^2leq ||f||^2$$
thus $||I-AR_{m}||leq 1$. My idea is to try and show that $||I-AR_{m}||geq 1$, since that would lead to the desired result, but I'm struggling with that, so if anyone can help that would be great.
Also, if you feel like more information is needed, please say so, and I will happily provide more information.
Thanks in advance!
functional-analysis
asked Jan 4 at 17:45
JamesJames
11910
$endgroup$
1
$begingroup$
Choose $f=g_n$ for some $n < m$ to show $|(I-AR_m)f|=|f|$ and conclude $|(I-AR_m)| = 1$.
$endgroup$
– DisintegratingByParts
Jan 4 at 18:31
$begingroup$
@DisintegratingByParts Oh, so I could for example pick $f=g_{k}$ such that it is not included in the sequence ${g_n}$ in the sum, i.e., picking a corresponding $g_k$ for a singular value that is not included in the sum?
$endgroup$
– James
Jan 6 at 14:14
add a comment |
$begingroup$
I'm working with the following operator $R_{m}:Yrightarrow X$ defined as
$$R_{m}f:=sum_{mu_ngeqmu_m}frac{1}{mu_n}langle f,g_nrangle varphi_n$$
where $X$ and $Y$ are Hilbert spaces, $mu_n$ are singular values of an injective compact linear operator $A$, and ${g_n},{varphi_n}$ are orthonormal sequences of eigenelements. I'm supposed to show that
$$||(I-AR_{m})f||=1text{ for all }minmathbb{N}$$
and I'm having some trouble doing so. I've managed to establish an upper bound, since I have shown that
$$||(I-AR_{m})f||^2=sum_{mu_n<mu_m}|langle f,g_nrangle|^2leq ||f||^2$$
thus $||I-AR_{m}||leq 1$. My idea is to try and show that $||I-AR_{m}||geq 1$, since that would lead to the desired result, but I'm struggling with that, so if anyone can help that would be great.
Also, if you feel like more information is needed, please say so, and I will happily provide more information.
Thanks in advance!
functional-analysis
asked Jan 4 at 17:45
JamesJames
11910
$endgroup$
I'm working with the following operator $R_{m}:Yrightarrow X$ defined as
$$R_{m}f:=sum_{mu_ngeqmu_m}frac{1}{mu_n}langle f,g_nrangle varphi_n$$
where $X$ and $Y$ are Hilbert spaces, $mu_n$ are singular values of an injective compact linear operator $A$, and ${g_n},{varphi_n}$ are orthonormal sequences of eigenelements. I'm supposed to show that
$$||(I-AR_{m})f||=1text{ for all }minmathbb{N}$$
and I'm having some trouble doing so. I've managed to establish an upper bound, since I have shown that
$$||(I-AR_{m})f||^2=sum_{mu_n<mu_m}|langle f,g_nrangle|^2leq ||f||^2$$
thus $||I-AR_{m}||leq 1$. My idea is to try and show that $||I-AR_{m}||geq 1$, since that would lead to the desired result, but I'm struggling with that, so if anyone can help that would be great.
Also, if you feel like more information is needed, please say so, and I will happily provide more information.
Thanks in advance!
functional-analysis
functional-analysis
asked Jan 4 at 17:45
JamesJames
11910
asked Jan 4 at 17:45
JamesJames
11910
asked Jan 4 at 17:45
JamesJames
11910
asked Jan 4 at 17:45
JamesJames
11910
asked Jan 4 at 17:45
JamesJames
11910
11910
1
$begingroup$
Choose $f=g_n$ for some $n < m$ to show $|(I-AR_m)f|=|f|$ and conclude $|(I-AR_m)| = 1$.
$endgroup$
– DisintegratingByParts
Jan 4 at 18:31
$begingroup$
@DisintegratingByParts Oh, so I could for example pick $f=g_{k}$ such that it is not included in the sequence ${g_n}$ in the sum, i.e., picking a corresponding $g_k$ for a singular value that is not included in the sum?
$endgroup$
– James
Jan 6 at 14:14
add a comment |
1
$begingroup$
Choose $f=g_n$ for some $n < m$ to show $|(I-AR_m)f|=|f|$ and conclude $|(I-AR_m)| = 1$.
$endgroup$
– DisintegratingByParts
Jan 4 at 18:31
$begingroup$
@DisintegratingByParts Oh, so I could for example pick $f=g_{k}$ such that it is not included in the sequence ${g_n}$ in the sum, i.e., picking a corresponding $g_k$ for a singular value that is not included in the sum?
$endgroup$
– James
Jan 6 at 14:14
1
1
$begingroup$
Choose $f=g_n$ for some $n < m$ to show $|(I-AR_m)f|=|f|$ and conclude $|(I-AR_m)| = 1$.
$endgroup$
– DisintegratingByParts
Jan 4 at 18:31
$begingroup$
Choose $f=g_n$ for some $n < m$ to show $|(I-AR_m)f|=|f|$ and conclude $|(I-AR_m)| = 1$.
$endgroup$
– DisintegratingByParts
Jan 4 at 18:31
$begingroup$
@DisintegratingByParts Oh, so I could for example pick $f=g_{k}$ such that it is not included in the sequence ${g_n}$ in the sum, i.e., picking a corresponding $g_k$ for a singular value that is not included in the sum?
$endgroup$
– James
Jan 6 at 14:14
$begingroup$
@DisintegratingByParts Oh, so I could for example pick $f=g_{k}$ such that it is not included in the sequence ${g_n}$ in the sum, i.e., picking a corresponding $g_k$ for a singular value that is not included in the sum?
$endgroup$
– James
Jan 6 at 14:14
add a comment |
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$begingroup$
Choose $f=g_n$ for some $n < m$ to show $|(I-AR_m)f|=|f|$ and conclude $|(I-AR_m)| = 1$.
$endgroup$
– DisintegratingByParts
Jan 4 at 18:31
$begingroup$
@DisintegratingByParts Oh, so I could for example pick $f=g_{k}$ such that it is not included in the sequence ${g_n}$ in the sum, i.e., picking a corresponding $g_k$ for a singular value that is not included in the sum?
$endgroup$
– James
Jan 6 at 14:14