Lemma 3.9 Rudin's functional analysis
$begingroup$
I need help in understanding the proof of the following
Suppose $Lambda_1,ldots,Lambda_n$ and $Lambda$ are linear
functionals on a vector space $X$. Let
$$ N = left{x : Lambda_1x =
ldots = Lambda_n = 0 right} $$ The following three are equivalent
a) There are scalars $alpha_1,ldots,alpha_n$ such that
$$ Lambda =
alpha_1 Lambda_1 + ldots + alpha_n Lambda_n $$
b) There's $gamma < infty$ such that $$ |Lambda x | leq gamma
> max_{1leq i leq n} |Lambda_i x| ;;,(xin X) $$
c) $Lambda x = 0$ for every $x in N$.
Proof: It is clear that (a) implies (b) and that (b) implies (c)...etc.
That's the bit I don't get. How does (a) implies (b)? I believe I get why (b) implies (c). Since if $x in N$ then for each $i$ we have $Lambda_i x = 0$, but then by (b) we have $|Lambda x = 0|$ and this implies $Lambda x = 0$. (correct?)
So again
Why (a) implies (b)? Is my proof of (b) implies (c) correct?
functional-analysis topological-vector-spaces
asked Dec 10 '18 at 10:02
user8469759user8469759
1,4031618
$endgroup$
add a comment |
$begingroup$
I need help in understanding the proof of the following
Suppose $Lambda_1,ldots,Lambda_n$ and $Lambda$ are linear
functionals on a vector space $X$. Let
$$ N = left{x : Lambda_1x =
ldots = Lambda_n = 0 right} $$ The following three are equivalent
a) There are scalars $alpha_1,ldots,alpha_n$ such that
$$ Lambda =
alpha_1 Lambda_1 + ldots + alpha_n Lambda_n $$
b) There's $gamma < infty$ such that $$ |Lambda x | leq gamma
> max_{1leq i leq n} |Lambda_i x| ;;,(xin X) $$
c) $Lambda x = 0$ for every $x in N$.
Proof: It is clear that (a) implies (b) and that (b) implies (c)...etc.
That's the bit I don't get. How does (a) implies (b)? I believe I get why (b) implies (c). Since if $x in N$ then for each $i$ we have $Lambda_i x = 0$, but then by (b) we have $|Lambda x = 0|$ and this implies $Lambda x = 0$. (correct?)
So again
Why (a) implies (b)? Is my proof of (b) implies (c) correct?
functional-analysis topological-vector-spaces
asked Dec 10 '18 at 10:02
user8469759user8469759
1,4031618
$endgroup$
add a comment |
$begingroup$
I need help in understanding the proof of the following
Suppose $Lambda_1,ldots,Lambda_n$ and $Lambda$ are linear
functionals on a vector space $X$. Let
$$ N = left{x : Lambda_1x =
ldots = Lambda_n = 0 right} $$ The following three are equivalent
a) There are scalars $alpha_1,ldots,alpha_n$ such that
$$ Lambda =
alpha_1 Lambda_1 + ldots + alpha_n Lambda_n $$
b) There's $gamma < infty$ such that $$ |Lambda x | leq gamma
> max_{1leq i leq n} |Lambda_i x| ;;,(xin X) $$
c) $Lambda x = 0$ for every $x in N$.
Proof: It is clear that (a) implies (b) and that (b) implies (c)...etc.
That's the bit I don't get. How does (a) implies (b)? I believe I get why (b) implies (c). Since if $x in N$ then for each $i$ we have $Lambda_i x = 0$, but then by (b) we have $|Lambda x = 0|$ and this implies $Lambda x = 0$. (correct?)
So again
Why (a) implies (b)? Is my proof of (b) implies (c) correct?
functional-analysis topological-vector-spaces
asked Dec 10 '18 at 10:02
user8469759user8469759
1,4031618
$endgroup$
I need help in understanding the proof of the following
Suppose $Lambda_1,ldots,Lambda_n$ and $Lambda$ are linear
functionals on a vector space $X$. Let
$$ N = left{x : Lambda_1x =
ldots = Lambda_n = 0 right} $$ The following three are equivalent
a) There are scalars $alpha_1,ldots,alpha_n$ such that
$$ Lambda =
alpha_1 Lambda_1 + ldots + alpha_n Lambda_n $$
b) There's $gamma < infty$ such that $$ |Lambda x | leq gamma
> max_{1leq i leq n} |Lambda_i x| ;;,(xin X) $$
c) $Lambda x = 0$ for every $x in N$.
Proof: It is clear that (a) implies (b) and that (b) implies (c)...etc.
That's the bit I don't get. How does (a) implies (b)? I believe I get why (b) implies (c). Since if $x in N$ then for each $i$ we have $Lambda_i x = 0$, but then by (b) we have $|Lambda x = 0|$ and this implies $Lambda x = 0$. (correct?)
So again
Why (a) implies (b)? Is my proof of (b) implies (c) correct?
functional-analysis topological-vector-spaces
functional-analysis topological-vector-spaces
asked Dec 10 '18 at 10:02
user8469759user8469759
1,4031618
asked Dec 10 '18 at 10:02
user8469759user8469759
1,4031618
asked Dec 10 '18 at 10:02
user8469759user8469759
1,4031618
asked Dec 10 '18 at 10:02
user8469759user8469759
1,4031618
asked Dec 10 '18 at 10:02
user8469759user8469759
1,4031618
1,4031618
add a comment |
add a comment |
2 Answers
2
active
oldest
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$begingroup$
Your proof of (b) implies (c) is correct.
As for (a) implies (b):
We have
$$
| Lambda x |
= | alpha_1Lambda_1 x +dots + alpha_nLambda_n x |
\
leq | alpha_1|cdot |Lambda_1 x | +dots + |alpha_n|cdot |Lambda_n x|
\
leq (|alpha_1|+dots + |alpha_n|) max_{1leq i leq n} |Lambda_i x|.
$$
This means we can choose
$gamma=|alpha_1|+dots + |alpha_n|$.
answered Dec 10 '18 at 10:17
supinfsupinf
6,3111028
$endgroup$
add a comment |
$begingroup$
a) implies b): $|Lambda x| leq sum |alpha_i| | Lambda_i x|leq max{|Lambda_i(x)|} sum |alpha_i|$ so you can take $gamma =sum |alpha_i|$ Your argument for b) implies c) is correct.
answered Dec 10 '18 at 10:13
Kavi Rama MurthyKavi Rama Murthy
57.5k42160
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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votes
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oldest
votes
$begingroup$
Your proof of (b) implies (c) is correct.
As for (a) implies (b):
We have
$$
| Lambda x |
= | alpha_1Lambda_1 x +dots + alpha_nLambda_n x |
\
leq | alpha_1|cdot |Lambda_1 x | +dots + |alpha_n|cdot |Lambda_n x|
\
leq (|alpha_1|+dots + |alpha_n|) max_{1leq i leq n} |Lambda_i x|.
$$
This means we can choose
$gamma=|alpha_1|+dots + |alpha_n|$.
answered Dec 10 '18 at 10:17
supinfsupinf
6,3111028
$endgroup$
add a comment |
$begingroup$
Your proof of (b) implies (c) is correct.
As for (a) implies (b):
We have
$$
| Lambda x |
= | alpha_1Lambda_1 x +dots + alpha_nLambda_n x |
\
leq | alpha_1|cdot |Lambda_1 x | +dots + |alpha_n|cdot |Lambda_n x|
\
leq (|alpha_1|+dots + |alpha_n|) max_{1leq i leq n} |Lambda_i x|.
$$
This means we can choose
$gamma=|alpha_1|+dots + |alpha_n|$.
answered Dec 10 '18 at 10:17
supinfsupinf
6,3111028
$endgroup$
add a comment |
$begingroup$
Your proof of (b) implies (c) is correct.
As for (a) implies (b):
We have
$$
| Lambda x |
= | alpha_1Lambda_1 x +dots + alpha_nLambda_n x |
\
leq | alpha_1|cdot |Lambda_1 x | +dots + |alpha_n|cdot |Lambda_n x|
\
leq (|alpha_1|+dots + |alpha_n|) max_{1leq i leq n} |Lambda_i x|.
$$
This means we can choose
$gamma=|alpha_1|+dots + |alpha_n|$.
answered Dec 10 '18 at 10:17
supinfsupinf
6,3111028
$endgroup$
Your proof of (b) implies (c) is correct.
As for (a) implies (b):
We have
$$
| Lambda x |
= | alpha_1Lambda_1 x +dots + alpha_nLambda_n x |
\
leq | alpha_1|cdot |Lambda_1 x | +dots + |alpha_n|cdot |Lambda_n x|
\
leq (|alpha_1|+dots + |alpha_n|) max_{1leq i leq n} |Lambda_i x|.
$$
This means we can choose
$gamma=|alpha_1|+dots + |alpha_n|$.
answered Dec 10 '18 at 10:17
supinfsupinf
6,3111028
answered Dec 10 '18 at 10:17
supinfsupinf
6,3111028
answered Dec 10 '18 at 10:17
supinfsupinf
6,3111028
answered Dec 10 '18 at 10:17
supinfsupinf
6,3111028
6,3111028
add a comment |
add a comment |
$begingroup$
a) implies b): $|Lambda x| leq sum |alpha_i| | Lambda_i x|leq max{|Lambda_i(x)|} sum |alpha_i|$ so you can take $gamma =sum |alpha_i|$ Your argument for b) implies c) is correct.
answered Dec 10 '18 at 10:13
Kavi Rama MurthyKavi Rama Murthy
57.5k42160
$endgroup$
add a comment |
$begingroup$
a) implies b): $|Lambda x| leq sum |alpha_i| | Lambda_i x|leq max{|Lambda_i(x)|} sum |alpha_i|$ so you can take $gamma =sum |alpha_i|$ Your argument for b) implies c) is correct.
answered Dec 10 '18 at 10:13
Kavi Rama MurthyKavi Rama Murthy
57.5k42160
$endgroup$
add a comment |
$begingroup$
a) implies b): $|Lambda x| leq sum |alpha_i| | Lambda_i x|leq max{|Lambda_i(x)|} sum |alpha_i|$ so you can take $gamma =sum |alpha_i|$ Your argument for b) implies c) is correct.
answered Dec 10 '18 at 10:13
Kavi Rama MurthyKavi Rama Murthy
57.5k42160
$endgroup$
a) implies b): $|Lambda x| leq sum |alpha_i| | Lambda_i x|leq max{|Lambda_i(x)|} sum |alpha_i|$ so you can take $gamma =sum |alpha_i|$ Your argument for b) implies c) is correct.
answered Dec 10 '18 at 10:13
Kavi Rama MurthyKavi Rama Murthy
57.5k42160
answered Dec 10 '18 at 10:13
Kavi Rama MurthyKavi Rama Murthy
57.5k42160
answered Dec 10 '18 at 10:13
Kavi Rama MurthyKavi Rama Murthy
57.5k42160
answered Dec 10 '18 at 10:13
Kavi Rama MurthyKavi Rama Murthy
57.5k42160
57.5k42160
add a comment |
add a comment |
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