Mixing
Prove that the dual space of $R^n$ is $R^n$
$begingroup$
I was going through the proof of this question in "Introductory Functional Analysis with Applications" by Kreyszig.
(Apologies for not being able to type the whole of it)
I got stuck after the line where we took the supremum over all x of norm 1.
How did we arrive at the step of equality in the Cauchy-Schwarz inequality?
Any good explanation would be helpful.
functional-analysis proof-explanation
asked Jan 27 at 9:57
Shatabdi SinhaShatabdi Sinha
18612
$endgroup$
add a comment |
$begingroup$
I was going through the proof of this question in "Introductory Functional Analysis with Applications" by Kreyszig.
(Apologies for not being able to type the whole of it)
I got stuck after the line where we took the supremum over all x of norm 1.
How did we arrive at the step of equality in the Cauchy-Schwarz inequality?
Any good explanation would be helpful.
functional-analysis proof-explanation
asked Jan 27 at 9:57
Shatabdi SinhaShatabdi Sinha
18612
$endgroup$
add a comment |
$begingroup$
I was going through the proof of this question in "Introductory Functional Analysis with Applications" by Kreyszig.
(Apologies for not being able to type the whole of it)
I got stuck after the line where we took the supremum over all x of norm 1.
How did we arrive at the step of equality in the Cauchy-Schwarz inequality?
Any good explanation would be helpful.
functional-analysis proof-explanation
asked Jan 27 at 9:57
Shatabdi SinhaShatabdi Sinha
18612
$endgroup$
I was going through the proof of this question in "Introductory Functional Analysis with Applications" by Kreyszig.
(Apologies for not being able to type the whole of it)
I got stuck after the line where we took the supremum over all x of norm 1.
How did we arrive at the step of equality in the Cauchy-Schwarz inequality?
Any good explanation would be helpful.
functional-analysis proof-explanation
functional-analysis proof-explanation
asked Jan 27 at 9:57
Shatabdi SinhaShatabdi Sinha
18612
asked Jan 27 at 9:57
Shatabdi SinhaShatabdi Sinha
18612
edited Jan 27 at 10:16
Shatabdi Sinha
asked Jan 27 at 9:57
Shatabdi SinhaShatabdi Sinha
18612
asked Jan 27 at 9:57
Shatabdi SinhaShatabdi Sinha
18612
asked Jan 27 at 9:57
Shatabdi SinhaShatabdi Sinha
18612
18612
add a comment |
add a comment |
1 Answer
1
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$begingroup$
Just calculate it directly;
$$f(x)=sum_k xi_kgamma_k = sum_k gamma_kgamma_k = sum_k gamma_k^2=left(sum_k gamma_k^2right)^{frac12}cdotleft(sum_k gamma_k^2right)^{frac12}=|x|cdotleft(sum_k gamma_k^2right)^{frac12}$$
since $|x|=left(sum_kxi_k^2right)^{frac12} = left(sum_kgamma_k^2right)^{frac12}$ when the $xi$ and $gamma$ are equal. It's not norm $1$, but we can just multiply by a constant for that.
answered Jan 27 at 10:30
jmerryjmerry
16.1k1633
$endgroup$
add a comment |
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1 Answer
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1 Answer
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oldest
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$begingroup$
Just calculate it directly;
$$f(x)=sum_k xi_kgamma_k = sum_k gamma_kgamma_k = sum_k gamma_k^2=left(sum_k gamma_k^2right)^{frac12}cdotleft(sum_k gamma_k^2right)^{frac12}=|x|cdotleft(sum_k gamma_k^2right)^{frac12}$$
since $|x|=left(sum_kxi_k^2right)^{frac12} = left(sum_kgamma_k^2right)^{frac12}$ when the $xi$ and $gamma$ are equal. It's not norm $1$, but we can just multiply by a constant for that.
answered Jan 27 at 10:30
jmerryjmerry
16.1k1633
$endgroup$
add a comment |
$begingroup$
Just calculate it directly;
$$f(x)=sum_k xi_kgamma_k = sum_k gamma_kgamma_k = sum_k gamma_k^2=left(sum_k gamma_k^2right)^{frac12}cdotleft(sum_k gamma_k^2right)^{frac12}=|x|cdotleft(sum_k gamma_k^2right)^{frac12}$$
since $|x|=left(sum_kxi_k^2right)^{frac12} = left(sum_kgamma_k^2right)^{frac12}$ when the $xi$ and $gamma$ are equal. It's not norm $1$, but we can just multiply by a constant for that.
answered Jan 27 at 10:30
jmerryjmerry
16.1k1633
$endgroup$
add a comment |
$begingroup$
Just calculate it directly;
$$f(x)=sum_k xi_kgamma_k = sum_k gamma_kgamma_k = sum_k gamma_k^2=left(sum_k gamma_k^2right)^{frac12}cdotleft(sum_k gamma_k^2right)^{frac12}=|x|cdotleft(sum_k gamma_k^2right)^{frac12}$$
since $|x|=left(sum_kxi_k^2right)^{frac12} = left(sum_kgamma_k^2right)^{frac12}$ when the $xi$ and $gamma$ are equal. It's not norm $1$, but we can just multiply by a constant for that.
answered Jan 27 at 10:30
jmerryjmerry
16.1k1633
$endgroup$
Just calculate it directly;
$$f(x)=sum_k xi_kgamma_k = sum_k gamma_kgamma_k = sum_k gamma_k^2=left(sum_k gamma_k^2right)^{frac12}cdotleft(sum_k gamma_k^2right)^{frac12}=|x|cdotleft(sum_k gamma_k^2right)^{frac12}$$
since $|x|=left(sum_kxi_k^2right)^{frac12} = left(sum_kgamma_k^2right)^{frac12}$ when the $xi$ and $gamma$ are equal. It's not norm $1$, but we can just multiply by a constant for that.
answered Jan 27 at 10:30
jmerryjmerry
16.1k1633
answered Jan 27 at 10:30
jmerryjmerry
16.1k1633
answered Jan 27 at 10:30
jmerryjmerry
16.1k1633
answered Jan 27 at 10:30
jmerryjmerry
16.1k1633
16.1k1633
add a comment |
add a comment |
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