Show that $| x |^2_2 | x|_{2k-2}^{2k-2} leq n^{1-2/k} | x |^{2k}_k$
$begingroup$
Let $x$ be in $R^n$. Show that:
$$| x |^2_2 cdot | x|_{2k-2}^{2k-2} leq n^{1-2/k} cdot | x |^{2k}_k~.$$
inequality
asked Jan 10 at 5:52
HTVHTV
1017
$endgroup$
add a comment |
$begingroup$
Let $x$ be in $R^n$. Show that:
$$| x |^2_2 cdot | x|_{2k-2}^{2k-2} leq n^{1-2/k} cdot | x |^{2k}_k~.$$
inequality
asked Jan 10 at 5:52
HTVHTV
1017
$endgroup$
add a comment |
$begingroup$
Let $x$ be in $R^n$. Show that:
$$| x |^2_2 cdot | x|_{2k-2}^{2k-2} leq n^{1-2/k} cdot | x |^{2k}_k~.$$
inequality
asked Jan 10 at 5:52
HTVHTV
1017
$endgroup$
Let $x$ be in $R^n$. Show that:
$$| x |^2_2 cdot | x|_{2k-2}^{2k-2} leq n^{1-2/k} cdot | x |^{2k}_k~.$$
inequality
inequality
asked Jan 10 at 5:52
HTVHTV
1017
asked Jan 10 at 5:52
HTVHTV
1017
asked Jan 10 at 5:52
HTVHTV
1017
asked Jan 10 at 5:52
HTVHTV
1017
asked Jan 10 at 5:52
HTVHTV
1017
1017
add a comment |
add a comment |
1 Answer
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$begingroup$
Ok, I figured out the answer.
We have
$$| x |_2 leq n^{1/2-1/k} | x |_k implies | x |_2^2 leq n^{1-2/k} | x |_k^2~.$$
See Relations between p norms for the first inequality. Also,
$$| x|_{2k-2} leq | x |_k implies | x_{2k-2} |^{2k-2}_{2k-2} leq | x |_k^{2k-2} ~$$
And we are done.
answered Jan 10 at 6:28
HTVHTV
1017
$endgroup$
add a comment |
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$begingroup$
Ok, I figured out the answer.
We have
$$| x |_2 leq n^{1/2-1/k} | x |_k implies | x |_2^2 leq n^{1-2/k} | x |_k^2~.$$
See Relations between p norms for the first inequality. Also,
$$| x|_{2k-2} leq | x |_k implies | x_{2k-2} |^{2k-2}_{2k-2} leq | x |_k^{2k-2} ~$$
And we are done.
answered Jan 10 at 6:28
HTVHTV
1017
$endgroup$
add a comment |
$begingroup$
Ok, I figured out the answer.
We have
$$| x |_2 leq n^{1/2-1/k} | x |_k implies | x |_2^2 leq n^{1-2/k} | x |_k^2~.$$
See Relations between p norms for the first inequality. Also,
$$| x|_{2k-2} leq | x |_k implies | x_{2k-2} |^{2k-2}_{2k-2} leq | x |_k^{2k-2} ~$$
And we are done.
answered Jan 10 at 6:28
HTVHTV
1017
$endgroup$
add a comment |
$begingroup$
Ok, I figured out the answer.
We have
$$| x |_2 leq n^{1/2-1/k} | x |_k implies | x |_2^2 leq n^{1-2/k} | x |_k^2~.$$
See Relations between p norms for the first inequality. Also,
$$| x|_{2k-2} leq | x |_k implies | x_{2k-2} |^{2k-2}_{2k-2} leq | x |_k^{2k-2} ~$$
And we are done.
answered Jan 10 at 6:28
HTVHTV
1017
$endgroup$
Ok, I figured out the answer.
We have
$$| x |_2 leq n^{1/2-1/k} | x |_k implies | x |_2^2 leq n^{1-2/k} | x |_k^2~.$$
See Relations between p norms for the first inequality. Also,
$$| x|_{2k-2} leq | x |_k implies | x_{2k-2} |^{2k-2}_{2k-2} leq | x |_k^{2k-2} ~$$
And we are done.
answered Jan 10 at 6:28
HTVHTV
1017
edited Jan 10 at 6:33
answered Jan 10 at 6:28
HTVHTV
1017
answered Jan 10 at 6:28
HTVHTV
1017
answered Jan 10 at 6:28
HTVHTV
1017
1017
add a comment |
add a comment |
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