- #1
ENgez
- 75
- 0
problem statement:
need to show:
[tex]||w||_p^2+||u||_p^2-2(||u||_p^p)^{\frac{2}{p}-1}\Sigma_i(u(i)^{p-1}w(i)) [/tex]
can be bounded as a function of
[tex] ||w-u||_p^2 [/tex]
where [tex] p\in[2,\infty) [/tex]
work done:
the expressions are equal for p=2, and i suspect that
[tex] ||w||_p^2+||u||_p^2-2(||u||_p^p)^{\frac{2}{p}-1}\Sigma_i(u(i)^{p-1}w(i)) \leq||w-u||_p^2 [/tex]
but i get stuck here. Is there some kind of p-norm inequality I can apply here?Thank you!
need to show:
[tex]||w||_p^2+||u||_p^2-2(||u||_p^p)^{\frac{2}{p}-1}\Sigma_i(u(i)^{p-1}w(i)) [/tex]
can be bounded as a function of
[tex] ||w-u||_p^2 [/tex]
where [tex] p\in[2,\infty) [/tex]
work done:
the expressions are equal for p=2, and i suspect that
[tex] ||w||_p^2+||u||_p^2-2(||u||_p^p)^{\frac{2}{p}-1}\Sigma_i(u(i)^{p-1}w(i)) \leq||w-u||_p^2 [/tex]
but i get stuck here. Is there some kind of p-norm inequality I can apply here?Thank you!