0
$\begingroup$

Let $X=l_p^{(3)}$, where $1\lt p \lt \infty$, and $\phi(x) = x_1-2x_2+3x_3$.

Decide whether $\phi$ is bounded, and if so, find $||\phi||$.

So by marking $y=(1,-2,3)$, we can see that $\phi(x)=\sum_{j=1}^3x_iy_i$.

Therefore, by Riesz representation theorem, we get that $||\phi||=||y||=(1+2^p+3^p)^{\frac{1}{p}}$.

However, in the book that I study, the final answer is $(1+2^q+3^q)^{\frac{1}{q}}$, where $1\lt q\lt \infty$ and $\frac{1}{p}+\frac{1}{q}=1$.

I know that $l_p^*=l_q$... But why $y\in l_q$ and not $y\in l_p$?

$\endgroup$
-1
$\begingroup$

(Assume the underlying field is real.) Riesz representation theorem says that if $1/p+1/q=1$, $p\in [1,\infty)$, then there is an (isometric) isomorphism$$ l^q \ni y\longleftrightarrow \phi_y(\cdot)\in [l^p]^* $$ between $l^q$ and $[l^p]^*$ where $\phi_y(x) = \sum_{i=1}^3 x_i y_i$ for all $x\in l^p$. In our case, $y=(1,-2,3)'$ and $\phi=\phi_y$. Hence we should have $$ \|\phi\|_{[l^p]^*}=\|\phi_y\|_{[l^p]^*}=\|y\|_{l^q} = (1+2^q+3^q)^{1/q}. $$

$\endgroup$
  • $\begingroup$ But isn't $l_p^*=l_q$? $\endgroup$ – ChikChak Jan 12 at 21:39
  • $\begingroup$ @ChikChak Yes, but more precisely, as I said $(l^p)^*\cong l^q$. $\endgroup$ – Song Jan 13 at 14:40
  • $\begingroup$ But still, why $y\in l^q$? $\endgroup$ – ChikChak Jan 14 at 21:16
  • $\begingroup$ @ChikChak Actually $y\in l^r$ for all $r\in [1,\infty]$. It is quite obvious for a finite sequence. $\endgroup$ – Song Jan 14 at 21:19

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.