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 $y\in l_p$?


(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 $$ \|\phi\|_{[l^p]^*}=\|\phi_y\|_{[l^p]^*}=\|y\|_{l^q} = (1+2^q+3^q)^{1/q}. $$

  • $\begingroup$ 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

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