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 $$\|\phi\|_{[l^p]^*}=\|\phi_y\|_{[l^p]^*}=\|y\|_{l^q} = (1+2^q+3^q)^{1/q}.$$
• But isn't $l_p^*=l_q$? – ChikChak Jan 12 at 21:39
• @ChikChak Yes, but more precisely, as I said $(l^p)^*\cong l^q$. – Song Jan 13 at 14:40
• But still, why $y\in l^q$? – ChikChak Jan 14 at 21:16
• @ChikChak Actually $y\in l^r$ for all $r\in [1,\infty]$. It is quite obvious for a finite sequence. – Song Jan 14 at 21:19