6
$\begingroup$

Let $1\leq p<\infty$ and let $B$ be the subspace of $(l_{p}(\mathbb{N}),\|\cdot\|_{p})$ given by $B=\{(x,0,z,0,0,...) : x,y \in \mathbb{C}\}$. Let $f : B \to \mathbb{C}$ be given by $f(x,0,z,0,0,...)=2x-z$.

I'd like to compute $\|f\|$.

My attempt: Assume $B$ is finite-dimensional. Write $\textbf{x}=(x,0,z,0,0,...,0)$. $\|f(\textbf{x})\| =|2x-z|\leq 2|x|+|z| \leq 2\|\textbf{x}\|_{1} \leq 2C\|\textbf{x}\|_{p}$ for some $C>0$, since any two norms are equivalent, when $B$ is finite-dimensional. Also, by Hölder's inequality, $\|\textbf{x}\|_{1}\leq 2^{1-1/p}\|\textbf{x}\|_{p}$ (is happy to provide the details if needed). Hence, $\|f(\textbf{x})\|\leq 2\|\textbf{x}\|_{1}\leq 2^{2-1/p}\|\textbf{x}\|_{p}$. Now, if I can find $\textbf{x}$ s.t. we get equality, then $\|f\|=2^{2-1/p}$. This, however, is where I am stuck. Perhaps there is a smaller bound on $\|f(\textbf{x})\|$ that I have missed?

$\endgroup$
2
  • $\begingroup$ You should indicate the range of $p$, possibly $1\leqslant p<\infty\,$? And maybe add the tag [dual-spaces] and use "\|" to produce nicer norm delimiters. $\endgroup$
    – Hanno
    Dec 7, 2019 at 15:15
  • $\begingroup$ Thanks for the comment. I've done what you suggested $\endgroup$
    – M. B.
    Dec 7, 2019 at 20:26

1 Answer 1

2
$\begingroup$

The answer is that $\|f\| = \|(2,-1)\|_q$, where $p$ and $q$ are conjugate indices, i.e. $$\|f\| = \begin{cases} 2, &p = 1 \\ (2^{p/(p-1)}+1)^{(p-1)/p}, &1 < p < \infty \end{cases}$$

To prove this, first use Hölder's inequality to show $\|f\| \leq \|(2,-1)\|_q$. Then consider $x = \|(2,-1)\|_q^{1-q}2^{q-1}$ and $z = -\|(2,-1)\|_q^{1-q}$ to show $\|f\| \geq \|(2,-1)\|_q$.

Here's a more general approach: Let $\mathbf{a} = (a_1,a_2) \in \mathbb{C}^2$ be given, and define $f_\mathbf{a}: B \to \mathbb{C}$ by $f_\mathbf{a}(x,0,z,0,0,\dots) = a_1x+a_2z$. Show $\|f_\mathbf{a}\| = \|\mathbf{a}\|_q$ using the same steps as above. The hardest part here will be picking $x$ and $z$ to work for any $\mathbf{a} \in \mathbb{C}^2$. Then $f = f_\mathbf{a}$ for $\mathbf{a} = (2,-1)$, and the result follows.

More elegantly, note that $(\ell_p)^* \cong \ell_q$ and use this to determine $\|f\|$. This will take some care since you will need to consider the extension $F: \ell_p \to \mathbb C$ of $f$ given by $F(x_1,x_2,x_3,\dots) = 2x_1-x_3$, show $\|F\| = \|(2,0,-1,0,0,\dots)\|_q$ and argue that $\|F\| = \|f\|$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .