# Computing $\|f\|$ when $f(x,0,z,0,0,...)=2x-z$.

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?

• 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. Dec 7, 2019 at 15:15
• Thanks for the comment. I've done what you suggested Dec 7, 2019 at 20:26

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 $$\|F\| = \|f\|$$.