Concrete example of non-norm-attaining bounded linear operator on disc algebra

Question

A bounded linear operator $T$ from a Banach space $X$ to a Banach space $Y$ is called norm-attaining, if there exists a vector $x\in X$ with $\|x\|=1$ such that $$\|Tx\|=\|T\|.$$ Let $\mathbb{D}=\{z\in \mathbb{C}: |z|<1\}$. Consider the disc algebra $$A(\mathbb{D}):=\{f\in\text{ Hol}(\mathbb{D}):f \text{ is continuous on }\overline{\mathbb{D}},$$ $$\|f\|_{\infty}=\sup_{z\in\mathbb{D}}|f(z)|<\infty\}.$$

Can we get a concrete example of non-norm-attaining bounded linear operator $T: A(\mathbb{D}) \longrightarrow A(\mathbb{D})$?

Answer 1

First take $x^*\in A(\mathbb{D})^*$ that is non-norm-attaining. Next pick a non-zero $g\in A(\mathbb{D})$, and define $T: A(\mathbb{D})\to A(\mathbb{D})$ by $Tf = x^*(f)\cdot g$. Thus $\|T\|=\|x^*\|\cdot\|g\|$ and $T$ is non-norm-attaining.

A concrete $x^*\in A(\mathbb{D})^*$ that is non-norm-attaining:

Take $z_n= e^{i(1/n)}$ and, for $f\in A(\mathbb{D})$, define $x^*(f)= \sum_{n=1}^\infty (-1)^n 2^{-n} f(z_n)$.

Denoting by $\mathbb{T}$ the boundary of $\mathbb{D}$, we have $A(\mathbb{D})^* =L_1/H^1_0\oplus_1 V_{sing}(\mathbb{T})$, where $V_{sing}(\mathbb{T})$ is the space of singular (with respect to the Lebesgue measure) Borel measures on $\mathbb{T}$. See Chapter 1 of [A. Pelczynski, Banach spaces of analytic functions and absolutely summing operators. AMS 1977].

Then $x^*\in V_{sing}(\mathbb{T})$, $\|x^*\|=\sum_{n=1}^\infty 2^{-n}=1$ and $x^*$ is non-norm-attaining because $f(z_n)\to f(1)$.