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Consider the complex Banach space $V=(\mathcal C(U(1), \mathbb C), \Vert \cdot \Vert_\infty)$ where $\mathcal C(U(1), \mathbb C)$ is the space of continuous complex maps defined on the unit circle endowed with the sup norm.

Define $G$ as the operator that maps an element $f \in V$ to the element $G(f)$ such that

$$G(f) : z \in U(1) \mapsto zf(z) \in \mathbb C.$$

Is $G$ in the same connected component that the identity in the subset of bounded linear invertible operators of $V$ denoted by $\mathcal{L}(V)^*$

Note: I moved this question @mathoverflow and deleted the math.stackexchange one.

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    $\begingroup$ In fact the general linear group of this Banach space is contractible. There is a proof in this article of Mityagin. A different article of his promises a simpler argument but I couldn't find an online copy. $\endgroup$
    – mme
    Commented Jul 30, 2020 at 14:42

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