Inverse closed subalgebra

Question

$a\in B\subset A$, $A$ is a Banach algebra with 1，$B$ is a closed subalgebra of $A$ with 1, if $a$ inverse in $A$，can you proof that $a$ inverse in $B$？

I think it is not true, but I try metric example all fail

Answer 1

A simple counterexample is to take $B=H^{\infty}$ the Hardy space (and Banach algebra with the supremum norm) of bounded analytic functions on the disc which is embedded naturally into $L^{\infty}(\mathbb T)$ the usual space of bounded measurable functions on the unit circle with the essential supremum norm - the functions in $H^{\infty}$ are precisely those in $L^{\infty}$ which do not have negative Fourier coefficients, as those extend as analytic functions to the unit disc in an obvious way.

(In $f \in L^1(\mathbb T)$ has Fourier series $\sum_{n \ge 0} a_ne^{in\theta}$ then $f(z)=\sum a_nz^n$ is analytic in the unit disc and its ae radial (nontangential) boundary limit exists and is precisely the original $f$, while if $f \in L^{\infty}$ the extension is in $H^{\infty}$ by maximum modulus)

Then $e^{\frac{z+1}{z-1}}$ is in $H^{\infty}$ but its inverse $e^{\frac{z+1}{1-z}}$ is not, being unbounded in the unit disc, though it is clearly invertible in $L^{\infty}$ as it has modulus $1$ ae on the unit circle.