Can there be a continuous linear injection of $L^\infty$ into one of its closed, separable subspaces? (Note: I am not requiring that injection to be surjective, nor to have closed range).
Specifically, I am looking at $L^\infty(D)$, where $D$ is the unit disk in the plane, and the subspace in question is the Bergman space $A^\infty$ of bounded holomorphic functions on $D$.
Yes, there can be. First, a simple example on the real line: $L^\infty([0,1])$ is injectively mapped into its separable subspace $C([0,1])$ by the Volterra operator $T$. $$Tf(x) = \int_0^x f(t)\,dt$$
Bergman space, dyadic construction
Let $Q_k$, $k=0,1,2,\dots$ be some enumeration of dyadic squares in the plane. Let $c_k=\frac{1}{|Q_k|}\int_{Q_k\cap D}f$ and define $$Tf(z) = \sum_{k=0}^\infty c_k (z/2)^k$$ Since $(c_k)\in \ell^\infty$, we have $Tf\in A^\infty(D)$. Also, $\sup_D|Tf|\le 2\sup|c_k| \le 2\sup|f|$. Finally, if $Tf\equiv 0$, then $c_k=0$ for every $k$, and the Lebesgue differentiation theorem implies $f=0$ a.e.
Bergman space, Cauchy transform
I am not confident that this construction works, but I'll state the idea: define $$Tf(z) = \iint_D \frac{f(\zeta)}{3+z-\zeta}$$ Thanks to $3$, the integral converges nicely for $z\in D$, and defines a bounded holomorphic function. Conceptually, this is just the Cauchy transform of $f$ restricted to the disk $D'$ of unit radius centered at $3$. It remains to consider injectivity. If the Cauchy transform vanishes identically on $D'$, then it vanishes everywhere outside of $D$, and that probably can't happen: there are uniqueness theorems for this transform.