Is the normed space of all bounded functions under the supremum norm isomorphic to $L^ \infty$?

We are familiar with the space $L^ \infty$, i.e., the space of all essentially bounded measurable functions, i.e., $$L^ \infty(X,\mathcal{M},\mu)=\{f: X \rightarrow \mathbb{R}| f \text{ is measurable and } \exists M>0 \ni: \mu \{x \in \mathbb{R} | |f(x)> M\}=0 \}$$ under the essential supremum norm, $$||f||_ \infty=\text{esssup}_{x \in X} |f(x)|:= \inf \{M>0 |\text{ } \mu \{x \in X \text{ }|\text{ } |f(x)|> M\}=0\}$$ (i.e., the infimum of all essential bounds). This is known to be a Banach space. This is also the natural analogue of the usual $L^p(X,\mathcal{M},\mu)$, $1\leq p<\infty$ for $p=\infty$, in the sense that $$\lim_{p \rightarrow \infty} ||f||_p =||f||_\infty$$ whenever there exist $q<\infty$ such that $f \in L^\infty(X,\mathcal{M},\mu) \bigcap L^q(X,\mathcal{M},\mu)$.

What can also be considered, and in fact what is more natural in a sense, is the space of all (actually) bounded functions $$\mathscr{L}^\infty(X)=\{f: X \rightarrow \mathbb{R}| \exists M>0 \ni: |f(x)| \leq M \forall x \in X\}$$ under the (actual) supremum norm, $$||f||=\sup_{x \in X} |f(x)|$$ Unlike the analogous cases of $1\leq p<\infty$, it can easily be seen that $||.||$ itself is a norm (and not just a semi-norm) on $\mathscr{L}^\infty(X)$. As such there is no real necessity to go for equivalence classes here. It can also be seen that this is a Banach space (see Is the space of bounded functions with the Supremum norm a Banach Algebra? for example). But is this different from our $L^\infty (X,\mathcal{M},\mu)$? The latter depends on the measure, while the former doesn't. On the one hand, there are more functions than there are measurable functions (w.r.t. any measure), while on the other, there are more essentially bounded functions than there are bounded functions. So, my questions are:

Is $\mathscr{L}^\infty(X)$ isomorphic to $L^\infty (X,\mathcal{M},\mu)$ as vector spaces?

Is $\mathscr{L}^\infty(X)$ under ||.|| isometric or isomorphic to $L^\infty (X,\mathcal{M},\mu)$ under $||.||_\infty$ as Banach spaces?

If not, is there any interesting relation between them?

We know that $$L^\infty (X,\mathcal{M},\mu)=\mathscr{L}^\infty(X)/\mathcal{N}(\mu)$$ as vector spaces, where $\mathcal{N}(\mu)=\ker(||.||_\infty)=\{f \in \mathscr{L}^\infty(X) | f=0 \text{ }\mu.a.e.\}$.

• I vaguely remember result of Bourgain that $\mathscr{L}^\infty([0,1])$ is not injective as Banach space while $L_\infty([0,1])$ is. So these spaces are not even isomorphic. As for isomorphism of vector spaces I think you can find the answer in Fremlin's monography, but I don't even remember the volume you should look at. The book is freely available. – Norbert Sep 10 '16 at 13:33
• It might be worth pointing out that $\mathscr{L}^\infty$ is just $L^\infty$ with respect to the counting measure on $X$. So more generally we could ask when $L^\infty(X, \mu)$ and $L^\infty(X, \nu)$ are isomorphic. – Josh Keneda Sep 10 '16 at 13:39
• @Norbert What do you mean by a Banach space being injective? – N Unnikrishnan Sep 10 '16 at 13:43
• Injective as an object in the category of Banach spaces with continuous linear maps between them. Equivalently, see here. If one Banach space is injective and the other isn't, they can't be isomorphic. – Josh Keneda Sep 10 '16 at 13:49
• @JoshKeneda That was a new concept for me, thanks. – N Unnikrishnan Sep 10 '16 at 14:16

No, in general they are not even isomorphic as vector spaces. Indeed, take the Dieudonné measure $\mu$ on $[0, \omega_1]$. Then $L_\infty(\mu)$ is one-dimensional, yet $\ell_\infty([0,\omega_1])$ is infinite-dimensional. (This is the standard symbol for what you call $\mathscr{L}_\infty$.)
Moreover, for any atomless finite measure space $(\Omega, \mu)$, $L_\infty(\mu)$ is not isomorphic to $\ell_\infty(\Omega)$ as a Banach space.
• @NUnnikrishnan, this isn't immediate. It uses some facts concerning the maximal ideal space of $L_\infty$; the key point is that the maximal ideal space of $L_\infty(\mu)$ satisfies ccc for any finite measure $\mu$, but $\ell_\infty(\Gamma)$ does not (for $\Gamma$ uncountable) and this property is preserved by Banach-space isomorphisms. – Tomek Kania Sep 10 '16 at 22:09