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Some basics for vector-valued L^p spaces, when p is finite

23 June, 2019

There is a well-established way to extend the familiar definition of the Lebesgue spaces L^p[0,1] (for 1 ≤ p ≤ ∞) to the setting of E-valued functions on [0,1], where E is a complex-valued Banach space. I say “well-established” but it was only relatively recently (c. 2013/2014) that I found myself needing to look up some of the technical subtleties, and even more recently I realised I had remained ignorant of some fairly important foundational aspects of this process or construction.

When 1≤ p < ∞ the natural guess is that we should take equivalence classes of those “measurable” functions f: [0,1]\to E that satisfy

(1)\qquad\qquad \displaystyle \int_0^1 \Vert f(t) \Vert^p \,dt < \infty

But we run into an immediate issue: which functions [0,1]\to E do we declare to be "measurable"? It is clearly necessary for t\mapsto \Vert f(t)\Vert to be measurable as a function [0,1]\to \mathbb R (as well as p-integrable), but is this a good definition or should it be treated as a consequence of some “better” definition?

Moreover: how should we treat the case p=∞ ?

Digression on Borel versus Lebesgue

We shall work with the Borel sigma-algebra \mathcal B on [0,1] rather than the Lebesgue sigma-algebra \mathcal L, though this is largely a matter of preference/convenience: see this EOM entry for some of the relevant context. In the sense of these lecture notes by J. van Neerven, we are working with (strongly or weakly) \mathcal B-measurable functions, rather than (strongly or weakly) μ-measurable functions where μ is taken to be Lebesgue measure.

The key definitions

Definition 1. A function f:[0,1]\to E is called a simple function if there are (Borel) measurable subsets X_1,\dotsm X_n and \xi_1, \dots, \xi_n in E such that

(2)\qquad \qquad \displaystyle f = \sum_{j=1}^n \xi_j 1_{X_j}

Definition 2. A function f:[0,1]\to E is said to be strongly measurable if there is a sequence (f_n) of simple E-valued functions such that \Vert f(t) - f_n(t) \Vert \to 0 for all t\in [0,1].

Definition 3. A function f:[0,1]\to E is said to be weakly measurable if, for each \psi\in E^*, the scalar-valued function \psi\circ f:[0,1]\to \mathbb C is (Borel-to-Borel) measurable.

Definition 4. A function f:[0,1]\to E is said to be Borel measurable if f^{-1}(S) is Borel for every Borel subset S\subseteq E.

It is a straightforward exercise to show that if (f_n) is a sequence of Borel measurable functions [0,1]\to\mathbb R and f_n \to f pointwise, then f is Borel measurable. See e.g. this MO answer. The same MO answer also explains why strongly measurable E-valued functions are both weakly measurable and Borel measurable. (Essentially, one verifies this for simple functions, and then invokes the result mentioned above.)

The image of any strongly measurable function is always separable (with respect to the norm topology of E), and so if f:[0,1]\to E is strongly measurable, it is weakly measurable and takes values in a separable subspace of E. An important theorem of Pettis tells us that the converse is true.

Theorem 5 (Pettis measurability theorem). Let f:[0,1]\to E be a weakly measurable function which takes values in a separable subspace of E. Then f is strongly measurable.

(One can weaken the hypotheses slightly: it is enough to require separable range and "σ(E,F)-measurability", where F is a weak-star dense subspace of E*. See the notes of van Neerven that are linked above.)

Note that if \phi: E\to F is any function between two Banach spaces, not necessarily linear, and f:[0,1]\to E is a simple function, then \phi\circ f : [0,1]\to F is also a simple function. With this in mind, the following result is an almost immediate consequence of Definition 1.

Proposition 6. Let E and F be Banach spaces and let \phi:E \to F be a continuous function. If f:[0,1]\to E is strongly measurable, then so is \phi\circ f:[0,1]\to F.

In particular, if f:[0,1]\to E is strongly measurable, then t\mapsto \Vert f(t)\Vert is (Borel-to-Borel) measurable.

Remark 7. Proposition 6 may seem unsurprising, given that the corresponding statement where “strongly” is replaced by “Borel” is extremely easy to prove. However, it’s not clear to me right now if the result still holds when “strongly” is replaced by “weakly”.

Lebesgue-Bochner spaces for finite p

In view of the comment following Proposition 6, we can now define E-valued L^1[0,1] in a natural way.

Definition 8. Let E be a Banach space. We say that a strongly measurable function f:[0,1]\to E is Bochner integrable if \int_0^1 \Vert f(t)\Vert \,dt < \infty. The vector space V of all such f carries an obvious/natural seminorm, and we define L^1([0,1];E) to be the quotient of V by the kernel of this seminorm.

We view elements of L^1([0,1];E) to be equivalence classes of Bochner integrable E-valued functions, with the equivalence relation being “differs only on a set of measure zero”.

Loosely speaking, for 1≤p< ∞, we can then define L^p([0,1];E) to be the space of all (equivalence classes of) strongly measurable functions f:[0,1]\to E for which t\mapsto \Vert f(t)\Vert^p is integrable. Equipping L^[([0,1];E) with the natural norm, one can show it is complete by adapting the usual proofs from the scalar-valued case.

It is not hard to prove that the simple functions [0,1]\to E form a norm-dense subspace of L^p([0,1];E), and so if one likes tensor product formalism one can view L^p([0,1];E) as a completion of the algebraic tensor product L^p[0,1] \otimes E.

But what about p=∞?

This will be handled in the next blog post.

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Continuity of a radial square-root map for complex numbers (take 2)

15 June, 2019

Following on from the previous blog post, it is worth giving a cleaned-up version of the proof. While putting this together, I also found a way to obtain the optimal constant of 2, although I am not entirely happy with the trick needed to achieve this.


Restating the problem (with the sharp constant)

As in the previous post, we define a function \phi: \mathbb C \to \mathbb C by \phi(z) = |z|^{-1/2}z. Alternatively: \phi(re^{i\theta}) = r^{1/2}e^{i\theta} for all r\geq 0 and all \theta\in\mathbb R. Our goal is now to prove the following inequality

Proposition 1. |\phi(w)-\phi(z)|^2 \leq 2 |w-z| for all w,z\in \mathbb C.

Reducing the problem to one with fewer parameters

We can simplify/reduce the problem slightly using symmetry arguments. To maintain a more formal style than the previous post, let’s spell this out more explicitly.

Lemma 2. Let t\geq 1. Then |t-e^{i\theta}|^2 \leq 2 |t^2-e^{i\theta}| for all \theta\in [0,\pi].

Proof of Proposition 1, using Lemma 2. The desired inequality is trivial if either w or z is zero.
Let w,z\in \mathbb C\setminus\{0\}, and assume without loss of generality that |w|\leq |z|.

Since \phi is angle-preserving (and orientation-preserving) we may apply a rotation or reflection of the complex plane which maps z to the real-line and w to a point in the closed upper-half plane. We then let
t= |z|^{1/2}|w|^{-1/2} \geq 1 and choose \theta\in [0,\pi] such that w=|w|e^{i\theta}.

Then

(1)\qquad\qquad \displaystyle |\phi(w)-\phi(z)|^2 = | |w|^{1/2} e^{i\theta} - |z|^{1/2} |^2 = |w| | e^{i\theta} - t |^2

which, by Lemma 2, is bounded above by

(2)\qquad\qquad \displaystyle 2|w|  |t^2-e^{i\theta}| = 2 |w| \left\vert |z| |w|^{-1} - e^{i\theta} \right\vert = 2|z -w|

as required. Q.E.D.

The proof of Lemma 2

The proof of Lemma 2 will be divided into two cases.

Case 1. 0\leq\theta \leq \pi/2.

In this case |t- e^{i\theta}| \leq |t+e^{i\theta}|; this is evident from the geometry of this configuration, but also follows from squaring both sides and using \cos\theta\geq 0. Therefore

(3)\qquad\qquad \displaystyle | t-e^{i\theta}|^2 \leq |t-e^{i\theta}| |t+e^{i\theta}| = |t^2-e^{2i\theta} |.

Also, observe that

(4)\qquad\qquad \displaystyle |e^{i\theta} -e^{2i\theta} | = |1-e^{i\theta} | \leq |t^2-e^{i\theta} |\;.

The last inequality is once again evident from a geometrical argument. To give a more algebraic/Cartesian argument: the two complex numbers t^2-e^{i\theta} and 1-e^{i\theta} have the same imaginary part; their real parts are t^2-\cos\theta and 1-\cos\theta respectively; and t^2-\cos\theta\geq 1-\cos\theta \geq 0 since t\geq 1.

Combining (3) and (4) we obtain

(5)\qquad\qquad \displaystyle |t-e^{i\theta}|^2 \leq |t^2-e^{2i\theta} \leq |t^2-e^{i\theta} | + |e^{i\theta}- e^{2i\theta} | \leq 2|t^2-e^{i\theta}|

as required. Q.E.D.

Case 2. \pi/2 \leq\theta \leq \pi.

Let d= -\cos\theta\in [0,1], so that

(6)\qquad\qquad\displaystyle |t-e^{i\theta}|^2 = t^2+1 +2td \quad\hbox{ and } \quad |t^2-e^{i\theta}| = (t^4+1+2t^2d)^{1/2}\;.

Note that as a special case of the Cauchy–Schwarz inequality,

(7)\qquad\quad\displaystyle x_1+x_2+x_3+x_4 \leq 2 \sqrt{x_1^2+x_2^2+x_3^2+x_4^2} \quad\hbox{ for all }x_1,x_2,x_3,x_4 \geq 0.

Applying (7) with x_1=t^2, x_2 = 1, x_3 =x_4 = td^{1/2}, we obtain

(8)\qquad\qquad\displaystyle t^2+1+2td^{1/2} \leq 2 ( t^4+1+2t^2d)^{1/2}\;.

Since 0\leq d\leq 1 and t\geq 0, 2td \leq 2td^{1/2}. Combining this with (6) and (8), we conclude that

(9)\qquad\qquad\displaystyle |t-e^{i\theta}|^2 \leq t^2+1+2d^{1/2} \leq 2 |t^2-e^{i\theta}|

as required. Q.E.D.

Remarks

  1. In comparison with the approaches taken in the previous blog post, note that in Case 2 we have improved the constant from 2\sqrt{2} to 2, but with a slightly more ad hoc argument that is less geometric (although we are able to sidestep any messy calculus with Cauchy–Schwarz).
  2. It is also worth noting that for much of the proof, the formulation in terms of complex numbers is merely for convenience, and that most of the arguments merely take place in \mathbb R^2 with its usual norm — or, to be more abstract and less co-ordinate dependent, in 2-dimensional Euclidean space. The exception is one part of Case 1, namely the inequality (3), where we made use of the algebraic structure in \mathbb C and the multiplicative property of | \cdot |. Can we find an alternative approach, which only uses the Euclidean geometry of \mathbb R^2?
  3. On a related note: the definition of the map \phi can be extended to any normed space V, as a radial square-root map: we define \phi(\mathbf{0})=\mathbf{0} and \phi(v) = \Vert v \Vert^{-1/2} v for any non-zero vetor v\in V. If V is Euclidean (i.e. its norm satisfies the parallelogram law) then, since 2-dimensional subspaces of any Euclidean real vector space are Euclidean, the previous remarks and Proposition 1 show that \Vert \phi(v_1)-\phi(v_2)\Vert^2 \leq 2 \Vert v_1-v_2\Vert for all v_1,v_2\in V. What happens for more general norms?

Continuity of a radial square-root map for complex numbers

12 June, 2019

I don’t remember where I first saw the following inequality (and operator-theoretic generalizations):

Lemma 1. If r and s are non-negative real numbers, then (r-s)^2 \leq |r^2-s^2|.

The lemma is most easily/intuitively proved by noticing that \vert r-s\vert \leq \max(r,s) \leq r+s (draw a picture!) and then multiplying both sides of this inequality by |r-s|. The lemma also shows, without need for any calculus, that the square-root function on [0,\infty) is Hölder continuous with exponent 1/2.

I was recently reminded of the inequality in Lemma 1 by some discussions with Matt Daws, who had pointed out to me that it gives an easy proof that the square-root function \Phi_+: L^1[0,1]_+ \to L^2[0,1]_+ is continuous. (See also comments by Matt and others on this old MO question.) In some ongoing discussions with Matt and Jon Bannon, I found myself wanting a version of this for not-necessarily real-valued functions in L^1[0,1].

Specifically, consider the function \phi: \mathbb C \to \mathbb C defined by

\displaystyle \phi(0)=0\quad\hbox{and}\quad \phi(z) = |z|^{-1/2}z

and then define \Phi: L^1[0,1] \to L^2[0,1] by

\Phi(f) = \phi\circ f

Note that the restriction of \Phi to the positive cone L^1[0,1]_+ agrees with the square root map \Phi_+ already mentioned above.

A direct calculation shows that \Phi is norm-preserving, but since it is nonlinear this does not automatically ensure continuity; and it is continuity of \Phi which we wanted to know/check. Continuity of \Phi, or very similar maps, is presumably folklore, but having failed in some half-hearted attempts to find a reference for this result, I decided it would be easiest to try and come up with a proof with bare hands. Guided by the easy proof that \Phi_+ is continuous, it is natural to try and prove some 2-dimensional version of the inequality at the start of this post, perhaps at the expense of worse constants. Specifically, we would like to know that the following result holds.

Claim 2. There exists a constant c\geq 1 such that |\phi(w)-\phi(z)|^2 \leq c|w-z| for all w,z\in \mathbb C.

By the same argument used to deduce continuity of \Phi_+ from Lemma 1, one sees that Claim 2 implies continuity of \Phi; in fact, we get Hölder continuity with exponent 1/2.

In what follows, I will sketch one possible way to prove this claim, not necessarily with the best constant c. It is meant as a tidied version of a train of thought, rather than an attempt to present the cleanest and most polished approach. (I should acknowledge the influence, in spirit if not the fine detail, of these old blogposts-before-weblogs-existed writings by Timothy Gowers, which were quite influential on me during my years as a PhD student and postdoctoral researcher.)


The first thing to notice is that \phi preserves arguments of complex numbers; its effect is merely radial (hence the title of this blog post). This means that in trying to prove Claim 2, we are always free to rotate w and z by a fixed angle, so that either can be assumed real if this is convenient. Introducing the change of variables r=|w|^{1/2}, s=|z|^{1/2} and rotating to make w real, we see that the claim is equivalent to

\displaystyle |r-se^{i\theta} |^2 \overset{\hbox{\bf?}}{\leq} c |r^2- s^2e^{i\theta} | \quad\hbox{for all }r,s\geq 0\hbox{ and }\theta\in{\mathbb R}\quad\quad(*)

Next: what happens if we expand out both sides of the desired inequality (*)? The left hand side is r^2+s^2 -2rs \cos\theta — which you can also see by drawing a picture and using the cosine rule from school trigonometry — and the right hand side is then \sqrt{r^4+s^4-2r^2s^2\cos\theta}. At this point it looks unappealing to square both sides again and attempt to compare terms; this might work, but it looked messy when I tried it, so let’s take a step back and think again.

Recall that the case θ=0 is covered by Lemma 1. How or why does the proof break down for other values of θ? Well, for θ=0 the left hand side of (*) is |r-s|^2 and the right hand side is c|r^2-s^2| = c|r-s| |r+s|, and we won in this case because |r-s|\leq |r+s|, so we can take c=1. But for general θ we don’t have the same convenient factorization of the right-hand side.

What would be nice is if we had c|r^2-s^2e^{2i\theta}| on the right hand side of (*). For this does factor as c|r-se^{i\theta}||r+se^{i\theta}|, and then we would be hoping to dominate |r-se^{i\theta}| by a multiple of |r+se^{i\theta}|. Of course e^{2i\theta} is not the same as e^{i\theta} for general θ, but perhaps for small values of θ we can control the discrepancy with some crude bound, using calculus and the Mean-Value theorem if necessary?

Putting this thought on one side for the moment, let us think what to do when e^{i\theta} is far from 1. In fact, as a “stress-test”, what happens when e^{i\theta}=-1? (With hindsight we should have thought of this case sooner, since it corresponds to looking at Lemma 1 and wondering whether it applies to all r,s\in\mathbb R, not just positive values.) In this case the left hand side of (*) is |r+s|^2 and the right hand side is c|r^2+s^2|, so clearly we cannot take c=1 any more; nevertheless, the AM-GM inequality shows that we could take c=2 in this case.

At this point we have two separate working arguments for θ=0 and θ=π, so how can we handle intermediate cases? We already had a brief look at what happens for small values of θ, so let’s look at what happens when θ is close to π. Drawing a picture of the appropriate obtuse-angled triangle, we realise that for the right-hand side of (*) to be small, both r^2 and s^2 must be small. In fact, as we let θ vary between π/2 and 3π/2, |r^2-s^2e^{i\theta}| is minimized at the endpoints (i.e. when we have a right-angled triangle), and so

\displaystyle \min_{\pi/2\leq \theta\leq\pi/2} |r^2-s^2e^{i\theta}| = \sqrt{r^4+s^4}

(The geometric intuition can be backed up by an appeal to the cosine rule; recall that on this interval, \cos\theta \leq 0.)
What about the left hand side of (*)? Well, we may as well replace |r-se^{i\theta}|^2 with its “worst-case scenario”, namely (r+s)^2, and we already saw this is bounded above by 2r^2+2s^2. Applying Cauchy-Schwarz, we see that this in turn is bounded above by 2\sqrt{2}(r^4+s^4)^{1/2}. So putting things together, we have established

Lemma 3. For \pi/2 \leq\theta\leq 3\pi/2, and any r,s\geq 0, we have |r-se^{i\theta}|^2 \leq 2\sqrt{2} |r^2-s^2e^{i\theta} |.

Let’s turn back to the case of acute-angled triangles, i.e. -\pi/2 \leq \theta\leq \pi/2, or equivalently the region where \cos\theta\geq 0. Can we make good on the earlier hopes that

  1. |r-se^{i\theta}| is dominated by (a multiple of) |r+se^{i\theta}| ?
  2. |r^2-s^2e^{2i\theta}| is dominated by (a multiple of) |r^2-s^2e^{i\theta}| ?

Recall that both of these statements do hold when θ=0. In fact, drawing a picture, we see that |r-se^{i\theta}|\leq |r+se^{i\theta}| for any θ in this range; once again, the geometric intuition from drawing triangles can be backed up with explicit expansion of both sides using the cosine formula and the fact that \cos\theta\geq 0. So we do indeed have |r-se^{i\theta}|^2 \leq |r^2-s^2e^{2i\theta}|, and it only remains to prove that the second of our two statements holds.

We seem to be doing well drawing pictures, so let’s do this. While we’re at it, let us rescale the second statement to reduce notational clutter, so that we are aiming to prove

\displaystyle |t-e^{2i\theta}| \overset{\hbox{\bf?}}{\lesssim} |t-e^{i\theta}| \quad\hbox{ for all } \theta\in [-\pi/2,\pi/2] \hbox{ and all } t\geq 0\quad\quad(**)

(Clearly, if we can prove this then the 2nd statement above will follow.)

Now, in drawing pictures, it seems that we should distinguish between the cases 0\leq t\leq 1 and t\geq 1. Let’s look at the second case, and consider the triangle formed by the three points t,e^{i\theta},e^{2i\theta}. Then |t-e^{2i\theta}| is bounded above by the sum of the other two side lengths of the triangle, but it is clear from the picture that |e^{2i\theta}-e^{i\theta}| \leq \hbox{arclength}(e^{i\theta},e^{2i\theta}) while |t-e^{i\theta}|\geq \hbox{arclength}(e^{i\theta},e^{2i\theta}). Hence we have proved that

\displaystyle |t-e^{2i\theta}| \leq 2 |t-e^{i\theta}| \quad\hbox{ for all } \theta\in [-\pi/2,\pi/2] \hbox{ and all } t\geq 1\quad\quad(**)

(Actually, now that we have written this down, we can simplify the proof of (**) slightly to get rid of the arguments using arc length. Observe that |t-e^{2i\theta}| \leq |t-e^{i\theta} | + |e^{i\theta}-e^{2i\theta}| =  |t-e^{i\theta} | + |1-e^{i\theta}| and then observe that since t\geq 1, we have |1-e^{i\theta}|\leq |t-e^{i\theta}|.)

But what if 0\leq t <1? Well, in this case we can apply (**) with t replaced by t^{-1} and θ replaced by -θ, to get

\displaystyle |t^{-1}-e^{-2i\theta}| \leq 2 |t^{-1}-e^{-i\theta}|

and then we have

\displaystyle |e^{2i\theta}-t| = |t| |t^{-1}-e^{-2i\theta}| \leq 2|t| |t^{-1}-e^{-i\theta}| = 2 |e^{i\theta}-t |

as required. Hence we have established the inequality (**) with a constant 2 on the right-hand side, and therefore by rescaling back up again, we have proved the following lemma.

Lemma 4. For -\pi/2 \leq\theta\leq \pi/2, and any r,s\geq 0, we have |r-se^{i\theta}|^2 \leq 2 |r^2-s^2e^{i\theta} |.

Combining Lemma 3 and Lemma 4, we have proved Claim 2, with a value of c=2\sqrt{2}.


Looking back on this, it is clear that the arguments above could be written up in a more concise and more formal way, but this will be left for a shorter follow-up blog post, which might say more about the motivation for the original problem. However, at present I don’t see how to avoid the division into two cases (Lemma 3 and Lemma 4). It should be possible to improve the constant in Lemma 3, since on one side of the inequality we were using worst-case behaviour of the left hand side at θ=π while using worst-case behaviour of the right hand side at θ=π/2.


Update: after writing the bulk of this post, I learned from Matt Daws that he can get c=2. His argument is more calculus-based and less explicitly geometric. Both Matt and I would be happy to hear of any explicit references for this inequality, either in the sharp form c=2 or even just for some explicit value of c.

the crossfire of childhood and stardom

6 April, 2019
tags:

And we’ll bask in the shadow

Of yesterday’s triumph

Sail on the steel breeze

Come on you boy child

You winner and loser

Come on you miner for truth and delusion

and shine

None, save the undone years

16 March, 2019
tags:

As time went on we saw less and less of Teddy and Vern until eventually they became just two more faces in the halls. That happens sometimes. Friends come in and out of your life like busboys in a restaurant.

you’re on the wire and can’t get back

how could you go and die

what a lonely thing to do

dona eis sempiternam requiem

Mais ne vous reveillez pas

30 December, 2018

Via Terry Tao’s blog, I just learned that Jean Bourgain passed away just before Christmas, at the relatively young age of 64.

I never met Bourgain, and have never studied in any depth many of the areas of analysis where he broke new ground and had lasting impact, but several of his papers – not even his deepest or hardest work, just ones that happened to touch on areas of interest to me – have intrigued me with varying degrees of enlightement and bafflement. Here is a non-comprehensive selection (representing only my own interests) based on a few bookmarks and some reflection off the top of my head.

A counterexample to a complementation problem
Compositio Math. (1981)

New Banach space properties of the disc algebra and H^\infty
Acta Math. (1984)

Translation invariant forms on L^p(G)(1<p<\infty )
Annales Institut Fourier (1986)

On the similarity problem for polynomially bounded operators on Hilbert space
Isr. J. Math. (1986)

A problem of Douglas and Rudin on factorization
Pacific J. Math. (1986)

On the dichotomy problem for tensor algebras
Trans. Amer. Math. Soc. (1986)

Bounded orthogonal systems and the Λ(p)-set problem
Acta Math. (1989)

Sidonicity and variants of Kaczmarz’s problem (with M. Lewko)
Annales Institut Fourier (2017)

I expect that other bloggers who are more au fait with Bourgain’s work in harmonic analysis, PDE, and additive combinatorics will say more about his impact in those areas; and those who have met him and had deeper involvement with his work will be able to offer more fitting tributes. A start is the blog post of Terry Tao which I mentioned at the start

Cette cathédrale en pierre
Traînez-la à travers bois
Jusqu’où vient fleurir la mer
Mais ne vous reveillez pas

We know they’re isomorphic, but how are they isomorphic?

14 October, 2018

The given information

  1. L \cong L \oplus L, in a very nice way
  2. m \cong m \oplus m, in a very nice way
  3. L\cong m \oplus Q_m, in a nice way
  4. m \cong L \oplus Q_L, in a faintly dodgy way

Wizardry from Warsaw

m \cong L\oplus Q_L \cong (L\oplus L) \oplus Q_L \cong L \oplus (L\oplus Q_L) \cong L \oplus m

L \cong m\oplus Q_m \cong (m\oplus m) \oplus Q_m \cong m\oplus (m\oplus Q_m) \cong m\oplus L

Conclusion: m \cong L\oplus m \cong m\oplus L \cong L.

A nagging question

Just how nice or dodgy is our final isomorphism m\cong L?