# Linear operators which you can write down are continuous

This is a follow-up to a comment I made on Tim Gowers’s post concerning the use of Zorn’s lemma (which I encourage students, in particular, to read if they have not done so yet). The issue was whether it is possible to write down concretely an unbounded (or in other words, not continuous) linear operator on a Banach space. I mentioned that I had been explained a few years ago that this is not in fact possible, in the same sense that it is not possible to “write down” a non-measurable function: indeed, any measurable linear map

$T\ :\ U\rightarrow V$

where U is a Banach space and V is a normed vector space, is automatically continuous.

Incidentally, I had asked this to my colleague É. Matheron in Bordeaux because the question had arisen while translating W. Appel’s book of mathematics for physicists: in the chapters on unbounded linear operators (which are important in Quantum Mechanics), he had observed that those operators could often be written down, but only in the sense of a partially defined operator, defined on a dense subspace, and we wondered if the dichotomy “either unbounded and not everywhere defined, or everywhere defined, and continuous”, was a real theorem or not. In the sense that measurability is much weaker than the (not well defined) notion of “concretely given”, it is indeed a theorem.

Not only did Matheron tell me of this automatic continuity result, he gave me a copy of a short note of his (“A useful lemma concerning subseries convergence”, Bull. Austral. Math. Soc. 63 (2001), no. 2, 273–277), where this result is proved very quickly, as a consequence of a simple lemma which also implies a number of other well-known facts of functional analysis (the Banach-Steinhaus theorem, Schur’s theorem on the coincidence of weak and norm convergence for series in l1, and a few others). On the other hand, I don’t know who first proved the continuity result (Matheron says it is well-known but gives no reference).

The proof is short enough that I will present it; it is a nice source of exercises for a first course in functional analysis, provided some integration theory has been seen before (which I guess is always the case).

Here is the main lemma, due to Matheron, in a probabilistic rephrasing, and a slightly weaker version:

Main Lemma: Let G be a topological abelian group, and let An be an arbitrary sequence of measurable (Borel) subsets of G, and (gn) a sequence of elements of G. Assume that for every n and every g in G, either g is in An, or g-gn is in An.

Let moreover be given a sequence of independent Bernoulli random variables (Xn), defined on some auxiliary probability space.

Then, under the condition that the series

$\sum_{n\geq 1}{X_n g_n}$

converges almost surely in G, there exists a subsequence (hn) of (gn) such that

$\sum_{n\geq 1}{h_n}$

converges and belongs to infinitely many An.

This is probably not easy to assimilate immediately, so let’s give the application to automatic continuity before sketching the proof. First, we recall that Bernoulli random variables are such that

$\mathbf{P}(X_n=0)=\mathbf{P}(X_n=1)=1/2.$

Now, let T be as above, measurable. We argue by contradiction, assuming that T is not continuous. This implies in particular that for all n, T is not bounded on the ball with radius 2-n, so there exists a sequence (xn) in U such that

$\|x_n\|<2^{-n},\ \text{and}\ \|T(x_n)\|>n.$

We apply the lemma with

$G=U,\text{ written additively},\ g_n=-x_n,\ A_n=\{x\in U\ |\ \|T(x)\|>n/2\}.$

The sets An are measurable, because T is assumed to be so. The triangle inequality shows that if x is not in An, then

$\|T(x-x_n)\|=\|T(x_n)-T(x)\|>\|T(x_n)\|-\|T(x)\|>n/2$

so that x-xn is in An. (This shows where sequences of sets satisfying the condition of the Lemma arise naturally).

Finally, the series formed with the xn is absolutely convergent by construction, so the series “twisted” with Bernoulli coefficients are also absolutely convergent. Hence, all the conditions of the Main Lemma are satisfied, and we can conclude that there is a subsequence (yn) of the (xn) such that

$y=\sum_{n\geq 1}{y_n}$

exists, and is in An infinitely often; this means that

$\|T(y)\|>n/2$

for infinitely many n. But this is impossible since T is defined everywhere!

Now here is the proof of the lemma. Consider the series

$Y=\sum_{n\geq 1}{X_ng_n}$

as a random variable, which is defined almost surely by assumption. Note that any value of Y is nothing but a sum of a subseries of the original series with terms gn. Let

$B_n=\{Y\in A_n\}$

so that the previous observation means that the desired conclusion is certainly implied by the condition

$\mathbf{P}(Y\text{ in infinitely many } A_n)>0.$

The event to study is

$I=\bigcap_{N\geq 1}{C_N}\ with\ C_N=\bigcup_{n\geq N}{B_n}$

The sets CN are decreasing, so their probability is the limit of the probability of CN, and each contains (hence has probability at least equal to that of) BN. So if we can show that

$\mathbf{P}(B_n)\geq 1/4\ \ \ \ \ \ \ \ \ \ \ \ \ (*)$

(or any other positive constant) for all n, we will get

$\mathbf{P}(C_N)\geq 1/4,\ \text{and hence}\ \mathbf{P}(I)\geq 1/4>0,$

which gives the desired result. (In other words, we argue from a particularly simple case of the “difficult” direction in the Borel-Cantelli lemma).

$\{\sum_{m}{X_mg_m}\in g_n+A_n\ and\ X_n=1\}=\{\sum_{m\not=n}{X_mg_m}\in A_n\ and\ X_n=1\}$

(for any n), which is a tautology. From the yet-unused assumption

$A_n\cup (g_n+A_n)=G,$

we then conclude that

$\{X_n=1\}\subset \{\sum_{m}{X_mg_m}\in A_n\}\cup \{\sum_{m\not=n}{X_mg_m}\in A_n\ and\ X_n=1\}=B_n\cup S_n,$

say. Therefore

$1/2=\mathbf{P}(X_n=1)\leq \mathbf{P}(B_n)+\mathbf{P}(S_n)$.

But we claim that

$\mathbf{P}(S_n)\leq\mathbf{P}(B_n).$

Indeed, consider the random variables defined by

$Z_m=X_m\ if\ m\not=n,\ \ \ Z_n=1-X_n$

Then we obtain

$S_n=\{\sum_{m}{Z_mg_m}\in A_n\ and\ Z_n=0\}$

but clearly the sequence (Zm) is also a sequence of independent Bernoulli random variables, so that

$\mathbf{P}(\sum_{m}{Z_mg_m}\in A_n\ and\ Z_n=0)=\mathbf{P}(\sum_{m}{X_mg_m}\in A_n\ and\ X_n=0)\leq\mathbf{P}(Y\in A_n)=\mathbf{P}(B_n)$

as desired. We are now done, since we have found that

$1/2\leq 2\mathbf{P}(B_n)$

which is (*).

(In probabilistic terms, I think the trick of using Zm has something to do with “exchangeable pairs”, but I’m not entirely sure; in analytic terms, it translates to an instance of the invariance of Haar measure by translation on the compact group (Z/2Z)N, as can be seen in the original write-up of Matheron).

### Kowalski

I am a professor of mathematics at ETH Zürich since 2008.

## 8 thoughts on “Linear operators which you can write down are continuous”

1. a student says:

In what sense is it not possible to “write down” a non-measurable function?

2. One can explain this in different ways (and add the implicit information that “measurable function” was meant there to refer to measurability with respect to the Borel or Lebesgue sigma-algebra; it is of course possible to select a much smaller sigma-algebra, and write explicit formulas for a function which is then not measurable).

The most intuitive maybe is that if you start with the basic functions of calculus, which are measurable, and then perform all the standard algebraic operations, or take limits in all the usual ways, the resulting function will remain measurable (because of the properties of measurable sets, in particular stability under countable set-theoretic operations).

In particular, any “computable” function is measurable.

3. Another meaning of the phrase “you can’t write down a nonmeasurable function” is that due to a famous theorem of Solovay you need the (uncountable) axiom of choice to construct a nonmeasurable function.

4. I’ve heard of Solovay’s result and it’s one of these theorems that I’d really like to look at more closely one day (for the beauty of it). I must say, I have no idea how difficult it really is (by which I mean, how difficult is it assuming one knows the basic techniques on which it is based, not how difficult “from scratch”).

It is certainly the most conclusive explanation possible for the intuitive statement that a function that can be written down is measurable.

5. Goldstern says:

You have to be careful to interpret “limits in the usual ways” correctly in kowalske’s statement. Countable limits are allowed.

However, one can easily define a very explicit function f(x,y) such that the function
g(x) := sup { f(x,y): y in R }
is not Borel measurable.

One can also define an explicit function F(x,y,z)
such that the function
G(x) = sup { inf{h(x,y,z): y in R }: z in R }
cannot be shown to be Lebesgue measurable (where “shown”=”proved in ZFC”). By Solovay’s result, it cannot be shown to be nonmeasurable, either.

6. Thanks for mentioning these facts, which I didn’t know. Can you give references for them (or similar things)?

My “the usual ways” really meant (in my mind) the fact that a pointwise limit of measurable functions is measurable, and the easy consequences that come out of it (inf and sups of sequences of functions, etc).

7. Goldstern says:

The functions g and G can be obtained from the following theorems:

There are analytic sets (also called Sigma^1_1-sets) which are not Borel.
(A subset of a Polish space is analytic iff it is the continuous image of a closed [or Borel] set in some other Polish space.)

It is not provable in ZFC that all Sigma^1_2 sets (continuous images of complements of analytic sets, also called PCA sets) are Lebesgue measurable.

Both theorems can be found in books on Descriptive Set Theory, such as the one by Moschovakis, or the one by Kechris. I think the first is due to Suslin, the second to Gödel.

A very explicitly given non-Borel set is the set of everywhere differentiable functions in the complete metric space C[0,1]. Its complement is analytic. Using a Borel isomorphism between C[0,1] and R, one obtains an analytic non-Borel subset of the reals.

In Gödel’s constructible universe, the set { (x,y): x is constructed before y } is a Sigma^1_2-set of reals which is not Lebesgue-measurable, by a Fubini argument.

8. Thanks for the references; the “everywhere differentiable function” example is particularly nice — I’ll certainly mention it the next time I teach integration.