Tricki

## Using the fact that a continuous function on a compact set attains its bounds

### Quick description

There are many circumstances in which it is very useful to be able to say that a continuous real-valued function attains its bounds. This article gives some examples, some simple and some less so.

### Example 1

Suppose one is dealing with a complicated function, for example the function on the interval . Due to the rather nasty nature of , the prospect of attempting to find the minimum of over this interval using calculus is not an inviting one. However is certainly continuous and is compact, and so attains its minimum. Manifestly is non-negative, but it is also never zero: indeed for it to vanish we would require that , which means that , but then the second term is , and this is not zero. This means that there must be some absolute constant such that for all .

### Example 2

The above example concerns a concrete function, but the principle that a function that takes positive values is, under appropriate circumstances, bounded below by a positive constant is very useful in more abstract arguments too. For example, let be an open subset of the complex plane and let be a closed curve in In complex analysis, it can be very helpful to know not just that is a subset of , but also that there is some such that for every point in , the ball of radius about is contained in . (For example, one sometimes likes to cover with a grid of squares in such a way that every square that intersects is fully contained in .)

This can be proved easily: define a function to be the distance from to the complement of . Since is open, for every . It is easy to check that is continuous. Indeed, one can show that . Moreover, , being the continuous image of a closed bounded interval, is compact. Therefore, the restriction of to attains its lower bound, and since is a subset of this lower bound cannot be zero.

### Example 3

The first example was rather artificial but the next (from the work of E. Breuillard A Height Gap Theorem For Finite Subsets Of and Non Amenable Subgroups) is not. Suppose one has a finite set of matrices over the complex numbers, where we think of as fixed. We may associate to two quantities. Firstly, we define to be the maximum norm of any matrix in (we define the norm of a matrix in a fairly standard way, by regarding matrices as operators from to ). Secondly, we define to be the modulus of the largest eigenvalue among the matrices in .

Now if is bounded (say at most 100) then so is ; this is a very easy exercise. The converse, however, is not true (consider for example a single matrix with 1's on the diagonal and some enormous number in the top-right corner). Remarkably, the converse is nearly true, and this is Breuillard's result.

Theorem 1 (Breuillard) Suppose that is a finite set of matrices such that all of are bounded by . Then there is an invertible matrix such that , where the constant depends only on .

Here means the set consisting of products of matrices in . The theorem basically says, very roughly, that up to conjugation the notions of matrix norm and maximal eigenvalue are very similar.

The first step in the proof is the one relevant here, in that it involves rephrasing the problem as one about the minimum of a continuous function on a compact set. As a first observation, note that if one multiplies all the matrices in by some scalar then both and get multiplied by . Thus our task is equivalent to showing that if then for some . Furthermore, conjugation does not change eigenvalues. Choosing a specific such that , we may conjugate by and rescale again, and thereby assume that and that .

Now the set of matrices with is certainly compact as a subset of . Here, however, we are dealing not with single matrices but with sets of matrices. However these sets of matrices are also members of a compact metric space.

There is, in fact, a quite general construction. Let be a compact metric space and suppose that is the collection of compact subsets of . This may be endowed with the structure of a metric space by defining the Hausdorff distance between sets and to be . It is a lengthy but not particularly difficult exercise to show that this is indeed a distance, and that the metric space is compact.

Furthermore, each of the functions , , is continuous on . By the principle that continuous functions on compact spaces attain their bounds, all we must do is rule out the existence of a compact set of matrices with and .

The fact that we have here rather than is hugely helpful, and it puts us squarely in the domain of algebra rather than analysis. The algebra is a little deep, and I refer the reader to Breuillard's article for references. Here is a sketch: suppose that and let be the -algebra generated by . By simple linear algebra this is spanned as a vector space over by , all of which consist entirely of nilpotent elements by assumption. By a theorem of Wedderburn, for some depending on . By a theorem of Engel, the matrices in may simultaneously be put in strictly upper triangular form, and hence a fortiori the same is true of the matrices in . But one may check directly, by calculation, that any compact set of strictly upper-triangular matrices may be conjugated so as to have arbitraryily small norm, contrary to the assumption that .

### General discussion

Breuillard notes that there is a drawback with this argument: it does not yield an explicit value for the constant . He says that he can obtain one, but only by finding "almost" versions of the results of Wedderburn and Engel. It is a common feature of compactness arguments that they do not give explicit bounds: obtaining such bounds is often an interesting problem.