What norm/metric are you using on the function space? Usually a bounded subset is one that is contained in a unit ball, and that requires a metric to make sense.

If X was compact I'd assume you're using the sup-norm. In that case a bounded subset B is one such that there is a constant N with |f(x)| < N for all x in X and f in B. But in that case the answer is trivial since we can take the constant function g(x) = N.

Since X might be noncompact we cannot use the sup-norm as it is undefined for unbounded functions.

Edit: If we're talking about the space of all compactly-supported functions then the constant function still works, but is not itself an element of the function space.

If you're asking is the set always dominated by an element of the function space itself, then the answer is no. Just set X equal the real line and define B ={f_1, f_2, . . . } where each f_n is equal to one over [ -n, n], equal to zero outside [-n - 1, n + 1], and linear over the intervals [-n - 1, -n] and [n, n + 1]. Then B is bounded above by one, but not dominated by any compactly-supported function.

Before thinking to an answer, I have to mention two related questions which should be clarified. 1) By C_c(X) one usually denotes the vector space of all continuous compactly supported functions on X (all continuous functions with compact support). This is not written in the statement of the question. 2) What do we mean by "compact open topology"? Note that if the notation C_c(X) has the sense recalled in 1) and if we endow this space with the sup-norm, then the answer given by Daron Anderson is correct.

Thanks dear Doran and dear Octav.

As it is stated in the question, C_c(X) is the whole space C(X) of all continuous functions on X. The subscript c refers to the topology \tau_c of uniform convergence on the compact subsets of X. Therefore C_c(X) is a locally convex space, but not a normed one. Its topology is given by the semi-norms

P_K(f) := \sum_{x \in K} |f(x)|, f \in C(X),

K runs over the compact subsets of X.

By the way, if X is the real line, the answer is in the affirmative, i.e., for every bounded set B in C_c(X), there exists a positive continuous function g \in C(X), such that |f(x)| \le g(x), for every x \in X and every f \in B.

Of course the answer is in the affirmative if X is compact as noticed Doran. The question is then, whether this still holds for arbitrary locally compact space.

Hi Lahbib,

Can you clarify what you mean by a bounded subset of C_c(X) please?

If X is second countable (probably some weaker conditions work too) then countably many semi-norms suffice to generate the topology. You can then combine those semi-norms into a metric that generates the topology. Then we could talk about bounded subsets with relation to that metric. That is another option for what boundedness might mean.

Except I have a suspicion which subsets are bounded would depend on the particular metric (where are many choices) and not the topology itself. . . .

Hi Daron,

A subset B of C(X) is said to be bounded if, for every compact K \subset X, there is a constant M>0 such that : |f(x)| \le M for every x \in K and every f \in B. This notion depends on the topology, since another definition of a bounded subset of a topological vector space is as follows: If A is a subset of some such space, we say that A is bounded if and only if it is absorbed by any neighborhood of zero. In case of $C_c(X) the two definitions coincide.

Okay, with that definition of boundedness the answer is yes for spaces that have ''exhaustion by compact neighborhoods'. Here is the proof for the real line:

In this case R = U [-n , n] is the union of compact intervals. Without loss of generality we can assume f(x) < n for all f in B and x in [-n,n]. In that case we can construct g recursively using the Tietze extension theorem. FIrst define g = 2 over [-1,1]. Then define f=4 over [-3,-2] U [2,3]. Then let G be the restriction of g to {-2,-1,1,2}. By definition G maps into [2,4]. So we can extend it to a map from [-2,-1]U[1,2] to [2,4]. Then combining g and the extension of G we get a map [-3,3] that dominates all f in B over its range. Continue in this manner to define g over all of X.

The standard example of spaces without this exhaustion property is the long line. However for this space the answer is still yes. Every function on the long line is eventually constant and hence extends to the one-point compactification. Then we're back compact X territory where we know the answer.

Have you though about X some infinite-dimensional Banach space?

Daron Anderson Dear Daron Anderson, an infinite-dimensional Banach space $X$ is not locally compact, consequently such an $X$ does not satisfy the conditions of the Lahbib Oubbi's question.

Lahbib Oubbi I don't know the answer, but it seems to me that at least such a collection B of functions has an upper semi-continuous majorant. Such a majorant F: X \to R can be defined as follows:

for every t in X denote F(t) the infimum of those C > 0 for which there exist a neighborhood U of t such that every f \in B satisfies f(x) < C for all x \in U.

So, on those locally compact spaces X, where every upper semi-continuous function F: X \to R is dominated by a continuous one the answer to your question is positive.

Enciclopedia of Math. https://www.encyclopediaofmath.org/index.php/Semicontinuous_function

mentions without a reference that " If X is a completely regular space and f an upper (resp. lower) semicontinuous function which does not take the value ∞ (resp. −∞), then f is the pointwise limit of a monotone nonincreasing (resp. nondecreasing) sequence of continuous functions", which implies that completely regular spaces have the property you need.

Dear Vladimir Kadets,

Thank you for the effort. Actually every locally compact space is already completely regular, and the property does not hold for completely regular spaces. The hope is that it holds for locally compact ones.

Lahbib Oubbi Could you please explain what property are you speaking about, when you write that it does not hold for completely regular spaces? Do you mean your original question with uniformly bounded on compacts collection of functions (let us call it Property 1), or the property that every upper semicontinuous real-valued function that takes only finite values has a continuous majorant (let us call it property 2)?

If the Property 2 holds true in every locally compact space, then the Property 1 you are interested in also holds true in every locally compact space, because in locally compact spaces every uniformly bounded on compacts collection B of functions possesses an upper semicontinuous majorant, as it is explained in my previous message.

If I understand correctly, Enciclopedia of Math. says that the Property 2 is valid in every completely regular space, so in particular in locally compact spaces. Unfortunately, they don't give a reference, so I cannot check it.

After some search in internet I see that the statement in Enciclopedia of Math. about existence of non-increasing sequence of continuous functions that converge pointvise to a given upper semicontinuous function having finite values contains a mistake. The result holds only in perfectly normal spaces, see

Hing Tong, Some characterizations of normal and perfectly normal spaces Duke Math. J., Vol. 19 (1952), pp. 289-292.

The weaker result about existence of a continuous majorant is closely related to theorem 2 of Katetov's paper http://matwbn.icm.edu.pl/ksiazki/fm/fm38/fm3819.pdf because if the topological space satisfies the conclusion of that theorem one can deduce what we need by some easy manipulations.