Fix a positive constant $K$, and let $T$ be the set of functions from $[0,1]$ to $[0,1]$ that are Lipschitz continuous with constant $K$ or less.
$T$ is a closed convex subset of $\mathcal{C}([0,1])$ - the space of continuous real-valued functions on $[0,1]$ with the uniform norm. Give $T$ the subspace topology.
a) Is $T$ compact ?
b) Is there a non-trivial Borel measure on $T$ that is translation-invariant ?
I think the answer to a) is yes. I have a construction showing that $T$ is totally bounded (I think), so since $T$ is closed and $\mathcal{C}([0,1])$ is complete, $T$ is compact.
I think the answer to b) is no, but cannot show it.
Translation-invariant in this context means, that if $m$ is the measure, and $A$ is a measurable subset of $T$, and $f + A$ is a translate of $A$ that is also a subset of $T$, then $m(f + A)$ = $m(A)$.
After thinking some more about my definition of translation-invariant, I realize that it severely restricts the functions $f$ that one can translate by. In fact, if $A$ is open, and both $A$ and $f + A$ are subsets of $T$, then $f$ must be a constant function. This because adding a non-constant function to an appropriate function with Lipschitz constant $K$ will increase the Lipschitz constant.
With this in mind, here is a measure that meets the conditions.
For c in $[0,1], $define the constant function $f_c$ by $f_c(x) = c$.
Define $m(A)$ to be the Lebesgue measure of $\{ y : f_y \in A\}$.
This is the push-forward measure of the map from $[0,1]$ to $T$ that takes $c$ to $f_c$.
Although this measure meets the conditions, it is concentrated on a 1-dimensional subset of $T$ - the constant functions. There are many open subsets of $T$ that have measure 0.
Since $T$ is compact, I was really hoping there might be one where every open subset has positive measure.