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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 $m(f + A)$ = $m(A)$.

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  • $\begingroup$ What is the definition of translation-invariant in this context? $\endgroup$ – Brent Kerby Feb 28 '15 at 7:09
  • $\begingroup$ Translation-invariant 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). $\endgroup$ – Glenn Davis Feb 28 '15 at 7:46
  • $\begingroup$ So, in other words, translation-invariance is only required on the codomain, not the domain. Couldn't we then define such a measure $m(A)$ to be the Lebesgue measure of $\{f(x_0) : f\in A\}$ for an arbitrary point $x_0\in[0,1]$? $\endgroup$ – Brent Kerby Feb 28 '15 at 7:59
  • $\begingroup$ Unfortunately that $m$ is not a measure. Suppose $K > 1$ and $x_0 = 1/2$. Let $f(x) = x$ and $g(x) = 1 - x$, and consider 2 balls in $T$ centered at $f$ and $g$ with the same small radius $r$. The balls are disjoint but $m()$ of both balls is $2r$. $\endgroup$ – Glenn Davis Feb 28 '15 at 8:27
  • $\begingroup$ Ah, you're right; that won't work. But how about $m(A) := \int_{-\infty}^{-\infty} \gamma_t(A)$, where $\gamma_t$ is the Wiener measure translated by $t$ in the codomain? $\endgroup$ – Brent Kerby Feb 28 '15 at 14:15
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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.

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