Is there an upper bound on the dimension for irreducible representations of a continuous trace $C^{*} $-algebra? The following are questions of Don Hadwin:
If $A$ is a unital continuous trace C*-algebra, is there an upper bound on the dimension of all the irreducible representations? 
It is known that all irreducible representations are finite-dimensional?
 A: There is an upper bound. Here is why (this is a simpler way of arguing than the one I used below):
In 4.5.2 of "C*-algebras", Dixmier defines a continuous trace C*-algebra as one such that
the set \[\{ a\in A^+ \mid \pi\mapsto\mathrm{Tr}(\pi(a))\mbox{ is continuous}\}\] spans a dense two-sided ideal of $A$. Here $\pi$ ranges through the spectrum of $A$. It is remarked in 4.5.2, that for every $x$ in this dense ideal the map $\pi\mapsto \pi(x)$ is continuous. If $A$ is unital, then this dense ideal must in fact be $A$ (since by getting very close to 1 it will contain invertible elements). It follows that $\pi\mapsto \mathrm{Tr}(\pi(1))$ is continuous, and thus bounded (since the spectrum of a unital C*-algebra is compact).
But $\mathrm{Tr}(\pi(1))$ is the dimension of $\pi$.

Longer argument:
Consider first the case where $A$ has a positive element $a$ such that each irreducible representation maps $a$  to a rank 1 operator. Then $a$ must be a full element (i.e., generate $A$ as a closed two-sided ideal). Thus, there exist finitely many $x_i$ and $y_i$ such that $1=\sum_{i=1}^n x_i a y_i$. This puts the bound $n$ on the rank of $\pi(1)$ for any irreducible $\pi$.
The general case is reduced to the previous case using the Fell condition. Since the spectrum of $A$ is compact it suffices to show that the dimension of its irreducible representations is locally bounded. The Fell condition for a continuous trace C*-algebra says that for each irreducible $\pi_0$ there exists a neighborhood $U(\pi_0)$ and a positive element $a\in A$ such that $\pi(a)$ has rank 1 for every $\pi\in U(\pi_0)$. We may choose $U(\pi_0)$ a closed neighborhood, since the spectrum of $A$ is Hausdorff. This entails the existence of a closed two-sided ideal $I$ such that $U(\pi_0)$ consists of the irreducible representations that factor through $A\to A/I$. Now the special case proved earlier can be applied to $A/I$ (and the image of $a$ in $A/I$).
