Shrinking and stretching of vector bundles Let $M$ be a manifold, $p:E\to M$ a rank $d$ vector bundle. Suppose that $U \subset E$ is an open subset such that $U \cap p^{-1}(x)$ is nonempty and convex for all $x \in M$. Is it true that $U \to M$ is a fiber bundle with fiber $\mathbb R^d$? And that $U \cong E$ as fiber bundles? We may assume with no loss of generality that $U$ contains the zero section.
This seems like a statement that could be a lemma in any number of textbooks (if true), e.g. in connection with the tubular neighborhood theorem, but I haven't seen it anywhere. Lang proves in his differential geometry book that any vector bundle over a manifold is what he calls compressible, meaning that any open neighborhood of the zero section of $E$ can be shrunk to a smaller open neighborhood which is diffeomorphic to $E$ as a bundle over $M$.
 A: Since there are no references so far, let me give a sketch proof along the lines of my comment. I'll assume that $M$ is compact.


*Let's show first that there is a smooth section of $E$ lying in $U$. Indeed, for any point $x\in M$ there is a neighbourhood $U_x$ with a section $s_x$. Take a  finite cover $U_i$  of $M$ by such neighbourhoods and take the corresponding partition $1=\sum f_i$ of unity. Then by convexity $\sum s_i f_i$ is a smooth section lying in $U$.

Clearly we can assume that $s$ is the zero section (by taking an appropriate fiberwise diffeo), we will assume this from now on.
Now we will construct an exhaustion of $U$ by an increasing sequence of  fiber-wise compact convex subsets $0\subset {\cal B_1}\subset ... \subset {\cal B_i}\subset ...$ so that $U=\cup_i {\cal B_i}$.

*

*Let me show first how to construct one such subset ${\cal B_1}\subset U$.

For every point $p\in M$ let us choose some covex compact subset $B_p$  with smooth boundary in the fiber $U_p$.  Then, since $U$ is open, there is an open neighbourhood $V_p$ of $p$ in $M$ such that over this neighbourhood there is a smoothly varying family of $B_x$ ($x\in V_p$), such that $B_x\subset U_x$. Take a finite cover of $M$ by such $V_i's$, let $\phi_i$ be the partition of unity. Then the sum
$${\cal B_1}=\sum_i \phi_i B_i (x)$$
is the desired subset $B\subset U$. Here by sum I mean the Minkowski sum.


*It is clear that the interior of $\cal B_1$ is diffeomorphic to the bundle of vectors of length less than $1$ in $E$ (for some fiber-wise Euclidean metric). So the only  need to construct a family of $\cal B_i$ that will exhaust $U$. This can be done as in 1).

