Can you see smoothness of the boundary of a convex body from its shadow?

Let $d \geq 3$ and suppose that $K \subset \mathbb{R}^{d}$ is a convex body (compact, convex, non-empty interior). Is the following true?

The boundary $\partial K$ is a $C^1$-manifold if and only if for each projection $\pi:K\rightarrow H$ to a hyperplane $H$ has the property that $\partial \pi(K)$ is a $C^1$-manifold.

I suspect the answer is true. I am most interested in the case $d=3$.

If it is true, does it hold true for $C^1$ replaced with $C^k$, $k \in \mathbb{N} \cup \{\infty\}$?

I have no familiarity with this field so I have no idea about the difficulty of the question. I would appreciate any references.

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The statement for $C^1$ regularity is true, but with "dimension-2 projections" instead of "codimension-1 projections''. This is even stronger, if $d\ge 3$. On the other hand, for $d=2$ the statement with "hyperplane projections" fails, since $1$ dimensional projections are just closed intervals, whose boundary is certainly smooth. Also, an analogous statement holds with "plane sections" instead of "plane projections", for analogous reasons. Here's a sketch of the proof.

A convex function is differentiable if and only if its subdifferential is a single point. By a compactness argument, this implies that a convex function everywhere differentiable on a convex domain of $\mathbb{R}^n$ is automatically $C^1$.

The boundary of a convex body is locally the graph of a convex function. The above property then implies: if the boundary of $K$ is not $C^1$, on some point $x$ of $\partial K$ it admits two supporting hyperplanes. If we project on the plane section orthogonal to the intersection of these two hyperplanes, we find a 2-dimensional projection of $K$ with boundary that is not smooth at $\pi x$ (because it has two supporting lines at that point). And conversely.

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Very nice argument! Presumably this proves the other implication in $d=3$ as well by pulling back the supporting lines from the projection. –  Bati Nov 26 '12 at 17:46
Exact, that's what I meant by "and conversely", in any dimension $d$. –  Pietro Majer Nov 26 '12 at 17:58
Sorry, missed that. –  Bati Nov 27 '12 at 12:58

According to this article, in the case $d=3$, the boundary in the projection need not be $C^2$ even if the original boundary is $C^\infty$. This seems to leave open the $C^1$ case, though.

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Thanks for the reference. I did think that the boundary of the projection would be at least as smooth as the original boundary! –  Bati Nov 26 '12 at 16:16
Actually the $C^1$ case is treated (Thm 2.1). But in any case, only one implication is considered, leaving open the question: if all plane projections of $K$ have a boundary of class $C^r$, how smooth is $\partial K$? (Maybe this part is obvious?). –  Pietro Majer Nov 26 '12 at 17:07