# Rotating the graph of a function

I think I have answered the following puzzle:

## Rotations

Under what conditions can you rotate the graph of a function about the origin, and still have the resulting graph being the graph of a function? If the graph of a function cannot be rotated about the origin without ceasing to be the graph of a function, might there be other points which could act as centre of rotation and preserve the property of being the graph of a function?

I think the answer is that the function has to be a line. Otherwise we can find an angle which lets us rotate the graph so that one $x$ maps to two $y$s.

Question: Is this correct? And if yes, how can I prove it?

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## 2 Answers

For any set of points on the plane $\mathbb R^2$, there's at least one angle by which it can be rotated so that the result is not a graph of a function from $\mathbb R$ to $\mathbb R$.

Specifically, let $A$ and $B$ be any two distinct points in the set, and rotate the set so that the line $AB$ becomes vertical. Now the images of $A$ and $B$ share the same $x$ coordinate but have different $y$ coordinates, which contradicts the requirement that a function should map each $x$ value to a single $y$ value.

(If the set of points does not contain two distinct points, then it is either a singleton or empty; neither of these can be the graph of a function from $\mathbb R$ to $\mathbb R$ in any orientation.)

However, the question you quote doesn't explicitly say that the graph would have to remain a graph of a function under any rotation; instead, I'd read it as asking about whether, given a function $f: \mathbb R \to \mathbb R$ and an angle $\alpha$, the graph of $f$ rotated by $\alpha$ is a graph of some function $g: \mathbb R \to \mathbb R$.

This seems a more complicated question, although it's easy to find sufficient conditions: for example, $\alpha \equiv 0 \pmod{2\pi}$ obviously works for any $f$, and, as you note, for functions of the form $f(x) = ax + b$ any value of $\alpha$ except $\alpha \equiv \frac\pi 2 - \arctan a \pmod\pi$ will produce the graph of another function of the same form.

More generally, if $f$ is Lipschitz continuous with Lipschitz constant $K$, then its graph can be rotated by any angle of the form $\alpha + n\pi$, where $|\alpha| < \frac \pi 2 - \arctan K$ and $n$ is an integer, so that the result is still the graph of a function. In particular, the function $f(x) = \sin x$ mentioned by Henning is Lipschitz continuous with $K = 1$.

In fact, we can generalize this slightly:

Definition: We'll say a function $f: \mathbb R \to \mathbb R$ is $(J,K)$ skew Lipschitz continuous if there exist constants $J,K \in \mathbb R$ such that, for all distinct $x_1,x_2 \in \mathbb R$, $$J \le \frac{f(x_1)-f(x_2)}{x_1-x_2} \le K.$$ (In particular, any $K$-Lipschitz continuous function $f: \mathbb R \to \mathbb R$ is $(-K,K)$ skew Lipschitz continuous.)

Claim: Rotating the graph of a $(J,K)$ skew Lipschitz continuous function by an angle $\alpha + n\pi$, where $$-\frac\pi2-\eta < \alpha < \frac\pi2-\theta$$ $\eta=\arctan J$, $\theta=\arctan K$ and $n$ is an integer, yields the graph of a $(J',K')$ skew Lipschitz function, where $J' = \tan(\alpha+n\pi+\eta)$ and $K'=\tan(\alpha+n\pi+\theta)$.

(Proof left as an exercise.)

I believe this is getting close to a sufficient and necessary condition, at least for continuous and differentiable functions, although there remain some edge cases to be taken care of: for instance, rotating the graph of a $(J,K)$ skew Lipschitz function $f$ by exactly $\alpha = \frac\pi2-\arctan K$ may or may not yield the graph of a function, even if $K$ is the smallest upper skew Lipschitz constant for $f$; it depends on whether the equality $\frac{f(x_1)-f(x_2)}{x_1-x_2}=K$ is ever actually attained, or only approached from below.

The second part of the question is easy to answer, though: the center of rotation does not matter. To see why, note that rotation around any point $A$ can be decomposed into translation of $A$ to the origin, rotation around the origin and translation of the origin back to $A$. And since the graph of a function remains the graph of a function under translation, only the rotation matters.

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Another interpretation of the question is whether there is some nontrivial angle the graph can be rotated and still stay a graph. This is somewhat more interesting than whether all angles work -- for example the graph of $\sin x$ can be rotated by up to $\pi/4$ and still stay a graph of something.

For the second part of the question note that rotation about a non-origin can always be decomposed into a rotation about the origin and a translation in the plane. Translations clearly don't change whether a set of points is the graph of some function.

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