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# How can we show the complex function f is an affine mapping?

I'm a bit lost at this, couldn't give a formal proof:

Let f be an entire function. Show that f is Lipschitz if, and only if, f is an affine mapping. The if part is immediate, but I'm stuck at the only if part.

Is this true if, instead of entire, f is holomorphic in a proper subset of the complex plane?

Thank you.

### 3 Answers

- SteinerLv 710 years agoFavorite Answer
I think there's a mistake in David's answer. Actually, being Lipschitz doesn't imply that |f(z)| ≤ k|z|. If f is defined at 0, then this last condtion implies that f(0) = 0. But there are many Lipschitz functions defined at 0 with f(0) ≠ 0. A simple exemple is just the affine mappings f(z) = az + b with b ≠ 0. Anyway, he and Gianlino (as always) brought about an interesting approach.

I'd like to complement Gianlino's answer with a somewhat different approach. The result you cited for the entire case can be seen as a consequence of the following fact, very similar to a well known one that holds for real functions defined on intervals.

Suppose f is holomorhic in an convex open set V. Then, f is Lipschitz if, and only if, f' is bounded in V. I'll sketch a proof, leaving the details to you.

If f is Lipschitz, then, for every distinct z1 and z2 in V, there is a constant k > 0 such that |f(z2) - f(z1)|/|z2 - z1| ≤ k. Keeping z1 fixed and letting z2 → z1, we get |f'(z1)| ≤ k. Since z1 is arbitrary, f' is bounded by k in V.

If, on the other hand, f' is bounded in V, then there is k > 0 such that |f'(z)| ≤ k for every z in V. If z1 and z2 are in V, then, since V is convex, the line segment [z1, z2] joining z1 and z2 lies in V. Since f is holomorphic, it follows from the properties of the integral that

|f(z2) - f(z1)| = |∫ [z1, z2] f'(z) dz| ≤ int; [z1, z2] |f'(z)| |dz| ≤ ∫ [z1, z2] k |dz| = k|z2 - z1|. Hence, f is Lipschitz with constant k.

An interesting curiosity: the number s = supremum {|f'(z)| : z ∈ V} is the smallest "Lipschitz constant" of f in V.

Now, if f is entire and Lipschitz, then V = C and, since C is open and convex, the result we proved shows that f' is bounded in C. Since derivatives of entire functions are themselves entire, it follows from Liouville Theorem that f' is constant. This, in turn, implies that f is an affine mapping: for every z, f(z) = az + b, where and be are complex constants.

As for your second question, the answer is, in general, no. Take the restriction of any entire function f that's not an affine mapping to the closure cV of a convex, open and bounded subset V of the plane. Then, f' is continuous in cV and, since cV is compact, f' is bounded in cV and, therefore, in V. From the result we proved, it follows f is Lipschitz in V, though it's not an affine mapping (by continuity, f is Lipschitz in cV).

For example, take f(z) = sin(z), clearly not an affine mapping, and V = the open unit disk. f is Lipschitz in V.

I'm not sure about unbounded subsets of the plane.

- DavidLv 710 years ago
if f is Lipschitz, then |f(z)| ≤ K|z| for all z.

this, in turn, implies that f is a polynomial of degree at most 1, that is, f is an affine function

(for any entire function f, |f(z)| ≤ M|z^n| implies that f is a polynomial of degree at most n (extended Liouville Theorem)).

unfortunately, my brain fails me on the second part. i do have some questions: is f assumed to exist on all of C, and also Lipschitz on all of C, or only on the subset U on which it is holomorphic?

- gianlinoLv 710 years ago
If f is bounded, g defined by g(z) = (f(z) - f(0)) / z is bounded holomorphic, hence constant.

so g'z) = C and f(z) = Cz + f(0).

For the subset q, this is not true in general. Take f(z) = e^{-z} on the right half plane.