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Statement

Write (C) for Cauchy's equation $f(x+y)=f(x)+f(y)$ and (C') for the condition

$$f\Bigl(\frac1x\Bigr)=\frac1{x^2}\,f(x)\qquad(\text{all }x\ne0).$$

Theorem II (p. 685). Let $f$ be real-valued and defined for every real $x$, and suppose that (C) holds for all pairs $(x,y)$ together with (C'). Then $f(x)=xf(1)$ for every real $x$.

No regularity of $f$ is assumed. The paper credits the question to I. Halperin (p. 683), and its added-in-proof note (p. 686) records independent solutions by S. Kurepa and by S. L. Segal.

Source. W. B. Jurkat, On Cauchy's functional equation, Proc. Amer. Math. Soc. 16 (1965), 683--686, Theorem II on p. 685, proof on pp. 685--686; the edition and read status are recorded on the source card.

Read depth. Claims checked: the statement was read clause by clause against the print. The proof was read through but not verified by a second reader.

Proof pointer

Proof on pp. 685--686. Applying (C') to the partial-fraction identity for $1/(x(x-1))$ gives $f(x^2)=2xf(x)-x^2f(1)$ for every real $x$. Polarizing with $4xy=(x+y)^2-(x-y)^2$ gives $f(xy)=xf(y)+yf(x)-xyf(1)$ (the paper's equation (3), p. 686), and putting $y=1/x$ and using (C') once more gives $f(x)=xf(1)$.

Dependencies

None beyond (C) and (C').

Bears on

No Erdős problem in the corpus. The paper's other result, Theorem I, is the one that bears on Problem 1126.