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It is well-known that the only continuous functions $f\colon\mathbb R\to\mathbb R^+$ satisfying $f(x+y)=f(x)f(y)$ for all $x,y\in\mathbb R$ are the familiar exponential functions. (Prove $f(x)=f(1)^x$ successively for integers $x$, rationals $x$, and then use continuity to get all reals.)
The usual example to show that the identity $f(x+y)=f(x)f(y)$ alone doesn't characterize the exponentials requires the axiom of choice. (Define $f$ arbitrarily on the elements of a Hamel basis for $\mathbb R$ over $\mathbb Q$, then extend to satisfy the identity.)
Is there an explicit construction of a discontinuous function satisfying the identity? On the other hand, does the existence of such a function imply the axiom of choice or some relative?