Timeline for Does there exist another form of the derivative for polynomials?

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Jun 27, 2019 at 15:52	comment	added	Jan-Christoph Schlage-Puchta		@user44191: If $\alpha>0$ and $\delta<0$, then $\lambda\mapsto \lambda\alpha+\frac{\delta}{\lambda}$ is a continuous function tending to $\infty$ for $\lambda\rightarrow\infty$ and to $-\infty$ for $\lambda\rightarrow 0$. So for any given $\beta, \gamma$, there is some $\lambda$, such that $\lambda\alpha+\beta+\gamma+\frac{\delta}{\lambda}=1$. So for any such tuple there is some function $F$ (namely $\lambda\mathrm{id}$ for a suitable $\lambda$), for which this particular $H$ satisfies the functional equation.
Jun 16, 2019 at 17:50	comment	added	user44191		I think I'm missing something; how do you get that all tuples following the inequality are possible? It seems to me all you proved is that tuples following the equality are possible (for $F = \lambda Id$).
Jun 16, 2019 at 15:54	history	answered	Jan-Christoph Schlage-Puchta	CC BY-SA 4.0