Matematické Fórum

Skrytý text:

First of all, g has to be zero. Indeed, by substitution (for , otherwise ), we get
,
hence
and by taking the derivative, we obtain for all .

Thus, the exercise can be reformulated as a problem of finding a minimum of
.
Note that is a continuous function that satisfies . Moreover, since

for all and all , where are finite sets (the roots), we deduce that is continusously differentiable (the integral does not see the finite sets ) and
.
This can be zero only if the integrand changes sign in the middle of the interval, that is, if
, i.e., . Due to the coercivity and continuous differentiability of , the point is the global minimum. Hence the answer is
.

Remark: the only property of the polynomial that is used is