Since $a^2 > 0$, we can conclude that:
$B_n((f a)^2)(x) > B_n(f^2)(x)$
Now, let's square $B_n(f)(x)$:
$B_n(f)^2(x) = \left(\sum_{k=0}^n {n \choose k} f\left(\frac{k}{n}\right) x^k (1-x)^{n-k}\right)^2$
By expanding the square, we can see that:
$B_n(f)^2(x) =
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