Show that there is is no nonnegative function in the inner product space $C[0,1]$ that satisfies three inner product relations

Question

I am working on this question for homework and am very lost.

Let $V = C[0, 1]$ be the real inner product space of continuous real-valued function on $[0, 1].$ Show that there is no nonnegative function $f \in V$ such that

$$\int_0^1 f(t)dt =1,$$ $$\int_0^1 tf(t)dt =a,$$ and $$\int_0^1 t^2 f(t)dt =a^2,$$

where the inner product is given by $\int_0^1 f(t)\overline{g(t)}dt.$

I have tried finding various contradictions using the Cauchy-Schwarz inequality, but haven't come up with anything. Any help would be great appreciated! Thank you so much!

Answer 1

Expand:

$$I=\int_0^1 (t-a)^2 f(t)dt$$

$$I=\int_0^1 t^2 f(t)dt-2a\int_0^1 t f(t)dt+a^2\int_0^1 f(t)dt$$

$$I=a^2-2a^2+a^2=0$$

Can you conclude from there ?

Remark: I have been guided into the choice of $I$ by a probabilisitic interpretation: $f$ can be the pdf of a certain random variable (first integral) having mean $a$ (second integral); therefore $I$ was a natural choice (interpretation as the variance).