find $f$ such that $f(\frac{1}{2})=\int_{0}^{1}f(t)g(t)dt$

Question

Let $V=P_2(X)$ with inner product $<f,g>=\int_{0}^{1}f(t)g(t) \ dt$.
Find $f \in V$ such that $f(\frac{1}{2})=\int_{0}^{1}f(t)g(t) \ dt$ for every $g \in V$.

I tried considering $g(t)=at^2+bt+c$ to find something about $f $ but that didn't work.

I tried to somehow use the orthogonal basis but I don't know how.

Any hint? I am stuck.

Answer 1

Clearly $f=0$ is a solution.

Assume $f$ is a solution. Picking $g=0$, we get $f(1/2)= \langle f, 0 \rangle =0$. However, picking $ g=f$ yields $$ 0=f(1/2)= \langle f, f\rangle.$$ Thus, $f=0$.

This implies that the zero function is the unique solution in our space.

Answer 2

$$g(t) \equiv f(t) \equiv 1.$$