Let $g$ be a continuous function on $[0,1]$ with certain conditions. Show that $g(x) = x$

Question

Let $g$ be a continuous function on $[0,1]$ s.t. $$\int_{0}^{1} g^2(x) dx = \frac{1}{3}$$ $$\int_{0}^{1} xg(x)dx = \frac{1}{3}$$

Show that $g(x) = x \forall x \in [0,1]$.

My work so far. I wrote it as $$\int_{0}^{1} (g^2(x) - xg(x))dx = 0$$ and $$\int_{0}^{1} (xg(x) - g^2(x)) dx = 0$$

So, if one of these is negative, then the other must be positive. I know that if integral of a non-negative function is $0$, then the function is $0$. However, I am not quite sure how to show that the function is non-negative. In fact I'm not even convinced it has to be. I believe this could work if the function is $g(x) = -x$. I must be forgetting an important property here since I'm not seeing it. Can anyone point me in the right direction?

Answer 1

We know that $$\int_0^1g^2(x)dx=\dfrac{1}{3};\quad \int_0^1g(x)\cdot x dx=\dfrac{1}{3}\Longrightarrow\int_0^12g(x)\cdot x dx=\dfrac{2}{3};\quad \int_0^1x^2dx=\dfrac{1}{3}.$$ Then $$\int_0^1\underbrace{g^2(x)-2xg(x)+x^2}_{(g(x)-x)^2}dx=0\Longrightarrow\int_0^1(g(x)-x)^2dx=0.$$ Since $(g(x)-x)^2$ is a positive function whose integral is equal to $0$, then we conclude that $$g(x)-x=0\quad a.e. x\in[0,1]\Longrightarrow g(x)=x\quad a.e. x\in[0,1].$$ Finally, thanks to $g$ being continous, we get that $$g(x)=x\quad \forall x\in[0,1].$$

Answer 2

Notice that $(g(x)-x)^2\geq 0$ for all $x$, so:\begin{align}0&\leq \int_{0}^{1} (g(x)-x)^2 dx \\ &= \int_{0}^{1} g^2(x) dx-2\int_{0}^{1} xg(x) dx +\int_{0}^{1} x^2 dx \\& = \frac{1}{3} -2\frac{1}{3} +\frac{1}{3}\\ &=0\end{align}

so $$\int_{0}^{1} (g(x)-x)^2 dx =0$$

so $g(x)=x$, for all $x$, since $x$ is continuous.