for $g(x)\in\,P_3(\mathbb{R})$ such that $g(0)=0$, $g'(0)=0$ and $\int_{0}^{1}|x+3x-g(x)|^2dx$, how to find $g(x)$ make the integral as small as possible

I start to find a subspace contain all polynomial satisfies $g(0)=0,g'(0)=0$, $x^2,x^3$ both satisfy the conditions. After this, i am lost

@Simple I guess there might be a typo here. (I would expect you to write $4x$ instead of $x+3x$. So maybe you meant $x+3$ or $x+3x^2$ or something similar.

We are working here with the normed space where the norm of $f(x)=a_0+a_1x+a_2x^2+a_3x^3$ is $\int_0^1 |f(x)|^2 dx$. (This probably can be expressed in terms of $a_0,\dots,a_3$, although I am not sure whether it helps us that much.

The norm is given by the inner product $\langle f(x),g(x)\rangle = \int_0^1 f(x) dx$.

So basically we have a (closed) subspace $S$ and a point $a$ in some inner product space (Hilbert space). We want to find the point $a_0\in S$ which is closest to $a$, i.e., the point which where the minimal distance is attained $\min\{\|a-a_0\|; a_0\in S\}$.

There should be a few methods how to solve such problems in inner product spaces (Hilbert spaces).

Sorry for the typos, in the definition of the inner product I wanted to write $$\langle f(x),g(x) \rangle = \int_0^1 f(x) g(x) dx.$$ This leads to the norm $\|f\|=\sqrt{\int_0^1 |f(x)|^2 dx}$, i.e., $\|f\|^2=\int_0^1 |f(x)|^2 dx$.