Show that there exists a polynomial $P > 0$ such that for all $f$ in a linear subspace of $C([0,1],\mathbb R)$ : $\int_{0}^1 f(x)P(x)dx = 0$ Let $V$ be a finite dimensional linear subspace of $C([0,1],\mathbb R)$.
It is assumed that for any non-zero $f$ in $V$, there exists $u \in [0,1]$ such that $f(u) > 0$.
Show that there exists a polynomial $P > 0$ on $[0,1]$ (P is not necessarily $> 0$ on all $\mathbb R$) such that
$$\int_{0}^1 f(x)P(x)dx = 0 \\ \forall f \in V$$

If someone has a hint, even just for the one dimension, it will be appreciate.
 A: The case of dimension (of $V$) one is not so difficult.
Let us consider $V = \{\lambda f;\ \lambda\in\mathbb{R}\}$. By assumption $f$ must assume both positive and negative values, hence there exists $x_1, y_1\in (0,1)$ such that $f(x_1) > 0$, $f(y_1) < 0$.
Let us consider the polynomials
$$
q_k(t) := c_k (1- t^2)^k,
\qquad k\in\mathbb{N},
$$
where the constant $c_k$ is chosen such that $\int_0^1 q_k(t) \, dt = 1$.
It can be proved that, for every $\delta > 0$, the sequence $(q_k)$ converges uniformly to $0$ in $\{\delta \leq |t| \leq 1\}$, and that
$$
\int_0^1 f(x) q_k(x-x_1) \, dx \to f(x_1),
\quad
\int_0^1 f(x) q_k(x-y_1) \, dx \to f(y_1),
\qquad \text{for}\ k\to +\infty.
$$
In particular, we can choose $k$ large enough so that
$$
I_1 := \int_0^1 f(x) q_k(x-x_1) \, dx > 0,
\quad
J_1 := \int_0^1 f(x) q_k(x-y_1) \, dx < 0.
$$
Now, it is easy to see that the polynomial
$$
P(x) := -J_1 q_k(x-x_1) + I_1 q_k(x-y_1)
$$
is positive in $[0,1]$ and
$$
\int_0^1 f(x)\, P(x)\, dx = -J_1 I_1 + I_1 J_1 = 0.
$$
As far as I can see, the general $n$-dimensional case is trickier.
If $V$ is generated by the functions $f_1, \ldots, f_n$, we can find distinct points $x_1, y_1, x_2, y_2, \ldots, x_n, y_n \in (0,1)$ and $k$ large enough such that
$$
I_j := \int_0^1 f_j(x) q_k(x-x_j) \, dx > 0,
\quad
J_j := \int_0^1 f_j(x) q_k(x-y_j) \, dx < 0,
\qquad j=1,\ldots,n.
$$
Moreover, since the points are distinct, given $\epsilon > 0$ we can also assume that
$$
\left|\int_0^1 f_j(x) q_k(x-x_i) \, dx\right| < \epsilon,
\quad
\left|\int_0^1 f_j(x) q_k(x-y_i) \, dx\right| < \epsilon,
\quad\forall i\neq j.
$$
These information should be enough to construct the required polynomial.
