Why such minimum is attained? I don´t know exactly how to solve this question. I need to prove that there exists a polynomial $f(x) \in \mathbb{C}[x]$ of degree less or equal $m$ such that $f(0)= 1$ and  $||f ||_1 = \int_0^1| f(x)|dx$ is the minimum in the space of polynomials of degree less or equal $m$ such that $f(0) =1$. Of course , this space is a Banach space ( because it´s a finite dimensional subspace of $L_1([0, 1])$) and hence locally compact (because the dimension is finite), so I was trying to use the norm $||f|| = \sum_{k= 1}^m|a_k|$ to get some kind of equivalence between these two norms and then use the compacity.
Thanks in advance.
 A: Let $V_m = \{ f \in \mathbb{C}[x] : \deg f \leqslant m\}$ be the space of polynomials of degree at most $m$. That is a finite-dimensional complex vector space (of dimension $m+1$).
On $V_m$, we consider the norm
$$\lVert f\rVert_1 := \int_0^1 \lvert f(t)\rvert\,dt.$$
That it is indeed a norm follows from the fact that a polynomial of degree $d \geqslant 0$ has at most $d$ zeros, so a polynomial with $\lVert p\rVert_1 = 0$ must be the zero polynomial.
We use $\lVert\,\cdot\,\rVert_1$ to topologise $V_m$, and by the fact that every finite-dimensional (real or) complex vector space carries only one Hausdorff TVS topology (proof can be found for example here), conclude that $V_m$ is locally compact and all closed balls of finite radius are compact.
Since $V_m$ is finite-dimensional, all linear forms on $V_m$ are continuous, hence $A := \{ f \in V_m : f(0) = 1\}$ is closed. Therefore the intersection
$$K = A \cap \overline{B_2(0)}$$
is compact. Since the constant polynomial with value $1$ is in $A$, and $\lVert 1\rVert_1 = 1 < 2$, the set $K$ is not empty. The norm is continuous ($\bigl\lvert \lVert f\rVert_1 - \lVert g\rVert_1 \bigr\rvert \leqslant \lVert f-g\rVert_1$), hence it assumes its minimum on the non-empty compact set $K$, say in $f_0 \in K$. For $f \in A \setminus K$, we have $\lVert f\rVert_1 > 2$, hence
$$\lVert f_0\rVert_1 = \min \{\lVert f \rVert_1 : f \in K\} = \min \{\lVert f \rVert_1 : f \in A\}.$$
