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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.

user40276
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    You need some inhomogeneous constraint, such as leading coefficient 1 (monic polynomial), to make your problem sensible. If by degree $m$ you mean degree less than or equal to $m$, the zero polynomial attains a minimum of zero. If exactly degree $m$, polynomials can be arbitrarily close to zero without some additional constraint. – hardmath Aug 23 '13 at 13:02
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    @hardmath Ops, I forgot to put that $f(0) =1$. Sorry. – user40276 Aug 23 '13 at 13:08
  • Degree exactly $m$ or at most $m$? – Daniel Fischer Aug 23 '13 at 13:10
  • @DanielFischer At most $m$ – user40276 Aug 23 '13 at 13:13
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    Then you have what you need. Let $A = {f \in L^1 : f(0) = 1, f \in \mathbb{C}[x], \deg f \leqslant m}$. The norm $\lVert,\cdot,\rVert_1$ is continuous, $A \cap {g \in L^1 : \lVert g\rVert_1 \leqslant 2}$ is compact and non-empty. – Daniel Fischer Aug 23 '13 at 13:21
  • @DanielFischer Sorry, but I could not understand your intention. – user40276 Aug 23 '13 at 13:51

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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

$$\lVert f_0\rVert_1 = \min \{\lVert f \rVert_1 : f \in K\} = \min \{\lVert f \rVert_1 : f \in A\}.$$

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