Show that there exists $a^* \in \mathbb{R}^{k+1}$ such that $f(a^*) \leq f(a)$, $\forall a \in \mathbb{R}^{k+1}$ 
Let $h \in C[0,1]$. Show that there exists $a^* \in \mathbb{R}^{k+1}$
  such that $P_{a^*}$ best approximates $h$ in quadratic mean. Let $f :
 \mathbb{R}^{k+1} \to \mathbb{R}$ define as
  $f(a)=\int_0^1(h(t)-P_a(t))^2dt$. Show that there exists $a^* \in
 \mathbb{R}^{k+1}$ such that $f(a^*) \leq f(a)$, $\forall a \in
 \mathbb{R}^{k+1}$.

I think we have to use the following result to show that the function is coercive.

Result for the problem : For $a= (a_0, \dots , a_k) \in \mathbb{R}^{k+1}$, we define the polynomial of degree $k$ $$P_a :
 \mathbb{R} \to \mathbb{R}$$ $$P_a(t)=a_0+a_1t+ \dots + a_k t^k.$$ Let
  $$g :\mathbb{R}^{k+1} \to \mathbb{R}$$ $$g(a) =
 \int_0^1(P_a(t))^2dt.$$ A result explain that there exists $\alpha >
 0$ such that $g(a) \geq \alpha \|a\|^2.$

Well-known theorem : Let $C \subset \mathbb{R}^k$ a close set, the continuous function $f : C \to \mathbb{R}$ and $\lim_{\|x\| \to \infty} f(x) = \infty$. Then $\exists x_n \in C$ such $f(x_n) \leq f(x)$, $\forall x \in C$.
Is anyone could resolve this problem? I don't think I have all the skills to do that question?
 A: What you are missing is the Cauchy Schwarz inequality: Note
$$\begin{split}
f(a) &= \int_0^1 (h(t) - P_a(t))^2 dt \\
&= \int_0^1 P_a(t)^2 dt + \int_0^1 h(t)^2 dt - 2\int_0^1 h(t) P_a(t) dt \\
&\ge\int_0^1 P_a(t)^2 dt + \int_0^1 h(t)^2 dt - 2\sqrt{\int_0^1 h(t)^2 dt} \sqrt{\int_0^1 P_a(t) ^2 dt} \\
&= g(a) + C^2 - 2C\sqrt{g(a)}
\end{split}
$$
where we write $C = \sqrt{\int_0^1 h(t)^2 dt}$, which is fixed independent of $a$. Try to use this and your results to show that $f$ is coercive. 
Edit: Note that the last term on the right is 
$$ (\sqrt {g(a)} -C)^2.$$
Now, to show that $f$ is coercive, let $a\in \mathbb R^{n+1}$ be big, so that $\|a\| > \frac{C}{\sqrt\alpha}$. So $\sqrt{g(a)} -C \ge \sqrt{\alpha} \| a\| -C >0$. Thus 
$$f(a) \ge (\sqrt{g(t)} -C)^2 > (\sqrt{\alpha} \|a\| - C)^2.$$
If $\|a\|\to \infty$, the right hand side also goes to infinity. So 
$$\lim_{a\to \infty} f(a) = \infty.$$
A: The space of all polynomial coefficients up to degree $k$ are of finite dimension $k+1$, hence it is closed. The function $f$ can be written as
\begin{equation}
f(a) = \|h-P_{a}\|_{L^2(0,1)}^2. 
\end{equation}
I think we can show that $\lim_{\|a\| \to \infty}f(a) = \infty$ by taking $a_0\to \infty$ while setting the others to 0.
