Determining a given set related to inner product on continuous function. 
Let $\alpha>0$ and $f\in \mathcal{C}^0([0,1],\mathbb{R})$.
$$I(\alpha) = \left\{\int_0^1f(t) P(t) \, \mathrm{d}t \; \bigg| P \in \mathbb{R}[X] \, \mathrm{and} \int_0^1P(t)^2  
 = \alpha\right\}$$
Show $I(\alpha)$ is an interval and determine it.


What I have done so far:
I immediately thought of using the inner product on $\mathcal{C}^0([0,1],\mathbb{R})$ defined by,
$$\langle f |g \rangle = \int_0^1f(t) g(t) \, \mathrm{d}t $$
And $\lVert  \cdot \rVert = \sqrt{\langle \cdot | \cdot\rangle}$
Therefore,
$$I(\alpha) = \left\{ \langle f \;| \; P \rangle \,| P\in\mathbb{R}[X] \;\mathrm{and} \;\lVert P \rVert^2=\alpha  \right\}$$
Let $A = \left\{ P\in\mathbb{R}[X] \; \mathrm{s.t.} \;\lVert P \rVert=\sqrt{\alpha}  \right\}$ and $\varphi_f : P \in\mathbb{R}[X]\mapsto\langle f|P\rangle $.
As $\varphi_f$ is a linear map, I have shown using Cauchy-Schwarz inequality that $\varphi_f$ is continuous from $[\mathbb{R}[X],\lVert  \cdot \rVert] $ to $[\mathbb{R},|\cdot|]$. Moreover,
$$\varphi_f(A) = I(\alpha)$$
So in order to show $I(\alpha)$ is an interval I thought of showing that $I(\alpha)$ is path connected (because the only path connected subsets of $\mathbb{R}$ are its intervals) by showing that $A$ is path connected as $\varphi_f$ is continuous.
To show $A$ is path-connected, I identify $A$ to be the sphere in $\mathbb{R}[X]$ with radius $\sqrt{\alpha}$ for the norm $\lVert \cdot \rVert$, i.e. $A = \mathcal{S}^{\lVert \cdot \rVert}(0,\sqrt{\alpha})$.
So given $P,Q$ in $A$ if $P = -Q$ then I will introduce the map,
$$\eta : t \in[0,1] \mapsto \sqrt{\alpha}\dfrac{(1-t)P+tQ}{\lVert(1-t)P+tQ\rVert} $$
which is well defined, continuous and verifying for all $t \in [0,1], \eta(t) \in A$, $\eta(0) = P$ and $\eta(1) = Q$ . If $P = -Q$ as $\dim(\mathbb{R}[X])=\infty\geq2$ we can find $T$ not colinear to $P$, so $R = \sqrt{\alpha}\frac{T}{\lVert T \rVert} \neq -P$ and belongs to $A$, so as previously we can find a connected path from $P$ to $R$ and a connected path from $R$ to $-P = Q$. So in any case we can find a connected path from $P$ to $Q$ and hence $A$ is path connected.
Finally $I(\alpha)$ is an interval.

In order to determine $I(\alpha)$ I wanted to find an upper bound and a lower bound. I can easily see that with the constant polynomial $\sqrt{\alpha}$ we have $\displaystyle \sqrt{\alpha} \int_0^1 f(t) \mathrm{d}t$
belongs to $I(\alpha)$.
Moreover by Cauchy-Schwarz inequality,
$$ \left| \int_0^1 f(t) P(t) \mathrm{d}t \right| \leq \sqrt{\alpha}\lVert f \rVert$$
On top of that, $I(a)$ is clearly symmetric centered on $0$.
And thus,
$$I(\alpha) \subset \big[-\sqrt{\alpha} \lVert f \rVert,\sqrt{\alpha} \lVert f \rVert \big]$$

I don't know if I am heading in the good direction in terms of method. If you have ideas to be maybe more efficient, and/or to find the upper bound of $I(\alpha)$ it will be appreciated.

Edit: Maybe I could use the density of real polynomial functions in $\mathcal{C}^0([0,1],\mathbb{R})$ for the $\lVert \cdot \rVert_{\infty}$ norm.
 A: I should maybe not have asked the question as I found an answer by myself.
I know according to the Stone-Weierstrass theorem that there exists a sequence $(P_n)_{n\in \mathbb{N}} \in {\mathbb{R}[X]}^{\mathbb{N}}$ that converges to $f$ for the $\lVert \cdot \rVert_{\infty}$ norm.
As we can clearly see that $\lVert \cdot \rVert \leq \lVert \cdot \rVert_{\infty}$ and that the linear map $\varphi_f$ is continuous from $[\mathbb{R}[X],\lVert \cdot \rVert ]$ to $[\mathbb{R},|\cdot|]$ then $\varphi_f$ is continuous from $[\mathbb{R}[X],\lVert \cdot \rVert_{\infty} ]$ to $[\mathbb{R},|\cdot|]$.
Then if $\lVert f\rVert \neq0$, $Q_n = \sqrt{\alpha}\dfrac{P_n}{\lVert P_n \rVert} \overset{\lVert \cdot \rVert_{\infty}}{\underset{n \rightarrow \infty}{\longrightarrow}} \sqrt{\alpha} \dfrac{f}{\lVert f\rVert}$ and $Q_n \in A$
Therefore,
$$ |\langle Q_n \, | \, f\rangle|\underset{n \rightarrow \infty}{\longrightarrow} \sqrt{\alpha} \lVert f \rVert$$
So we can conclude that,
$$I(\alpha) = [-\sqrt{\alpha} \lVert f \rVert,\sqrt{\alpha} \lVert f \rVert] $$ or
$$I(\alpha) = (-\sqrt{\alpha} \lVert f \rVert,\sqrt{\alpha} \lVert f \rVert)$$
depending on whether this bound is reached or not.
And this bound is reached if and only if we have equality in Cauchy-Schwarz inequality therefore if and only if $f$ is polynomial.
