Consider an interval $I = [t_0,t_1]$ and a finite dimensional Banach space $X$. Let further be $\mathcal{C}_0^1(I,X)$ be the set of $\mathcal{C}^1$ functions vanishing on the boundary. Show the following special case of the Fundamental Lemma of calculus of variations:
If a continuous $x^\ast: I \rightarrow X^\ast$ has the property $\int_I x(t)^\ast (x(t)) dt = 0$ for all $x \in \mathcal{C}_0^1(I,X)$, then $x^\ast(t) = 0$.
Remark: $X^\ast$ denotes the dual space of $X$.
I do not see how to prove this. Could you please give me a hint?
Edit:
I recognsie that by the Riesz Representation Theorem we may find a $y \in X$ such that $\langle y, \cdot \rangle = x^\ast (\cdot)$ that $\langle y,x \rangle =x(t)^\ast(x(t))$. We note that for any $x \in \mathcal{C}_0^1$ we got:
\begin{align} 0 &= \int_I \langle y(t),x(t)y(t) \rangle dt \\ &= \int_I x(t) \langle y(t),y(t) \rangle\\ \text{Using integration by parts yields:}\\ &= \bigg( x(t) \int_I \langle y(t),y(t) \rangle dt \bigg) \bigg\rvert_{t_0}^{t_1} - \int_I x^\prime(t) \langle y(t),y(t) \rangle dt \end{align}
I see that $\bigg( x(t) \int_I \langle y(t),y(t) \rangle dt \bigg) \bigg\rvert_{t_0}^{t_1} = 0$, but I do not get how to go on from here.