Dimension of space of continuous functions Please can you help me in this exercice.

Prove that the normed space of continuous functions $f: [0,t] \to \mathbb{R}$ with the fundamental norm $||\cdot||_2$ is infinite dimensional. 

Indication: We can use the sequence $(f_n)$ given by $f_n(x)= \sin(nx), n \in \mathbb{N}, x \in [0,t]$
 A: In a finite vector space of dimension $n$, any $n+1$ vectors are linearly dependent. Hence, to prove what you want, showing that for any $n \in \mathbb{N}$ you can find $n$ linearly independent vectors is sufficient.
It can be shown that, for integers $l \neq m$, $\int _0^{2\pi} \sin(lx)\sin(mx)\,dx = 0$, and $\int_0^{2\pi} \sin^2(lx)\,dx = \pi $. You can use this to prove that for all $n\in \mathbb{N}$, the functions $\{ f_k \}_{k=1}^n$ are linearly independent. 
To do so; suppose some linear combination $\Sigma _{i=1}^n a_i f_i$ is equal to the zero function. For any fixed $k \in \{1,\cdots,n \}$, multiplying by $f_k$ and integrating from $0$ to $2\pi$ tells us $ \pi a_k =0$, and we're done.
A: You can show that, for any $n$, you can find a set of $n+1$ linearly independent functions.
Divide your interval into $n$ equal segments by the points $x_0=0$, $x_1=t/n$, $x_2=2t/n$, $\dots$, $x_n=nt/n=t$. Consider the polynomial
$$
f(x)=(x-x_0)(x-x_1)\dots (x-x_n)
$$
and define
$$
f_k(x)=\frac{f(x)}{x-x_k}.
$$
Just do the formal division, so $f_k(x)$ is a polynomial of degree $n$ which is zero on all points $x_j$ except for $x_k$. For instance,
$$
f_0(x_0)=(x_0-x_1)(x_0-x_2)\dots(x_0-x_n)\ne0.
$$
Consider a linear combination
$$
g(x)=a_0f_0(x)+a_1f_1(x)+\dots+a_nf_n(x).
$$
Then $g(x_k)=a_kf(x_k)$ is zero only if $a_k=0$. Thus if the linear combination is zero, all coefficients are zero and so $\{f_0,f_1,\dots,f_n\}$ is a set of $n+1$ linearly independent vectors in your vector space.
Of course, we're assuming $t>0$; for the degenerate interval $[0,0]$, the space is one dimensional. The norm has no role in this.
