I would like to prove the following inequality: $$\langle f,\frac{|N.+1|}{N} \rangle^2 \leq \langle f,. \rangle^2$$
where $f$ is a step function of the form $f(s)=\sum\limits_{i=1}^N f_i \mathbb{1}_{I(i)}(s)$, $f_i\in \mathbb{R}$, and $I(i)=\left[\frac{i-1}{N};\frac{i}{N}\right]$ for $i\in{1,...,N}$.
$\langle . \rangle$ denotes the scalar product in $\mathbb{L}^2 ([0,1])$ i.e $\langle f,g \rangle = \int_0^1 f(x)g(x)dx$, and $|.|$ is the floor function.
I have already tried to use $\frac{|Ns+1|}{N}=\sum_{i=1}^N \frac{i}{N}\mathbb{1}_{I(i)}(s)$, and then the fact that the product of two step functions is a step function to compute the left side integral, but I don't know where to go from there . Any ideas would be much appreciated!