Let $\{a_n\}$ be a sequence of positive numbers. I am trying to show that if
$$f(x)=\frac{1}{x}\left[\sum_{n=1}^{\lfloor x\rfloor}a_n +(x-\lfloor x\rfloor)a_{\lceil x\rceil}\right]$$ then
$$||f||_p^p \geq \sum_{N=1}^{\infty}\left(\frac{1}{N}\sum_{n=1}^N a_n\right)^p \tag{1}$$
My attempt:
If $g=\left(\frac{1}{N}\sum_{n=1}^N a_n\right)\chi_{(N,N+1]}$, then $||g||_p^p = \sum_{N=1}^{\infty}\left(\frac{1}{N}\sum_{n=1}^N a_n\right)^p$. So $g$ is a step function which takes value $\left(\frac{1}{N}\sum_{n=1}^N a_n\right)$ on the interval $(N,N+1]$ whereas $f$ takes value $\frac{1}{x}\left[\sum_{n=1}^{N}a_n +(x-N)a_{N+1}\right]$ again on the interval $(N,N+1]$ where $x\in (N,N+1]$. So if we show the following we are done:
$$\begin{align} \frac{1}{x}\left[\sum_{n=1}^{N}a_n +(x-N)a_{N+1}\right] &\geq \left(\frac{1}{N}\sum_{n=1}^N a_n\right) \\ \iff \left( \frac{1}{x}-\frac{1}{N}\right) \sum_{n=1}^{N}a_n &\geq \left(\frac{N-x}{x}\right)a_{N+1} \end{align} $$
which would be true if $a_n$ is monotonically decreasing so that $\sum_{n=1}^{N}a_n \geq a_{N+1}$ and clearly $\left( \frac{1}{x}-\frac{1}{N}\right) \geq \left(\frac{N-x}{x}\right)$ for $x\in(N,N+1]$.
May I know why $(1)$ would be true for arbitrary $\{a_n\}$?