Inequality in Fourier analysis Let $2\leq p<\infty$. For $1\leq k\leq n$, let $f_k\in L^p(\mathbb R)$ and suppose $\hat {f_k}\in L^1(\mathbb R)$ and is supported on a bounded interval $I_k\subseteq [0,1]$. If $\{I_k\}_{k=1}^n$ have disjoint interiors, is it true that
$$
\sum_{k=1}^n \left\|f_k\right\|_p^2\lesssim  \left\|\sum_{k=1}^nf_k\right\|_p^2,
$$
where the constant may depend on $p$ but is independent of $n$?
 A: Since your functions are on disjoints intervals $I_k$, one can start from the right-hand side of your inequality and get
$$
\begin{align*}
\left(\int_{\mathbb{R}} \left|\sum_{j=1}^n f_j\right|^p\right)^\frac{2}{p} &=   \left(\sum_{k=1}^n\int_{I_k} \left|\sum_{j=1}^n f_j\right|^p\right)^\frac{2}{p}
\\
&= \left(\sum_{k=1}^n\int_{I_k} \left|f_k\right|^p\right)^\frac{2}{p}
\\
&= \left(\sum_{k=1}^n\int_{\mathbb{R}} \left|f_k\right|^p\right)^\frac{2}{p}
\end{align*}
$$
so that one always has
$$
\left\|\sum_{k=1}^n f_k\right\|_{L^p}^2 = \left(\sum_{k=1}^n \left\|f_k\right\|^p_{L^p}\right)^\frac{2}{p}
$$
Hence, if I write $u_k := \|f_k\|_{L^p}$, and $|u_k|_{p} := \left(\sum u_k^p\right)^\frac{1}{p}$ the usual $\ell^p$ norm of vectors, your question can be restated as: "Is it true that
$$
|u_k|_2^2 \lesssim |u_k|_p^2
$$
with a constant independent of $n$ ?" and this is purely a question about sequences. If $p≤2$, then yes, this is true since $(a+b)^\theta ≤ a^θ + b^\theta$ for any $\theta\in[0,1]$, and the same holds for the sum of $n$ numbers. In particular, taking $θ=p/2$ gives 
$$
|u_k|_2^2 = \sum u_k^2 = \left(\sum u_k^2\right)^\frac{2p}{2p} ≤ \left(\sum u_k^p\right)^\frac{2}{p} = |u_k|_p^2
$$
(and you actually do not need $\hat{f}\in L^1$). But you are asking for $p\geq 2$, which is false in general for sequences since taking $u_k = 1$ for all $k$ gives you
$$
\frac{|u_k|_2}{|u_k|_p} = \frac{n^\frac{1}{2}}{n^\frac{1}{p}} = n^{\frac{p-2}{2p}}
$$
and $\frac{p-2}{2p} > 0$. And one can easily find functions $f_k$ with disjoint supports verifying $u_k = \|f_k\|_{L^p} = 1$. Take for example the functions
$$
f_k(x) = n^{1/p}\,\mathbf{1}_{[\frac{k-1}{n},\frac{k}{n}]}(x).
$$
A: The triangle inequality is satisfied by the definition of the $L_p$ norm and gives the reverse inequality.
