How to check this $L^p$ inequality?

Question

Notation:

$I_{k,j}\subseteq (0,1)$ is an interval of length $2^{-k}$.

$\|f\|_p^p=\int_0^1 |f|^p$

$1/p+1/p'=1$

What I got is:

$\large \begin{align*} &2^{kp}\sum_{i=1}^{2^k}\int_{I_{k,j}}[(\int_{I_{k,j}}|f(t)|^p\,dt)2^{-kp/p'}]\,dx\\ &\leq\|f\|_p^p(2^{kp})(2^k)(2^{-k})2^{-kp/p'}\\ &=2^k\|f\|_p^p \end{align*} $

with an extra factor of $2^k$. Did I make any mistake?

I realised my mistake. In my context, $\bigcup_{j=1}^{2^k}I_{k,j}=(0,1)$ (I forgot to mention this). Gordon managed to solve it correctly though.

Answer 1

Note that \begin{align*} 2^{kp}\sum_{i=1}^{2^k}\int_{I_{k,j}}[(\int_{I_{k,j}}|f(t)|^p\,dt)2^{-kp/p'}]\,dx &= 2^{kp}\sum_{i=1}^{2^k} 2^{-k} 2^{-kp/p'}\int_{I_{k,j}}|f(t)|^p\,dt\\ &=2^{kp-k-kp/p'}\int_0^1|f(t)|^p\,dt\\ &=2^{kp-k-kp(1-1/p)}\int_0^1|f(t)|^p\,dt\\ &=\int_0^1|f(t)|^p\,dt. \end{align*}