# Prove that if $f \in L^3[0, 1]$ and $g \in L^7[0, 1]$, then $(fg) \in L^2[0, 1]$.

I'm hitting a wall with this problem, even though it seems easy. I found a similar-looking problem here, but that problem will help me only if $||g||_\infty$ is finite.

I know that, because [0, 1] is measurable with finite measure and $L^a[0, 1] \subset L^b[0, 1]$ whenever $a \geq b \geq 1$, we must have that $||f||_j < \infty$ for $1 \leq j \leq 3$ (and, on that same note, that $||g||_k < \infty$ for $1 \leq k \leq 7$). I've tried using Holder's inequality, but no matter what I do, I keep arriving at the fact that $(fg) \in L^1[0, 1]$.

Any guidance or corrections here would be greatly appreciated, since I'm still trying to grasp normed linear spaces. Thank you in advance!

$$\|fg\|_2^2 = \|f^2g^2\|_1 \le \|f^2\|_{3/2} \|g^2\|_3 = \|f\|_3^2 \|g\|_6^2 < \infty.$$
• I think I understand. So, for clarification: --First equality: turn the norms into integrals, interpret as norm for p = 1. --First inequality: Holder's inequality, with p = 3 and its conjugate q = 3/2 --Second equality: Playing with the numbers to get norms out of the integrals --Last inequality: $||g||_6 \leq c||g||_7$ for some c, and we know $||g||_7$ is finite (as is $||f||_3$). Is this correct? Commented Nov 28, 2016 at 9:03
• Small typos: That should be $\|g^2\|_{7/2}$ and $\|g\|^2_{7}.$....+1. Commented Nov 28, 2016 at 10:18
• @user254665 See d4rk_1nf1n1ty's summary; I need to use $3$ since it is the conjugate dimension to $3/2$ (needs to satisfy $1/p+1/q=1$). Commented Nov 28, 2016 at 17:26