# Proof of an inequality of $L^p$ norms

For a general measure space, we define : $\|f\|_p= \left(\int\vert f\vert^p du\right)^{1/p}$. Let $0 < a < b < c < \infty$ and prove the following: $$\|f\|_b \leqslant \max\{\|f\|_a, \|f\|_c\}.$$ Any help is appreciated because I dont understand the solution underneath

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 The idea here is that $||f||_p \leq ||f||_q$ if $1 \leq p \leq q \leq \infty$ . But the inequality reverses in $0 1 and when a<1 with b,c > 1 ? – user48931 Nov 19 '12 at 7:44 ## 1 Answer We have that $$\frac{1}{c}<\frac{1}{b}<\frac{1}{a}.$$ Therefore, there exists$\alpha\in (0,1)\$ such that $$\frac{1}{b}=\frac{\alpha}{c}+\frac{1-\alpha}{a}.$$ We claim that $$\Vert f\Vert_b\leqslant \Vert f\Vert_c^\alpha\Vert f\Vert_a^{1-\alpha} (\ast)$$ from where it follows that $$\Vert f\Vert_b\leqslant\max\{\Vert f\Vert_a, \Vert f\Vert_c\}$$ because $$\Vert f\Vert_b\leqslant \Vert f\Vert_c^\alpha\Vert f\Vert_a^{1-\alpha}\leqslant \max\{\Vert f\Vert_a,\Vert f\Vert_c\}^{\alpha}\max\{\Vert f\Vert_a,\Vert f\Vert_c\}^{1-\alpha}=\max\{\Vert f\Vert_a,\Vert f\Vert_c\}.$$ Now, (*) can be proved easily using Holder's Inequality as follows $$\int |f|^b=\int |f|^{\alpha b}|f|^{(1-\alpha)b}\leqslant \left(\int|f|^{\alpha b\frac{c}{\alpha b}}\right)^{\frac{\alpha b}{c}}\left(\int|f|^{(1-\alpha)b\frac{a}{(1-\alpha) b}}\right)^{\frac{(1-\alpha)b}{a}}=\left(\int|f|^{c}\right)^{\frac{\alpha b}{c}}\left(\int|f|^{a}\right)^{\frac{(1-\alpha)b}{a}}=\Vert f\Vert_c^{\alpha b}\Vert f\Vert_a^{(1-\alpha) b},$$ where we have used the fact that $$\frac{\alpha b}{c}+\frac{(1-\alpha)b}{a}=1.$$ Simiplifying we have the result.

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 Holder inequality states that for any 1