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Let $1\leq p<q<r<\infty$ and $f\in L ^p(\Omega)\cap L^r(\Omega)$. Then for $a=\frac{\frac{1}{p}-\frac{1}{q}}{\frac{1}{q}-\frac{1}{r}}$ and $c>0$ the inequaltiy $$\|f\|_q\leq c\|f\|_r+\frac{1}{c^a}\|f\|_p$$ holds.
I think I need to use the Young inequality to prove the desired inequality. However, I'm not sure how to choose the powers most skillfully. In particular, the q-th root at the beginning is giving me difficulty. I would appreciate any hints!

Edit using Umberto's suggestion:
For $\dfrac 1 q = \dfrac \lambda p + \dfrac{1-\lambda}r$ one gets $$\dfrac 1 p-\dfrac 1 q=(1-\lambda)\left(\dfrac 1 p -\dfrac 1 r\right)$$ $$\dfrac 1 q-\dfrac 1 r=\lambda\left(\dfrac 1 p -\dfrac 1 r\right).$$ It follows that $a=\frac{1-\lambda}{\lambda}$ holds. Using Young's inequality for $p=\frac{1}{\lambda}$ and $q=\frac{1}{1-\lambda}$ $$\|f\|_q\leq \|f\|_p^{\lambda}\cdot\|f\|_r^{1-\lambda} =\epsilon\|f\|_p^{\lambda}\cdot\frac{1}{\epsilon}\|f\|_r^{1-\lambda}\leq \lambda \epsilon^{\frac{1}{\lambda}}\|f\|_p+(1-\lambda)\left(\frac{1}{\epsilon}\right)^{\frac{1}{1-\lambda}}\|f\|_r.$$ Choosing $\epsilon=\left(\frac{1}{\lambda}c\right)^{\lambda}$: $$\|f\|_q\leq c\|f\|_p+(1-\lambda)\left(\frac{1}{\lambda}c\right)^{a^{-1}}\|f\|_r.$$ Since the exponent of $c$ and the factor of $\|f\|_r$ are not right: Am I missing something?

MaxwellDgt
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1 Answers1

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I'm not sure about the exact value of $a$, but you can try this. Since $p < q < r$ there exists $0 < \lambda < 1$ satisfying $\dfrac 1 q = \dfrac \lambda p + \dfrac{1-\lambda}r$. Thus $$ 1 = \frac{\lambda q}p + \frac{(1-\lambda)q}{r}$$ which suggests Holder's inequality: $$\int |f|^q = \int |f|^{\lambda q} |f|^{(1-\lambda)q} \le \left( \int |f|^p \right)^{\frac{\lambda q}{p}} \left( \int |f|^r \right)^{\frac{(1-\lambda)q}{r}}$$ This gives you $$\|f\|_q \le \|f\|_p^\lambda \|f\|_r^{1-\lambda}.$$

Young's inequality leads to $$\|f\|_q \le \epsilon^{1/\lambda} \|f\|_p + \frac{1}{\epsilon^{1/(1-\lambda)}} \|f\|_r$$ for any positive $\epsilon$. Try $c = \epsilon^{1/\lambda}$ and see what the other constant works out to be.

Umberto P.
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