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Let $E\subset\mathbb{R}^n$ denote a bounded domain and $f: E\to \mathbb{R}$. Further, let $1\leq s<t<u<\infty$ and $f\in L^u(E)$. I have recently seen the following interpolation inequality

$$\|f\|_{L^t(E)} \leq C_1 \|f\|_{L^s(E)} + C_2 \|f\|_{L^u(E)}$$

in a paper, where $C_1,C_2$ denote some positive constants depending on $n,s,t,u$. However, I can't quite verify this inequality myself. I assume that it involves an appropriate usage of Hölder's inequality and also Young's inequality. Unfortunately, the exponents I am getting do not really fit together and I cannot see how this inequality is obtained. I am grateful for any input and help or some references to look this idea up.

HelloEveryone
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2 Answers2

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In a bounded domain $E$, if $f\in L_u(E)$ then $f\in L_r(E)$ for all $0<r<u$ for $|f|^r\in L_{u/r}(E)$ and so $$\int_E|f|^r\leq\left(\int_E|f|^u\right)^{r/u}|E|^{1-\tfrac{r}{u}}$$ By Hölder's inequality. In particular, if $s<t<u$ $$\|f\|_t\leq|E|^{(u-t)/t}\|f\|_{L_u(E)}$$ and so, $$\|f\|_{L_t(E)}\leq \|f\|_{L_s(E)}+|E|^{u/t-1}\|f\|_{L_u(E)}$$

Edit:

My initial boundary were not optimal since they depend on the domain $E$. In fact, the inequality holds even in infinite domains with constants that depend only on $s,t, u$.

Let $\lambda\in(0,1)$ such that $t=\lambda u +(1-\lambda)s$, i.e. $\lambda=\frac{t-s}{u-s}$. If $f\in L_s(\mu)\cap L_u(\mu)$, then $|f|^{\lambda u}\in L_{1/\lambda}(\mu)$ and $|f|^{(1-\lambda)s}\in L_{1/(1-\lambda}(\mu)$ and by Hölder's inequality $$\int|f|^t=\int|f|^{\lambda u}|f|^{(1-\lambda)s}\leq\Big(\int|f|^u\Big)^{\lambda}\Big(\int|f|^s\Big)^{1-\lambda}$$ Since $\frac{\lambda u}{t}+\frac{(1-\lambda)s}{t}=1$ and $\frac{\lambda u}{t}$, $\frac{(1-\lambda)s}{t}\geq0$, the convexity of the exponential function ($\exp(\alpha y+(1-\alpha)x)\leq \alpha e^x+(1-\alpha)e^{y}$ for $0<\alpha<1$, and $x,y\in\mathbb{R}$) yields

\begin{align} \|f\|_t&\leq\big(\|f\|_u\big)^{\tfrac{\lambda u}{t}}\big(\|f\|_s\big)^{\tfrac{(1-\lambda)s}{t}}\leq \frac{\lambda u}{t} \|f\|_u +\frac{(1-\lambda)s}{t}\|f\|_s\\ &=\frac{t-s}{u-s}\frac{u}{t} \|f\|_u+\frac{u-t}{u-s}\frac{s}{t}\|f\|_s \end{align} \end{align}

When $\mu(E)<\infty$ it suffices to assume $f\in L_u(\mu)$.

Mittens
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In general, we know that, for $f \in L^{p_0}(X, \mu) \cap L^{p_1}(X, \mu)$ with any measure space $(X, \mu)$ and $0 < p_0 < p_1 < \infty$, the quantity $\|f\|_{L^p}$ is log-convex in $1/p$, i.e. if $1/p = \theta/p_0 + (1-\theta)/p_1$ then: $$\|f\|_{L^p} \leq \|f\|_{L^{p_0}}^{\theta} \|f\|_{L^{p_1}}^{1 - \theta}$$ See Proof of the log-convexity of $L^p$ norms or the forum post by Terence Tao referenced within for proofs of this fact, one of them being with Hölder's inequality as you expected.

Now, a log-convex function is convex (this is actually a part of a Characterization of log-convexity), so that, for the same $\theta$ as before: $$\|f\|_{L^p} \leq \theta \|f\|_{L^{p_0}} + (1 - \theta)\|f\|_{L^{p_1}}$$ and we are done since $\theta$ only depends on $p$, $p_0$ and $p_1$ (using the fact that $f \in L^u(E)$ implies $L^s(E)$ when $E$ is of finite measure in your case, and bounded implies finite measure for the Lebesgue measure).

Bruno B
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