It is well-known that if $1\leq p\leq q\leq \infty $ then

$$ L^p(X)\cap L^q(X)\subset L^r(X)\quad\quad \text{whenever $r\in [p,q]$}\tag{I}\label{Eq}$$.

Indeed let $u\in L^p(X)\cap L^q(X)$. For some $0\leq \theta\leq 1$ we can write $$\frac{1}{r}=\frac{1-\theta}{p}+\frac{\theta}{q} $$

So that, $|u|^{1-\theta}\in L^{p/1-\theta}$ and $|u|^{\theta}\in L^{q/\theta}$ and the Hölder's inequality implies,

$$\|u\|_{L^r(X)}\leq \|u^{1-\theta}\|_{L^{p/1-\theta}(X)}\|u^{\theta}\|_{L^{q/\theta}(X)}= \|u\|_{L^{p}(X)}^{1-\theta}\|u\|_{L^{q}(X)}^{\theta}$$

Replacing $\theta=\frac{\frac{1}{r}- \frac{1}{p}}{\frac{1}{q}- \frac{1}{p}}$ by its value gives

$$\|u\|_{L^r(X)}\leq \|u\|_{L^{p}(X)}^{\frac{\frac{1}{q}- \frac{1}{r}}{\frac{1}{q}- \frac{1}{p}}}\|u\|_{L^{q}(X)}^{\frac{\frac{1}{r}- \frac{1}{p}}{\frac{1}{q}- \frac{1}{p}}}$$

Question: What is the analogue of the property \eqref{Eq} for Orlicz space $L^\phi(\Bbb R^d)$? Any hint or references is welcome...

Recall the Orlicz space $L^\phi(\Bbb R^d)$: \begin{align*} L^\phi(\Bbb R^d)&= \Big\{u: \Bbb R^d\to \Bbb R\text{ meas.}:~ \int_{\Bbb R^d} \phi\Big(\frac{|u(x)|}{\lambda}\Big)d x<\infty ~~\text{for some $\lambda>0$}\Big\}. \end{align*} The space $L^\phi(\Bbb R^d)$ is equipped with the Luxemburg norm $\|\cdot\|_{L^\phi(\Bbb R^d)}$ by \begin{align} \|u\|_{L^\phi(\Bbb R^d)}=\inf \Big\{ \lambda>0~: \int_{\Bbb R^d} \phi\Big(\frac{|u(x)|}{\lambda}\Big)d x\leq 1\Big\}. \end{align}

Where $\phi$ is a sufficiently nice Young function, e.g., $\phi$ is continuous, increasing, convex and in addition, the mapping $x\mapsto \frac{\phi(x)}{x}$, $x>0$ is increasing and satisfies \begin{align*} &\lim_{x\to 0^+}\frac{\phi(x)}{x}= \lim_{x\to \infty}\frac{x}{\phi(x)}= 0. \end{align*}