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Let $f:\mathbb R^d \to \mathbb C$ be a measurable function, $d\geq 3.$

Can we expect the inequality:

$$\|f\|_{L^{\frac{2(d+2)}{d-2}}}\leq C \|f\|_{L^{\frac{2(d+2)}{d}}}^\frac{d-2}{2} \|f\|_{L^{\infty}}^{\frac{2}{d}}?$$

where $C$ is some constant and assume everything above make sense.

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Here's a simple counter-example: take $d=4$, in this case your inequality becomes $$\|f\|_6\leq C\|f\|_3\|f\|_\infty^{\frac{1}{2}}\quad (*).$$ Define $$f_n(x)=\chi_{[0,n^{-1}]^4}(x).$$ Then clearly $f_n\in L^p$ for all $1\leq p\leq\infty$ with $\|f_n\|_\infty=1$. Also by straight forward computation: $$\|f_n\|_6=n^{-\frac{2}{3}},\quad \|f_n\|_3=n^{-\frac{4}{3}}.$$ Plugin $f=f_n$ into $(*)$ we get $$n^{-\frac{2}{3}}\leq C n^{-\frac{4}{3}}\quad\Rightarrow 1\leq C n^{-\frac{2}{3}}.$$ Since $n$ is arbitrary, we see that if $(*)$ holds then $$1\leq Cn^{-\frac{2}{3}},\quad\forall n\in\mathbb{N}.$$ Clearly this is impossible.

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    @FL: Thanks: I know $m([1, \frac{1}{n}])=\frac{1}{n}$. But what is $m([1,1/n]^4)=?$ ($m$ is Lebesgue measure ) – 2017-02-13
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    For any $n$ dimensional interval $[a_1,b_1]\times\dots\times[a_n,b_n]$, its $n-$dimensional Lebesgue measure is $\prod_{i=1}^n(b_i-a_i)$. The measure will not change if we replace $[a_i,b_i]$ by the open or half open interval. – 2017-02-13