The <Holder inequality> interpolates between $L^2(U)$ and $L^6(U)$:
$$
\|v\|_{L^3}^3
=\int_U|v|^{3/2}|v|^{3/2}
\leq\|v\|_{L^2}^{3/2}\|v\|_{L^6}^{3/2}.
$$
Taking cube roots and applying the three-dimensional <Sobolev inequality> $H^1(U)\hookrightarrow L^6(U)$ gives
$$
\|v\|_{L^3(U)}
\leq C\|v\|_{L^2(U)}^{1/2}\|v\|_{H^1(U)}^{1/2}.
$$
This is the <H1 L3 interpolation inequality in three dimensions>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 105 / 2 / b / i / Solution

{c}
{title2=$\|u\|_3\leq C\|u\|_2^{1/2}\|u\|_{H^1}^{1/2}$}

On a bounded three-dimensional domain, interpolation between $L^2$ and the <Sobolev inequality> $H^1\hookrightarrow L^6$ gives
$$
\|u\|_{L^3}\leq C\|u\|_{L^2}^{1/2}\|u\|_{H^1}^{1/2}.
$$

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 Incoming links: H1 L3 interpolation inequality in three dimensions