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For conjugate exponents $p,q\in [1,\infty ]$, $1/p+1/q=1$, Hölder's inequality states that $\int |fg|\le \Vert f{\Vert }_p\Vert g{\Vert }_q$. If $r=\infty$ in the requested product estimate, every $p_i=\infty$ and multiplication of essential bounds proves it directly. Suppose $r<\infty$. Factors with $p_i=\infty$ can again be bounded pointwise by their essential suprema.

Here is an explicit finite-exponent proof of the remaining Generalized Holder inequality. Put $q_i=p_i/r$, so ${\sum }_{i}1/q_i=1$, and normalize $h_i=|f_i{|}^r/\Vert f_i{\Vert }_{p_i}^r$. If one norm is zero the product vanishes almost everywhere. Otherwise $\int h_i^{q_i}=1$. The weighted arithmetic-geometric mean inequality, applied to $h_i^{q_i}$ with weights $1/q_i$, givesIntegrating gives $\int {\prod }_i{h}_i\le 1$. If only one finite exponent remains, its weight is one and the same argument is an equality. Restoring the normalization and taking the $r$th root provesIn particular the product belongs to $L^r$.

For the projection estimate, use the three-dimensional functional Loomis-Whitney inequalityThis permitted three-dimensional result also follows directly by Cauchy-Schwarz inequality: integrate the square root of $g_1{g}_2$ in $x_3$, obtaining the square root of the product of their marginal integrals, and then apply Cauchy-Schwarz inequality in $(x_1,x_2)$ against $\sqrt{g_3}$.

Let $f_j$ be any nonnegative integrable function of the three coordinates excluding $x_j$. For $j=1,2,3$, define $G_j$ by integrating $f_j$ in $x_4$. First Hölder's inequality with three equal exponents yieldsThe fourth factor $f_4(x_1,x_2,x_3)$ is independent of $x_4$. A further Hölder's inequality with exponents $3$ and $3/2$, followed by the three-dimensional estimate, therefore givesThe marginal-integral identifications use Tonelli's theorem. Now take $f_j=1_{K_j}$. The projections are compact and hence measurable with finite measure. Every $x\in K$ has all its projected coordinates in the corresponding $K_j$, so $1_K(x)\le {\prod }_j{1}_{K_j}({\hat{x}}_j{)}^{1/3}$. ConsequentlyThis includes zero-measure projections and is sharp for axis-parallel rectangular boxes.

The condition ${\sum }_{i=1}^{n}1/p_i=1$ and positivity imply $p_i\ge 1$. We prove the Generalized Holder inequality by induction on $n$. The case $n=2$ is the usual Holder inequality. For the induction step, setThe exponents $q$ and $p_n$ are conjugate, so Hölder followed by the induction hypothesis givesThis proves the claim.

Since $U\subset {\mathbb{R}}^3$ is bounded and smooth, the Sobolev embedding theorem gives continuous embeddings $H^1(U)\hookrightarrow L^4(U)$ and $H^1(U)\hookrightarrow L^6(U)$. Because $|\sin s|\le 1$,Taking the $L^2$ norm in time gives

For $F(s)=s^2\sin s$, the mean value theorem and $|F^{\prime }(s)|\le 2|s|+s^2$ implyApply the Generalized Holder inequality in space, using $L^4$ for the quadratic products and $L^6\cdot L^3$ for the cubic products, and then use the two Sobolev embeddings. Pointwise in time this yieldsTaking the $L^2$ norm in time proves the required estimate with the factor $\gamma T^{1/2}$.