# Sobolev Mixed Norm Inequality

I have come across a problem which would simplify considerably if the following inequality

$$\left(\int |u|^2\,dx\right)^p \leq n(\left\Vert u \right\Vert_{1,2}^2 + \left\Vert u \right\Vert_{1,r}^r)$$

holds for some $$n \in \mathbb N$$, $$0, where $$u$$ ranges over $$W^{1,2}_0 (\Omega)$$ (zero trace) and $$1 is fixed.

My question is: Is it reasonable for this inequality to hold?

What I have so far: If we could further impose restrictions on dimension of $$\Omega$$, this would follow immediately by embedding results once we set $$p=\frac r 2$$. However, the general scenario seems to require some further estimates - and I would be curious if anyone could point me in the right direction.

• This cannot be true for scaling reasons unless $2p \in [r,2]$: Plug in $\lambda u$, where $\lambda > 0$ is arbitrary and consider $\lambda \to 0$ and $\lambda \to \infty$.
– gerw
May 21, 2021 at 5:52

First, we homogenize your inequality. I.e., we plug in $$\lambda u$$ with $$\lambda > 0$$ and consider $$\lambda \to 0$$ and $$\lambda \to \infty$$. This shows that the inequality cannot hold in case $$2p \not\in [r,2]$$.
In case $$2p = r$$, the first term on the rhs disappeards and your inequality is equivalent to $$\|u\|_2 \lesssim \|u\|_{1,r}.$$ This holds under some relation of $$r$$ and the dimension $$d$$ (Sobolev embedding).
In the remaining case $$2 p \in (r,2)$$, you can divide by $$\lambda^{2p}$$ and minimize the resulting rhs w.r.t. $$\lambda > 0$$. This results in the homogeneous inequality $$\|u\|_2 \lesssim \|u\|_{1,2}^{2\frac{1-r/(2p)}{2-r}} \|u\|_{1,r}^{r\frac{2/(2p)-1}{2-r}}.$$ Note that Young's inequality implies that this inequality is indeed equivalent to your original inequality; and the exponents on the right-hand side add up to one.
From Sobolev, we get $$\|u\|_q \lesssim \|u\|_{1,2}, \qquad \|u\|_s \lesssim \|u\|_{1,r}$$ for $$1/q = 1/2 - 1/d$$ and $$1/s = 1/r - 1/d$$. Thus, $$\|u\|_{q}^{2\frac{1-r/(2p)}{2-r}} \|u\|_{s}^{r\frac{2/(2p)-1}{2-r}} \lesssim \|u\|_{1,2}^{2\frac{1-r/(2p)}{2-r}} \|u\|_{1,r}^{r\frac{2/(2p)-1}{2-r}}.$$ Interpolation implies $$\|u\|_t \lesssim \|u\|_{1,2}^{2\frac{1-r/(2p)}{2-r}} \|u\|_{1,r}^{r\frac{2/(2p)-1}{2-r}}.$$ for $$\frac1t = \frac{2\frac{1-r/(2p)}{2-r}}{s} + \frac{r\frac{2/(2p)-1}{2-r}}{q} .$$ Finally, we have $$\|u\|_2 \le \|u\|_t$$ if $$2 \le t$$. Hence, in case $$\frac12 \ge \frac{2\frac{1-r/(2p)}{2-r}}{s} + \frac{r\frac{2/(2p)-1}{2-r}}{q} .$$ we have shown your inequality. (If this relation is not satisfied, I expect that your inequality does not hold)