As it well known, $C^\infty_0 (\mathbb{R}^n)$(the space of infinitely differentiable functions with compact support) is dense in $L^p (\mathbb{R)^n}$

Here I want to consider the same result with mixed norm. The mixed norm space is defined by $$ \Vert f \Vert_{L^p_t L^q_x}:=\left(\int_{\mathbb R} \left[\int_{\mathbb{R}^n} |f(t,x)|^p dx\right]^{\frac{q}{p}}dt\right)^{\frac{1}{p}}.$$

This norm can be viewed as $$ \Vert f \Vert_{L^p_t L^q_x}:=\left(\int_{\mathbb R} \Vert f(t,\cdot) \Vert_{L^q (\mathbb{R^n})}^p dt \right)^{\frac{1}{p}}.$$

So by this point of view, I can verify that the space $L^p_t L^q_x$ is actually $L^p (\mathbb{R}; L^q(\mathbb{R}^n)$) space, which is a special case of Bochner space.

Then in the theory of Banach-valued function space, $$ f(t) = \sum_{i=1}^n a_i \phi_i (t),$$ where $a_i$ is an element of Banach space $B$, $\phi_i$ is a characteristic function in $\mathbb{R}$ with finite measure, is dense in $L^p (\mathbb{R};B)$ when $1\leq p <\infty$. So by taking molification, we see that $C^\infty_0 (\mathbb{R};B)$ is dense in $L^p (\mathbb{R};B)$.

As $C^\infty_0$ is dense in $L^p(\mathbb{R}^n)$, $C^\infty_0 (\mathbb{R}^{n+1})$ is dense in $L^p_t L^q_x$.

My question is the following:

  • Can I proved it properly?

  • How can prove the fact without using Bochner integral?

    Thank you in advance.

  • $\begingroup$ Your idea looks good: just approximate these $a_i$'s by smooth functions, and you get a smooth approximation of $f\in L^p(L^q)$. $\endgroup$ – daw Jul 19 '16 at 7:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.