# $L^2$ and $L^\infty$ normed inequality for PDE solution: Which one is more informative and why?

I have the following inequalities

$$max_{t \in [0,T]} \lVert u_1(t, \cdot)-u_2(t, \cdot) \rVert_{L^2(\mathbb{R})} \leq C \lVert g_1(x) - g_2(x) \rVert_{L^2(\mathbb{R})}$$

and

$$max_{t \in [0,T]} \lVert U_1(t, \cdot)-U_2(t, \cdot) \rVert_{L^{\infty}(\mathbb{R})} \leq C \lVert g_1(x) - g_2(x) \rVert_{L^{\infty}(\mathbb{R})}$$

where for $x \in \mathbb{R}, t \in [0, T], T \in \mathbb{R}$ the function $U_j(t,x)$ is the solution to some PDE with different initial condition $g_j(x) = \mathbb{1}_{[K_j, \infty)}(x)$ with $K_1 < K_2$ and $x \in \mathbb{R}$ and $u_j(t,x)$ is the solution to the same PDE in weak formulation.

Which inequality is more informative and why? I am a bit confused because I read that on the one hand for $p<q$ that $$\lVert x \rVert _{L^q} \leq \lVert x \rVert _{L^p}$$ but also that $$\lVert x \rVert _{L^p} \leq n^{(\frac{1}{p}-\frac{1}{q})}\lVert x \rVert _{L^q}$$ Does that mean that both inequalities are equally informative in my case with $n=1$? And what about other cases with $n>1$?

EDIT Nr. 2 (the first edit concerned the whole question) Could it be that the first inequality is more informative because it tells us that $g(x) \in L^2(\mathbb{R})$ when g(x) is not known?

• I think the inequality $|x|_{L^p} \le n^{\frac{1}{p}-\frac{1}{q}}.|x|_{L^q}$ is not valid for any $n$. It must be saying there exists $n$ s.t. ... . And in your case, $n$ is not equal to 1. – crbah Mar 22 '16 at 10:16

The inequalities you cited only hold for the sequence version of the $L^p$-spaces, which are typically called $\ell^p$. In fact, the last one only holds for $x \in \mathbb{R}^n$.
For your function space $L^p(\mathbb{R})$ these inequalities do not hold: the function which is constant $1$ belongs to $L^\infty(\mathbb{R})$, but not to $L^2(\mathbb{R})$; and for a function with a "suitable" singularity it might be the other way round.
• @mwater: No, I mean the inequalities $\|x\|_{L^q} \le \|x\|_{L^p}$ and $\|x\|_{L^p} \le n^{1/p-1/q} \, \|x\|_{L^q}$. These inequalities do not hold for $L^r(\mathbb{R})$! – gerw Mar 22 '16 at 15:33