# Norm equivalences of $L_2$ and $L_\infty$ spaces of measurable function

For the $$l_p$$ spaces of sequence $$\{v_i\}_{i=1,...,n}$$ we have the following norm equivalences $$\|v\|_{l_\infty} \le \|v\|_{l_2} \le \sqrt{n} \|v\|_{l_\infty}.$$

For the $$L_p$$ spaces of measurable function $$f: \Omega \to \mathbb{R}$$ I think there still holds $$\|f\|_{L_2(\Omega)} \le |\Omega|^{\frac12} \|f\|_{L_\infty(\Omega)}$$ but do we have analogously to the $$l_p$$ norms $$\|f\|_{L_\infty(\Omega)} \le \|f\|_{L_2(\Omega)}$$ and why it holds?

Edit: I think the last inequality doesn't hold in general, but for finite element functions, say polynomials of degree $$k$$, $$f \in P^k(\Omega)$$, we do have the so-called inverse inequality $$\|f\|_{L_\infty(\Omega)} \le C |\Omega|^{-\frac12} \|f\|_{L_2(\Omega)}.$$ Am I right?

• $$\|f\|_{L_\infty(\Omega)} \le M\|f\|_{L_2(\Omega)}$$ is very very false. Nov 14, 2021 at 23:18

## 2 Answers

In general we can't even find a constant $$C$$ such that $$||f||_{\infty} \le C ||f||_2$$ for all $$f$$. As a counterexample, for any $$n \in \mathbb{N}$$ take $$f_n(x) = 1_{[-1/n^2,1/n^2]} \cdot n$$ Then $$||f_n||_{\infty} = n$$ but $$||f_n||_2 = \sqrt{2}$$

Edit I:

Even if you now restrict yourself to arbitrary polynomials or smooth functions, this does not work.

Consider $$\Omega = [-1,1]$$ and choose a suitable non-negative bump function $$g_n$$ that vanishes outside of $$(-1/n^2,1/n^2)$$, has its maximum at $$g_n(0) = n$$ and satisfies $$||g_n||^2_2 \le \frac{1}{n^2} \cdot n^2 = 1$$.

Again, this shows that we can't bound $$||g_n||_{\infty} = n \le C ||g_n||_2$$

The intuitive reason is that we can force smooth functions to have a spike going up to $$n$$ on a very small interval, which is so small that the spike does not influence the $$L^2$$-norm.

Note:

What you can do (for example) is bound the $$L^2$$-norm in terms of the Sobolev norm. If you consider functions vanishing at the boundary of your domain (which I assume is sufficiently "nice") then you can bound $$||f||_2$$ in terms of $$||Df||_2$$ for all appropriate Sobolev functions $$f$$

Edit II:

If you change it to polynomials of some fixed degree it works. See the other answer

For a fixed $$k$$ (and bounded $$\Omega$$), for polynomials $$f\in P^k(\Omega)$$ of degree at most $$k$$, we indeed have $$\|f\|_{L_\infty(\Omega)} \le C \|f\|_{L_2(\Omega)}\newcommand{\Le}{\text{Le}}$$ is correct, but $$C$$ will depend badly on $$k$$.

Specifically for the polynomials over say, $$\Omega = [-1,1]$$, any $$p\in P^k$$ can be written in terms of the Legendre polynomials $$\Le_i$$,

$$p = \sum_{i=0}^k p_i \Le_i,\quad p_0,\dots,p_k\in\mathbb R$$ Since any Legendre polynomial over $$[−1,1]$$ is bounded by one in absolute value, we have $$\|p\|_{L_\infty([-1,1])} \le \sum_{i=0}^k |p_i| = \sum_{i=0}^k |p_i|\le \sqrt{k}\sqrt{\sum_{i=0}^k|p_i|^2 }$$ On the other hand, by the orthogonality relation $$\int_{-1}^1\Le_i\Le_j dx = \frac{\delta_{ij}}{i+1/2}$$, $$\|\sum_{i=0}^k p_i \Le_i\|^2_{L_2} = \sum_{i=0}^k \frac{|p_i|^2}{i+1/2} \ge \frac1{k+1/2} \sum_{i=0}^k |p_i|^2\ge \frac1{k(k+1/2)}\|p\|_{L_\infty}^2$$

so $$\|p\|_{L_\infty} \le (k+1/2) \|p\|_{L^2}$$. And you can't improve this bound too much, because if we set $$p=\Le_k$$ then we find $$\|\Le_k\|_{L_\infty}=1, \|\Le_k\|_{L_2} = \frac1{\sqrt{k+1/2}}$$ i.e. $$\|\Le_k\|_{L_\infty} = \sqrt {k+1/2}\|\Le_k\|_{L_2}.$$

• If you don't care about the constants you can just use that any two norms on the finite dimensional space of polynomials of degree $\le k$ are equivalent. Jan 29, 2022 at 15:35
• @Jochen yes thats true and I should have added that somewhere. Jan 29, 2022 at 15:48