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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?

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

2 Answers 2

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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

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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. $$ \|\Le_k\|_{L_\infty} = \sqrt {k+1/2}\|\Le_k\|_{L_2}.$$

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  • $\begingroup$ 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. $\endgroup$
    – Jochen
    Jan 29, 2022 at 15:35
  • $\begingroup$ @Jochen yes thats true and I should have added that somewhere. $\endgroup$ Jan 29, 2022 at 15:48

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