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Let $\phi,\psi \in L^{p}([0,1],m)$ where $1\leq p < \infty$ and $m$ is the Lebesgue measure. Also, $||\phi||_{0}\dot{=}\sup_{x}|\phi(x)|$ and $||\phi||_{p}\dot{=}[\int_{0}^{1}|\phi(x)|^pdm]^{1/p}$. It`d be useful for me to work a proof out if:

$$||\phi-\psi||_{0}<\epsilon \Rightarrow ||\phi-\psi||_{p}<\epsilon$$ i.e. the sup norm wins. For $p=1$ it`s pretty straightforward but I couldn't work out the inequalities for $p>1$. Am I missing something quite clear? Also, sorry if it's a duplicate, I did a little research but maybe I just didn't use the right keywords.

Thanks in advance.

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  • $\begingroup$ Can you write down your proof for $p=1$ ? The general case shouldn't be too different. $\endgroup$
    – tristan
    Commented Jan 8, 2018 at 20:16
  • $\begingroup$ for $p=1, \int_{0}^{1}|\phi(x)-\psi(x)|dm \leq$ (interval length) $ \sup_{x}|\phi(x)-\psi(x)| $ Since the interval length is $1$, we have estabilished the inequality. $\endgroup$
    – Janov
    Commented Jan 8, 2018 at 20:21
  • $\begingroup$ In the same fashion, for $p>1$, $\int_0^1 \lvert \phi(x)-\psi(x) \rvert^p \mathrm{d}m(x) \leq \int_0^1 \lVert \phi-\psi \rVert^p \mathrm{d}m(x)$. $\endgroup$
    – tristan
    Commented Jan 8, 2018 at 20:23
  • $\begingroup$ Isn`t there a typo? $\endgroup$
    – Janov
    Commented Jan 8, 2018 at 20:29
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    $\begingroup$ Oh yes my bad, it should be $\int_0^1 \lvert \phi(x)-\psi(x)\rvert^p \,\mathrm{d}m(x) \leq \int_0^1 \lVert \phi-\psi \rVert_0^p \,\mathrm{d}m(x)$. $\endgroup$
    – tristan
    Commented Jan 8, 2018 at 20:44

1 Answer 1

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The standard notation for your $\|\cdot\|_0$ is $\|\cdot\|_\infty$, so I will use the latter.

You have $$ \|\phi-\psi\|_p^p=\int_0^1|\phi-\psi|^p\,dm\leq\int_0^1\|\phi-\psi\|_\infty^p\,dm\leq\|\phi-\psi\|_\infty^p\,\int_0^11\,dm=\|\phi-\psi\|_\infty^p. $$ Thus $$\|\phi-\psi\|_p\leq\|\phi-\psi\|_\infty.$$ The inequality is tight, as seen by taking constant functions.

Note that if you have an interval of different length, a constant would appear. And the inequality does not hold for $L^p(\mathbb R)$.

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  • $\begingroup$ Thanks! About the notation, in the proof I'm working on there's still $L^{\infty}$ involved so, despite I knew the standard notation for the sup norm is $||.||_{\infty}$, I thought it was more convenient to call the norm on $L^{\infty}$ by that notation. How does one refers to $L^{\infty}$ norm usually? $\endgroup$
    – Janov
    Commented Jan 8, 2018 at 22:41
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    $\begingroup$ Exactly by $\|\cdot\|_\infty$. You are right that there is a difference, with the sup not being essential. But the norm you call $\|c\dot\|_0$ is only used on continuous functions, where the sup agrees with the essential sup. The argument above works with the essential supremum norm, which is always below your $\|\cdot\|_0$. $\endgroup$ Commented Jan 8, 2018 at 22:47

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