We consider $1 \leq p < q < +\infty$.

How we can show that $L^{\infty}([0,1]) \subset L^q([0,1]) \subset L^p([0,1])$.

I really don't know how to prove these inclusions. I just know that $m([0,1])=1$ for Lebesgue measure, but it doesn't really help me.

Someone could help me ? Thank you in advance.

  • $\begingroup$ Do you know Holder's Inequality? $\endgroup$ Mar 8, 2018 at 21:37
  • $\begingroup$ Ah, yes ! If $f \in L^p$ and $g \in L^q$, $fg \in L^1$ and $||fg||_1 \leq ||f||_p ||g||_q$ with $1/p + 1/q =1$. But I don't see how to use it in order to prove these inclusions :/ $\endgroup$ Mar 8, 2018 at 21:46
  • 1
    $\begingroup$ For the first inclusion, you know that if a function is bounded so is it's q-th power, and since your over a set of finite measure, what can you say about it's $L^q$ norm? $\endgroup$ Mar 8, 2018 at 22:06

2 Answers 2



Using Hölder inequality

$$\|f\|_p^p= \int_0^1 |f|^p .1 d\mu \leq \left( \int_0^1 |f|^{p \frac{q}{p}}d\mu \right)^\frac{p}{q} \mu([0,1])^{1- \frac{p}{q}}= \left( \int_0^1 |f|^q d\mu \right)^\frac{p}{q}=\|f\|_q^p$$

  • $\begingroup$ More of a complete answer, then a hint $\endgroup$ Mar 8, 2018 at 22:17

No doubt there is some really clever Hoelder inequality trick to do this, but I can't remember it. On the other hand, a little common sense will do it. Let $f \in L^q$ and let $A = \{x : |f(x)| \le 1\}$, and $A' = [0,1]\setminus A$.
$$ \int |f|^p = \int_A |f|^p + \int_{A'} |f|^p \le \mu(A) + \int_{A'} |f|^q \le 1 + ||f||_q^q, $$ where the first inequality uses that $|f(x)| \le 1$ for $x \in A$ and $|f(x)|^p \le |f(x)|^q$ for $x \in A'$, since $p < q$.


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