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On a finite measure space, prove that $\Vert f \Vert_p \leq \mu(E)^{1-\frac 1p} \Vert f \Vert_\infty$ for any $f \in \mathcal{L}^\infty(E)$ and $\mu(E) < \infty$, $1\leq p < \infty$.

I managed to prove the following statement for finite measure spaces for $1 \leq p < r < \infty$, $f \in \mathcal{L}^r(E) \Rightarrow f \in \mathcal{L}^p(E)$ and that $\Vert f \Vert_p \leq \mu(E)^{\frac 1p - \frac 1r} \Vert f \Vert_r$ where $\mu(E) < \infty$. (Used Holders)

I wanted to know how if I can use this result to prove the given statement.

Any help is appreciated.

user62089
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    There is something wrong above for $p=1$ if $\mu E >1$. Take $f(x) = 1$, then $|f|1 = \mu E$, $|f|\infty = 1$. – copper.hat Mar 18 '13 at 16:44

1 Answers1

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As copper.hat already pointed out, the inequality $\|f\|_p \leq \mu(E)^{1-\frac{1}{p}} \cdot \|f\|_{\infty}$ does not hold for all $p \geq 1$. Instead, it should read $$\|f\|_p \leq \mu(E)^{\frac{1}{p}} \cdot \|f\|_{\infty}$$

This follows from $$\|f\|_p \leq \mu(E)^{\frac{1}{p}-\frac{1}{r}} \cdot \|f\|_r \qquad (r>p)$$by letting $r \to \infty$, using that $$\lim_{r \to \infty} \|f\|_r = \|f\|_{\infty}$$ (see for a proof here).

Edit: Actually, there is a rather quick (and direct) proof of the inequality: $$\begin{align} \|f\|_p^p &= \int_E \underbrace{|f|^p}_{\stackrel{p \geq 1}{\leq} \|f\|_{\infty}^p} \, d\mu \leq \mu(E) \cdot \|f\|_{\infty}^p \\ \Rightarrow \|f\|_p &\leq \mu(E)^{\frac{1}{p}} \cdot \|f\|_{\infty} \end{align}$$

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saz
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