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Let $1\le p_1 < p < p_2 \le \infty$ and $ f\in L^p(\mathbb R)$. Prove that there exist $f_1 \in L^{p_1} (\mathbb R)$ and $f_2 \in L^{p_2}(\mathbb R)$ such that $f = f_1 + f_2$.

I tried to make one of the function bounded but it leads nowhere. It seems to me I need to find function from $L^{p_2}$ first. Thank you in advance for any help.

treskov
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  • You do want one of the functions bounded, only the bound cannot be arbitrary for the problem to work nicely. – Dunham Nov 28 '18 at 23:20

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Let $D_1$ be the set where $|f|>1$ and $D_2$ be the set where $|f|\leq 1$. Let $f_1$ be $f$ restricted to $D_1$ and $f_2$ be $f$ restricted to $D_2$. Both functions are 0 outside of the defined domains. \begin{align*} \int_{\mathbb{R}}|f_1|^{p_1} + \int_{\mathbb{R}}|f_2|^{p_2} &= \int_{D_1}|f_1|^{p_1} + \int_{D_2}|f_2|^{p_2}\\ &\leq \int_{D_1}|f_1|^{p} + \int_{D_2}|f_2|^{p}\\ &= \int_{\mathbb{R}}|f|^{p} \end{align*}

Dunham
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  • You have proven that $f_i\in L^{p_i}$ by showing that $$\int_{\mathbb{R}}|f_1|^{p_1} + \int_{\mathbb{R}}|f_2|^{p_2} \leq\int_{\mathbb{R}}|f|^{p}$$Of course, here it is assumed that $p_2<\infty$. But if $p_2=\infty$, then $f_2\in L^{p_2}$ is trivial (since $|f_2|$ is bounded by $1$), so this is not a problem. – Filippo Feb 11 '22 at 18:09
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Let $A=f^{-1} ([-1,1]) , B=A=f^{-1} ((-\infty,-1)\cup (1,\infty ))$. Take $$f_1 (x)=\begin{cases} f(x) \mbox{ for } x\in B \\ 0 \mbox{ for } x\notin B\end{cases}$$

$$f_2 (x)=\begin{cases} f(x) \mbox{ for } x\in A \\ 0 \mbox{ for } x\notin A\end{cases}$$