Write an $L^p$ functions as a sum of an $L^2$ and $L^{\infty}$ functions

Question

It's written in a article that:

If $f\in L^p$, $p>2$, then $f=f_1+f_2$, where $f_1\in L^2$, $f_2\in L^\infty$, and $||f_1||_{L^2}\leq 2||f||_{L^p}$, $||f_2||_{L^\infty}\leq 2||f||_{L^p}$.

But I don't know how to prove it. Could I ask for some directions for proving it, or if it's really common, a reference for this theorem? Thanks a lot.

Answer 1

Actually you don't even need the factor of $2$ there.

Indeed, take $f_1=f \cdot 1_{\{x:\;|f(x)|> \| f\|_p\}}$ and $f_2=f \cdot 1_{\{x:\;|f(x)| \leq \|f\|_p\}}$.

Clearly, $\|f_2\|_{\infty} \leq \|f\|_p$ by definition.

On the other hand, if $a,b$ are positive real numbers with $a>b$ then $a/b>1$ so (since $p>2$) we see that $(a/b)^p>(a/b)^2$ which means that $a^2 < b^{2-p}a^p$. Thus if $|f(x)|>\|f\|_p$ then applying this inequality with $a=|f(x)|$ and $b=\|f\|_p$ we see that $|f(x)|^2 < \|f\|_p^{2-p} |f(x)|^p$. Hence $$\|f_1\|_2^2 = \int |f_1|^2 \; d\mu \leq \int \|f\|_p^{2-p} |f|^p \; d\mu = \|f\|_p^2$$ which implies that $\|f_1\|_2 \leq \|f\|_p$.

Answer 2

Suppose that $f\in L^p$ for some $p\in(2,\infty)$. (If $p=\infty$, simply take $f_1=0$ and $f_2=f$.) The case in which $\|f\|_p=0$ is trivial (take $f_1=f_2=0$), so suppose that $\|f\|_p>0$. Let $$A\equiv\{x\in X\,|\,|f(x)|>2\|f\|_p\}$$ and define $f_1\equiv f\times\mathsf I_A$ and $f_2\equiv f\times \mathsf I_{A^\mathsf c}$. Clearly, $f=f_1+f_2$.

Moreover, $$|f_2|=|f|\times\mathsf I_{A^\mathsf c}\leq 2\|f\|_p$$ pointwise, so that $\|f_2\|_{\infty}\leq 2\|f\|_p$.

Furthermore, one has that (remember that $p>2$, so no negative exponents are involved) $$|f|^p=|f|^{p-2}\times|f|^2\geq|f|^{p-2}\times|f|^2\times \mathsf I_{A}\geq 2^{p-2}\times\|f\|_p^{p-2}\times|f|^2\times \mathsf I_{A}=2^{p-2}\times\|f\|_p^{p-2}\times|f_1|^2.$$ Integrating both sides of this inequality yields: $$\|f\|_p^p\geq2^{p-2}\times\|f\|_p^{p-2}\times \|f_1\|_2^2.$$ Rearrange (division is possible since $\|f\|_p>0$) to obtain $$\|f_1\|_2^2\leq\frac{1}{2^{p-2}}\times \|f\|_p^2,$$ or, taking square roots, $$\|f_1\|_2\leq\frac{1}{2^{p/2-1}}\times \|f\|_p\leq2\|f\|_p.$$ The last inequality is implied by the fact that $p>2$.