Criteria to be in weak $L^{p}$ space

Question

Let $X$ be a $\sigma$-finite measure space. Let $f : X \rightarrow \mathbb{C}$ be a measurable function and $1 < p < \infty$. Suppose for $f$ there is a constant $C$ such that $|\int_{X}f\chi_{E}\, d\mu| \leq C\mu(E)^{1 - 1/p}$ for all sets $E$ of finite measure. Why must $f \in L^{p, \infty}(X)$?

Answer 1

EDIT: the answer is actually wrong as noticed Xiang, since $p'$ as defined here is negative and $(p,p')$ can't be used to apply Hölder's inequality. I misread the bounds on p. Please consider his answer below

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It looks like a good candidate for Hölder inequality, with $\frac{1}{p}$ and $p'$ such that $p + \frac{1}{p'}=1$, so $p'=\frac{1}{1-p}$ Let first assume that $f$ is positive and real-valued :

$\Gamma(E) = \int_X |f|^p 1_E d\mu \leq (\int_X(|f|^p)^{\frac{1}{p}} d\mu)^p(\int_X 1_E^{p'} d\mu)^{\frac{1}{p'}} = (\int_X|f| d\mu)^p(\int_X 1_E)^{1-p} $

With the hypothesis, and since $f$ is positive so $| \int \ldots | = \int \ldots $, we can derive : $\Gamma(E) \leq C^p\mu(E)^{p-1}\mu(E)^{1-p}=C^p$

So we just find that $\Gamma(E)$, the integral of $f$ over a finite-measure set $E$, is bounded independently from $E$. Since $X$ is $\sigma$-finite, you can find an increasing sequence of such $E$s that converges to $X$ and applying Monotonuous convergence theorem for positive functions $|f|1_E$ shows that $f$ is $L^p$ and $||f||_p \leq C$.

The previous proof works obviously for $f$ real negative, or $f=ig$ is imaginary valued with $g$ positive or negative (because, in each cases, $| \int \ldots |$ can be simply expressed wrt $\int \ldots$. For general case, I think you can decompose $f$ as $f=f1_{\{f \in \mathbb{R^+}\}} + f1_{\{f \in \mathbb{R^-}\}} + f1_{\{f \in i\mathbb{R^+}\}} + f1_{\{f \in i\mathbb{R^+}\}}$, see that the inequality hypothesis is still true for each subfunctions, apply the previous proof and thus deduce that $f$ is $L^p$ as a sum of $L^p$ elements.

Answer 2

By decomposing $f$ into real part and imaginary part and notice that $$|\int_X \mathrm{Re}(f)1_E\,d\mu|\leq |\int_Xf1_E\,d\mu|,$$ we may assume that $f$ is real. By decomposing $f$ into positive part and negative part and notice that $$|\int_X f^+1_E\,d\mu|=|\int_X f1_{f\geq 0}1_E|\leq C\mu(\{f\geq 0\}\cap E)^{1-1/p}\leq C\mu(E)^{1-1/p},$$ we may assume that $f$ is non-negative. Since $X$ is $\sigma$-finite, then we can write a set as a union of an increasing sequence of sets with finite measure, thus we see from the monotone convergence theorem that $$\int_X f1_E\,d\mu\leq C\mu(E)^{1-1/p}$$ for every measurable set $E$.

Now let $E=\{x\in X:f(x)\geq t\}$, from the above inequality, we see that $$t\int_X 1_{f\geq t}\,d\mu\leq \int_X f1_{f\geq t}\,d\mu\leq C\mu(E)^{1-1/p}.$$ Notice that $\mu(E)=\int_X 1_{f\geq t}\,d\mu$, thus we have $$t\mu(E)^{1/p}=t\lambda_f^{1/p}(t)\leq C,$$ where $\lambda_f(t):=\mu(E)$ is the distribution function of $f$. By the definiton of weak $L^{p,\infty}$-norm (i.e. $\|f\|_{L^{p,\infty}}:=\sup_{t>0}t\lambda_f(t)^{1/p}$), we see that $\|f\|_{L^{p,\infty}}\leq C$, the claim then follows.