Showing that an absolute integrable monotone decreasing function $f: [1,\infty[ \rightarrow \mathbb{R}$ is in $L^p([1,\infty[)$

Question

For an exercise in my analysis course, I have to show that: if $f: [1,\infty[ \rightarrow \mathbb{R}$ is monotone decreasing and $f \in L^1([1,\infty[)$, then $f \in L^p([1,\infty[)$ for every $p > 1$.

Define $f_+ = \max(f,0)$ and $f_{-} = \max(-f,0)$

I tried several things: My guess is that I have to use the integral of $|f|$ to provide an upper bound for the integral of $|f|^p$. I tried the following:

$|f| = f_+ + f_-$. Then $f_+$ is monotone decreasing and $f_-$ is monotone increasing. Moreover $|f|^p = f_+^p + f_{-}^p$ since $f^+$ and $f_{-}$ are never nonzero together. Hence $$\int_{[1,\infty[} |f|^p d\lambda = \int_{[1,\infty[} f_+^p d\lambda + \int_{[1,\infty[} f_+^p d\lambda$$

Now I want to try to get the monotonicity into play. Let $g_k = f_+ \cdot \chi_{[k,k+1[}$, then $f_+ = \sum^\infty_{k=0} g_k$. We can find a similar sequence for $f_{-}$, let's call that one $h_k$.

\begin{align*}\int_{[1,\infty[} f_+^p d\lambda + \int_{[1,\infty[} f_+^p d\lambda &= \int_{[1,\infty[}(\sum^\infty_{k=0} g_k)^p d\lambda+ \int_{[1,\infty[}(\sum^\infty_{k=0} h_k)^p d\lambda \\&= \int_{[1,\infty[} \sum^\infty_{k=0} g_k^p d\lambda+ \int_{[1,\infty[}\sum^\infty_{k=0}h_k^p d\lambda \\& = \sum^\infty_{k=0} \int_{[1,\infty[} g_k^p d\lambda+ \sum^\infty_{k=0}\int_{[1,\infty[}h_k^p d\lambda\end{align*} We are allowed to bring in the exponent because for every $x$ only one of them is nonzero.

\begin{align*} \sum^\infty_{k=0} \int_{[1,\infty[} g_k^p d\lambda+ \sum^\infty_{k=0}\int_{[1,\infty[}h_k^p d\lambda &\leq \sum^\infty_{k=0} \int_{[1,\infty[} f_{+}(k)^p\chi_{[k,k+1[} d\lambda+ \sum^\infty_{k=0}\int_{[1,\infty[}f_{-}(k+1)^p\chi_{[k,k+1[} d\lambda\\ &= \sum^\infty_{k=0}(f_{+}(k)^p + f_{-}(k+1)^p)\end{align*}

Now I am not sure how to proceed. Am I looking into the wrong direction? Please only provide hints!

Answer 1

First note that $f_- = 0$, as if $f_-(x^*) > 0$, then for each $x \ge x^*$, we have $f(x) \le f(x^*) < 0$, which implies that $f$ isn't integrable. Now choose $x^*$ such that for $x \ge x^*$ we have $f(x) \le 1$. Then \begin{align*} \int_1^\infty f(x)^p\, dx &= \int_1^{x^*} f(x)^p\, dx + \int_{x^*}^\infty f(x)^p\, dx\\ &\le \int_1^{x^*} f(1)^p\, dx + \int_{x^*}^\infty f(x)\, dx\\ &\le (x^*-1) \cdot f(1)^p + \int_1^\infty f(x) \, dx\\ &< \infty. \end{align*}

Answer 2

If your function is monotone and decreasing on $[1,\infty[$ then your function must be positive (if it's negative at any moment, it will not be integrable).

So I don't know (yet) the way to proceed, but I'm not sure you're on the right direction.

I hope this could help.