For every monotonically-decreasing non-negative function $ f $, does there exist a function $ g $ so that $ f g $ is integrable? [closed]

Question

Let $ f $ be a monotonically-decreasing non-negative function satisfying $ \displaystyle \lim_{x \to \infty} f(x) = 0 $. Is it true that the following claim holds?

Claim: There exists a function $ g $ such that $ \displaystyle \int_{0}^{\infty} g(x) ~ \mathrm{d}{x} = \infty $ and $ \displaystyle \int_{0}^{\infty} f(x) g(x) ~\mathrm{d}{x} < \infty $.

Note that we can assume w.l.o.g. that $ \displaystyle \int_{0}^{\infty} f(x) ~\mathrm{d}{x} = \infty $, as otherwise $ g \equiv 1 $ does the job. My idea for a proof was to split $ [0,\infty) $ into intervals $ I_{n} \stackrel{\text{df}}{=} [x_{n},x_{n + 1}) $, so that $ \displaystyle \int_{I_{n}} f(x) ~ \mathrm{d}{x} = \frac{1}{n} $, and define $ g $ by $ \displaystyle g(x) \stackrel{\text{df}}{=} \sum_{n = 0}^{\infty} \frac{1}{\operatorname{length}(I_{n}) \cdot n} \mathbf{1}_{[x_{n},x_{n + 1})} $, where $ \operatorname{length}(I_{n}) \stackrel{\text{df}}{=} x_{n + 1} - x_{n} $.

Unfortunately, this doesn’t seem to work as one can construct a piecewise-constant function $ f $ so that $ \operatorname{length}(I_{n}) \rightarrow \infty $, which seems to kill my original approach (which, roughly speaking, was to construct $ g $ so that $ \displaystyle \int_{0}^{\infty} f(x) g(x) ~ \mathrm{d}{x} \leq \sum_{n = 1}^{\infty} \frac{1}{n^2} $).

Answer 1

On the positive half of the $x$-axis, build mutually disjoint triangular "spikes" of base $1/n$ and height $2n$ $(n=1, 2, 3, ...)$ spreading them sufficiently far in the positive direction of the $x$-axis so that when multiplied by $f(x)$, the area of the $n$-th flattened spike becomes smaller than $2^{-n}$. This can be done since $f(x)$ converges to $0$ as $x\to\infty$. There is your $g(x)$.

Remark: I just edited this answer. The assumption of monotonicity of $f$ is redundant. The construction of $g$ does not require it. It suffices that $\displaystyle\lim_{x\to\infty}f(x)=0$.

Answer 2

The key is to choose $g$ so that its integral goes to infinity at a rate controlled by the decrease of $f$. If $f$ is differentiable, we can take $g(x) = - \frac{ df/dx}{f(x)}$ so $\int g dx = \int df/f = \infty$ and $\int gf dx = \int df \neq \infty$, and if not then the following slight modification works:

Let $g(x)= \log f(n-1) -\log f(n)$ for $x$ between $n$ and $n+1$ and let $g(x)=0$ for $x$ between $0$ and $1$. Then certainly the integral of $g$ is the telescoping sum $\log f(n-1) -\log f(n)$ which diverges. But

$$\int_0^\infty f(x) g(x) \leq \sum_{n=0}^\infty f(n) (\log { f(n-1)} - \log{ f(n)}) = \sum_{n=0}^\infty f(n) \int_{f(n-1)}^{f(n)} \frac{dt}{t} $$

$$\sum_{n=0}^\infty \int_{f(n-1)}^{f(n)} \frac{f(n) dt}{t} \leq \sum_{n=0}^\infty \int_{f(n-1)}^{f(n)} dt = f(0)$$.

Answer 3

We can even add the hypothesis that $f$ and $g$ must be continuous

We first observe as Will did that if $f$ is $C^{1}$ then $g=-\dfrac{f^{^{\prime }}}{f}$ solves the question. Indeed, we have $\int_{0}^{+\infty }g\left( x\right) dx=% \left[ -\ln f\right] _{0}^{+\infty }=+\infty $ and $\int_{0}^{+\infty }f\left( x\right) g\left( x\right) dx=\int_{0}^{+\infty }-f^{^{\prime }}\left( x\right) dx=f\left( 0\right) -f\left( +\infty \right) =f\left( 0\right) <+\infty $

If we manage to prove the existence of a $C^{1}$ function $\widetilde{f}$ such that $\forall x,$ $f\left( x\right) \leq \widetilde{f}\left( x\right) $ and $\widetilde{f}\left( x\right) \longrightarrow 0$ as $x\longrightarrow +\infty $ then the result follows easily by taking $g=-\dfrac{\widetilde{f}% ^{^{\prime }}}{\widetilde{f}},$ as before we have $\int_{0}^{+\infty }g\left( x\right) dx=\left[ -\ln \widetilde{f}\right] _{0}^{+\infty }=+\infty $ and $$ \int_{0}^{+\infty }f\left( x\right) g\left( x\right) dx\leq \int_{0}^{+\infty }\widetilde{f}\left( x\right) g\left( x\right) dx=\int_{0}^{+\infty }-\widetilde{f}^{^{\prime }}\left( x\right) dx<+\infty . $$

Let $\varphi _{n}$ a smooth kernel verifying $\forall t\in %TCIMACRO{\U{211d} }% %BeginExpansion \mathbb{R} %EndExpansion ,$ $\varphi _{n}\left( t\right) \geq 0,$ $\varphi _{n}\left( t\right) =0$ for $|t|\geq \frac{1}{n},$ and $\int_{-\infty }^{+\infty }\varphi _{n}\left( t\right) =1.$

We define $\tau f$ on $[1,+\infty \lbrack $by $\tau f\left( t\right) =f\left( t-1\right) $ $\forall t$ and we extend it to a continous function on $% %TCIMACRO{\U{211d} }% %BeginExpansion \mathbb{R} %EndExpansion $ null on $]-\infty ,0]$ that we continue to denote by $\tau f.$

let $f_{n}$ defined on $% %TCIMACRO{\U{211d} }% %BeginExpansion \mathbb{R} %EndExpansion $ by $$ f_{n}\left( x\right) =\left( \tau f\ast \varphi _{n}\right) \left( x\right) =\int_{-\infty }^{+\infty }\tau f\left( x-t\right) \varphi _{n}\left( t\right) $$

Then $f_{n}$ is $C^{1}$ and we have for $x\geq 1$ $$f_{n}\left( x\right) =\int_{-1/n}^{1/n}\tau f\left( x-t\right) \varphi _{n}\left( t\right) dt=\int_{x-1/n}^{x+1/n}\tau f\left( t\right) \varphi _{n}\left( x-t\right) dt=\int_{x-1/n}^{x+1/n}f\left( t-1\right) \varphi _{n}\left( x-t\right) dt $$

As $f$ is nonincreasing then $$ f\left( x+\frac{1}{n}-1\right) \int_{x-1/n}^{x+1/n}\varphi _{n}\left( x-t\right) dt\leq f_{n}\left( x\right) \leq f\left( x-\frac{1}{n}-1\right) \int_{x-1/n}^{x+1/n}\varphi _{n}\left( x-t\right) dt $$

ie $$ f\left( x+\frac{1}{n}-1\right) \leq f_{n}\left( x\right) \leq f\left( x-% \frac{1}{n}-1\right) $$

Let $\widetilde{f}\left( x\right) =f_{2}\left( x\right) $ then we have $% f\left( x\right) \leq f\left( x-\frac{1}{2}\right) \leq \widetilde{f}\left( x\right) \leq f\left( x-\frac{3}{2}\right) $$ . \widetilde{f}$ is positive and $\forall x\geq 2,$ $\widetilde{f}\left( x\right) \leq f\left( x-2\right) $ hence $\widetilde{f}\left( x\right) \longrightarrow 0$ as $x\longrightarrow +\infty $ and we have $\forall x,$ $% f\left( x\right) \leq \widetilde{f}\left( x\right) .$