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Suppose $g\in L^1([0,\infty))$ and $f$ is measurable. If there exists $M:=\lim_{n\to \infty } \int _0 ^n f(x)dx$, then $$\lim_{c\to0}\int _0 ^\infty f(x)g(cx)dx = Mg(0) $$

In order to use dominated convergence theorem, I tried to find $F \in L^1([0,\infty))$ s.t., $$\lvert f(x)(g(cx)-g(0)\rvert=\left\lvert f(x)\int _0 ^{cx} g'(t)dt\right\rvert\leq F.$$ But I couldn't find it.

davyjones
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B.T.M
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2 Answers2

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This is just a partial answer, since it is not clear that $f$ is integrable, see e.g. Functions that are Riemann integrable but not Lebesgue integrable. Hence it is not even clear that $f(x)g(cx)$ is integrable.

But if we additionally assume $f\in L^1([0,\infty))$ or equivalently $\int_0^\infty|f(x)|\,dx<\infty$, then we can do the following: $$|f(x)(g(cx)-g(0))|=|f(x)|\cdot\left\lvert\int_0^{cx}g'(t)\, dt\right\rvert\leq |f(x)|\cdot\int_0^{cx}|g'(t)|\, dt$$ $$\leq |f(x)|\cdot\int_0^{\infty}|g'(t)|\, dt=:F(x),$$ which is integrable because $f$ is integrable and well defined since $g'\in L^1$. Then you can employ the dominated convergence theorem.

davyjones
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humanStampedist
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  • For the first paragraph, notice that $f \in L^{\infty}$ and $g(c \cdot) \in L^1$, hence their product is also in $L^1$. – Sangchul Lee Apr 06 '18 at 14:36
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I will assume the followings on $f$ and $g$:

  1. $f$ is locally integrable on $[0, \infty)$ and $F(x) = \int_{0}^{x} f(t) \, dt$ converges to some $M \in \mathbb{R}$ as $x\to\infty$.

  2. There exists $g' \in L^1([0,\infty))$ such that $g(x) = g(0) + \int_{0}^{x} g'(t) \, dt$ for all $x \in [0,\infty)$.

(Both conditions follow from OP's assumption.) In particular, the second assumption tells that the limit $g(\infty) := \lim_{x\to\infty} g(x)$ exists. Then for each $c > 0$ and $R > 0$, integration by parts yields

\begin{align*} \int_{0}^{R} f(x) g(cx) \, dx &= \left[ F(x) g(cx) \right]_{0}^{R} - c \int_{0}^{R} F(x) g'(cx) \, dx \\ &= F(R) g(cR) - \int_{0}^{cR} F(x/c) g'(x) \, dx. \end{align*}

In particular, for each fixed $c > 0$ the above quantity converges as $R \to \infty$ to

$$ \lim_{R\to\infty} \int_{0}^{R} f(x) g(cx) \, dx = Mg(\infty) - \int_{0}^{\infty} F(x/c)g'(x) \, dx $$

where we utilized the fact that $F(x/c)g'(x)$ is dominated by the integrable function $\|F\|_{\infty}|g'(x)|$. Using this fact again, now by letting $c \to 0^+$, we have

$$ \lim_{c\to 0^+}\lim_{R\to\infty} \int_{0}^{R} f(x) g(cx) \, dx = Mg(\infty) - \int_{0}^{\infty} M g'(x) \, dx = Mg(0). \tag{*}$$

Remark. Using OP's assumption that $f$ is bounded and $g$ is integrable, we find that $f(x)g(cx)$ is also integrable and hence $\text{(*)}$ reduces to the desired statement

$$ \lim_{c\to 0^+} \int_{0}^{\infty} f(x) g(cx) \, dx = Mg(0).$$

Sangchul Lee
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