Let $f\colon\mathbb R\to\mathbb R$ be a non-negative and measurable function, and assume that both $$\int_{\mathbb R} f(t)dt<\infty\ \ \text{and}\ \ \int_{\mathbb R}e^tf(t)<\infty.$$

Show that the integral $G(x)=\int_{\mathbb R} e^{tx}f(t)dt$ is finite when $0\le x\le 1$. Then prove that the funtion $G(x)$ is continuous on $0\le x\le 1$, and differentiable on $0<x<1$.

My attempt:

It is trivial to show that $G(x)<\infty$ when $0\le x\le 1$. To show that $G(x)$ is continuous on $0\le x\le 1$, we need to consider the difference: \begin{align} G(x_2)-G(x_1)&=\int_{\mathbb R}e^{tx_2}f(t)dt-\int_{\mathbb R}e^{tx_1}f(t)dt\\ &=\int_{\mathbb R}(e^{tx_2}-e^{tx_1})f(t)dt \end{align} where $x_1,x_2\in [0,1]$.

Note that $$ |(e^{tx_2}-e^{tx_1})f(t)|\le\max\{2f(t), 2e^tf(t),f(t)+e^tf(t)\}\in L^1(\mathbb R), $$ it follows that $$ \lim_{x_2\to x_1} [G(x_2)-G(x_1)]=\int_{\mathbb R}0\cdot f(t)dt=0 $$ i.e., $G(x)$ is continuous on $0\le x\le 1$.

Next, we study the differentiability of $G(x)$ on $(0,1)$.

We have \begin{align} \frac{G(x_2)-G(x_1)}{x_2-x_1}=\int_{\mathbb R}\frac{e^{tx_2}-e^{tx_1}}{x_2-x_1}f(t)dt \end{align} and $$ \frac{e^{tx_2}-e^{tx_1}}{x_2-x_1}f(t)=\frac{e^{tx_2}-e^{tx_1}}{tx_2-tx_1}tf(t) .$$ If we let $x_2\to x_1$, then $$\lim_{x_2\to x_1}\frac{e^{tx_2}-e^{tx_1}}{x_2-x_1}f(t)=e^{tx_1}tf(t).$$ But this time, I cannot find a dominating integrable function $g(t)$ such that $|e^{tx_1}tf(t)|<g(t)$ on $\mathbb R$. Then how to prove the differentiability of $G(x)$ on $(0,1)$?


1 Answer 1


Try to use Lebesgue dominated convergence theorem instead of Leibniz's rule directly.

Here $x_1$ is fixed in $(0,1)$, and your bounding function "$g(t)$" can actually depend on $x_1$, (this is different from the statement of Leibniz's rule).

Let us use $x, x+h$ instead of $x_1, x_2$, what you are trying to show is for a fixed $x\in (0,1)$, we have $$\lim_{h\rightarrow 0} \int \frac{e^{(x+h)t} - e^{xt}}{h} f(t) dt = \int \lim_{h\rightarrow 0} \frac{e^{(x+h)t} - e^{xt}}{h} f(t) dt.$$ So we need to find a bounding function for $\frac{e^{(x+h)t} - e^{xt}}{h} f(t)$ that is independent of $h$, for all $h$ small.

For $t>0$ and $h>0$, by mean value theorem, $$\frac{e^{(x+h)t} - e^{xt}}{h} = te^{ct}$$ for some $c\in (x, x+h)$. From monotonicity of the exponential function $$te^{ct} \leq te^{(x+h)t}\leq te^{(x+h_0)t}$$ where $h_0$ is a fixed small positive number such that $x+h_0 < 1$.

For $t>0, h<0$, we have $$\frac{e^{(x+h)t} - e^{xt}}{h} \leq te^{xt}\leq te^{(x+h_0)t}.$$ So our bounding function on $t>0$ would be $$te^{(x+h_0)t}f(t) = (e^{t}f(t)) \frac{t}{e^{(1-(x+h_0))t}} \in L^1$$ which works for all $|h| \leq h_0$.

For $t<0$, we can apply the same argument, to obtain a bounding function.

  • $\begingroup$ I figured out actually we can directly bound $e^{tx_1}tf(t)$ by $e^{t}f(t)$ when $t>0$ and $f(t)$ when $t<-M$. hhhh $\endgroup$
    – Bach
    Jul 23, 2019 at 6:56
  • $\begingroup$ Yes, the upper bound can be done with a direct bound. When $t < -M$ note $x_1$ can be arbitrarily close to zero, so I am not sure how you would get $|e^{x_1t} t f(t)|\leq |f(t)|$ since $|e^{x_1t} t| $ might be large. (Unless your $M$ depends on $x_1$.) $\endgroup$
    – Xiao
    Jul 23, 2019 at 11:04
  • $\begingroup$ Basically the same method as yours. I mean bound the function separately when $t>0$ and $t<0$. $\endgroup$
    – Bach
    Jul 23, 2019 at 11:06

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