# Differentiability of $G(x)=\int_{\mathbb R} e^{tx}f(t)dt$ on $(0,1)$

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.

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 $$|e^{tx_1}tf(t)| on $$\mathbb R$$. Then how to prove the differentiability of $$G(x)$$ on $$(0,1)$$?

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.

• 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
– Bach
Commented Jul 23, 2019 at 6:56
• 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 $x_1$.)
– Xiao
Commented Jul 23, 2019 at 11:04
• Basically the same method as yours. I mean bound the function separately when $t>0$ and $t<0$.
– Bach
Commented Jul 23, 2019 at 11:06