# function is right-differentiable at zero iff certain integral finite

Suppose $f$ is nonnegative and integrable and $\int_\mathbb{R} f(x) \; dx < \infty$. For $t \ge 0$, define $$g(t) := \int_\mathbb{R} e^{-tx^4 \sin \left( \frac{1}{1+x^2} \right)} f(x) \; dx$$ I need to show (a) $g$ is continuous for all $t \in [0,\infty)$ and (b) $g$ is right-differentiable at $t=0 \iff \int_\mathbb{R} x^2 f(x) \; dx < \infty$.

For (a), I fix $t_0 \in [0,\infty)$. Then, since $|g| < \int f < \infty$, by LDCT we get $$\lim \limits_{t \to t_0} \int_\mathbb{R} e^{-tx^4 \sin \left( \frac{1}{1+x^2} \right)} f(x) \; dx = \int_\mathbb{R} e^{-t_0x^4 \sin \left( \frac{1}{1+x^2} \right)} f(x) \; dx = g(t_0).$$ so $g$ is continuous. But I'm pretty stuck on (b)... this isn't for coursework I don't mind if you give a full answer. Thank you.

## 1 Answer

Note first that we have $0\leq 1-\exp(-u)\leq u$ and $(u+1)\exp(-u)-1\leq 0$ for $u\geq 0$. Some hints:

a) Write for $t>0$ $G(t)=\frac{g(0)-g(t)}{t}=\int_{\mathbb{R}}h(x,t)dx$. Show that $$|h(x,t)|\leq x^4\sin(\frac{1}{1+x²})f(x)\leq x^2f(x)$$

b) Suppose that $g$ has a derivative at $0$. If $v>0$, show that $\displaystyle \frac{1-\exp(-vt)}{t}$ is decreasing. Hence, as $t$ decrease $\to 0$, $h(x,t)$ is increasing for fixed $x$. Now you can use the Monotone Convergence Theorem.

• This is very helpful, thank you. The first inequalities regarding exponentials were quite useful in particular. How did you recognize/prove them so quickly? – user288742 Jun 25 '16 at 3:41