Example for Sufficiency of Conditions for Differentiating Under Lebesgue Integral

Question

In the book I am reading*, sufficient conditions for differentiating under the Lebesgue integral are stated as follows:

$J\in \mathbb{R}$ is an interval and $\left(X,\mathcal{M},\mu\right)$ a measure space, and $f: X\times J \rightarrow [-\infty,\infty]$ s.t.

1) for each $t\in J$, $f(x,y)$ is a (Lebesgue) integrble function of $x$ on $X$,

2) for $\mu$-a.e. $x\in X$, $f(x,t)$ is differentiable for all $t\in J$,

3) and there exists a (Lebesgue) integrable function $h:X\rightarrow [-\infty,\infty]$ s.t. for $\mu$-a.e. $x\in X$ and for all $t\in J$, $$\left| \frac{df(x,t)}{dt}\right| \le h(x).$$

After proving the theorem, the author goes on to give the example of the function $g(x,t)=t^3e^{-t^2x}$ (for $x>0, -\infty <t<\infty$) which does not satisfy the conditions and indeed can not be differentiated under the integral sign because $$1=\frac{d}{dt}\left(\int_{0}^{\infty}g(x,t)d\lambda\right)(t=0)\neq \int_{0}^{\infty}\frac{dg}{dt}(x,t=0)d\lambda=0$$ where $\lambda$ is the Lebesgue Measure on $X=\mathbb{R}$.

Unfortunately I am struggling to understand exactly why $g$ does not satisfy the conditions stated, for instance for the interval $J=(-1,1)$.

Any help would be appreciated as I feel that I'm missing something both simple and important.

*The book is "Lebesgue Integratiton on Euclidian Space"/Frank Jones and the example mentioned appears on page 155.

Answer 1

The third condition is not satisfied. We can't find a dominating function that works for all $t$. There's no problem if $t$ stays away from $0$ (and $\pm \infty$), but if we allow $t$ to approach $0$, the smallest dominating function $$x \mapsto \sup_{t \in J} \:\biggl\lvert\frac{\partial g}{\partial t}(x,t)\biggr\rvert$$ is no longer integrable.

Computing $$\frac{\partial g}{\partial t}(x,t) = (3t^2 - 2t^4x)e^{-t^2x} = t^2(3-2t^2x)e^{-t^2x}$$ we see that $$\frac{\partial g}{\partial t}(x, \pm1/\sqrt{x}) = \frac{1}{ex}\,.$$ With $J = (-\varepsilon, \varepsilon)$, the parameter $1/\sqrt{x}$ lies in $J$ for $x > \varepsilon^{-2}$, so a dominating function must be $\geqslant c\cdot x^{-1}$ eventually, which means it cannot be integrable.

Since $\int_0^{\infty} g(x,t)\,d\lambda(x) = t$ for all $t$, we have $\int_0^{\infty} \frac{\partial g}{\partial t}(x,t)\,d\lambda(x) = 1$ for all $t \neq 0$ [leaving out the verification that for $t \neq 0$ differentiation under the integral is legitimate], but - in somewhat imprecise terms - as $t \to 0$ "the mass of $\frac{\partial g}{\partial t}$ escapes to $+\infty$". And the mass escaping implies that $\sup_t \bigl\lvert\frac{\partial g}{\partial t}\bigr\rvert$ decays too slowly to be integrable.

Answer 2

We want to prove that $g$ violates the third conditions, i.e. there is no function $h\in L^1(0,+\infty)$ such that $$\left|\frac{\partial g(x,t)}{\partial t}\right|\leq h(x),\qquad a.e.\, x\in X, t\in J $$ Where $J=(-a,a)$ is an open interval containing the origin.

Notice that such a function $h$, if it existed, must necessarily satisfy $$h(x)\geq \sup_{t\in J}\left|\frac{\partial g(x,t)}{\partial t}\right|,\qquad a.e.\,x\in X $$ So our goal is to find out the sup on the RHS.

Fix $x\in (0,+\infty)$ and let $$f(t)=\frac{\partial g(x,t)}{\partial t},\qquad t\in J $$

Let us start with some basic calculations \begin{align*}f(t)&=(-2t^4+3t^2)e^{-t^2x} \\ f'(t)&=(4t^5x^2-14t^3x+6t)e^{-t^2x}\\ \end{align*} The first derivative $f'(t)$ has always exactly three roots: $t_0=0$, $t_1=\frac{1}{\sqrt{2x}}$ and $t_2=\sqrt{\frac{3}{x}}$. Moreover $f'(t)>0$ in $(t_0,t_1)$, $f'(t)<0$ in $(t_1,t_2)$ and $f'(t)>0$ in $(t_2,+\infty)$. This shows that $t_1$ is the only local maximum in $(0,\infty)$. Notice that $t_1\leq a$ if $x\geq (2a^2)^{-1}$. Since $f(0)=0=\lim_{t \to +\infty}f(t)$, $t_1$ is the global maximum, and $$f(t_1)=\left(-\frac{1}{2x^2}+\frac{3}{2x}\right)e^{-1/2}\leq \sup_{t\in J}\left|\frac{\partial g(x,t)}{t}\right|\leq h(x) $$ at least when $x\geq (2a^2)^{-1}$. But since the function $$x\mapsto -\frac{1}{2x^2}+\frac{3}{2x} $$ is not integrable in $((2a^2)^{-1},+\infty)$, $h(x)$ cannot be integrable in $(0,+\infty)$.