Differentiability of an (improper) integral

Question

let $f: \mathbb{R}^{m+1} \to \mathbb{R}$ be $C^1$. For $x \in \mathbb{R}$, define $h_x(y) = f(x,y)$ and suppose $h$ is absolutely integrable in $\mathbb{R}^m$. Define $$F(x) = \int_{\mathbb{R}^m} h_x(y)dy$$ Further suppose that there exists an absolutely integrable function $g:\mathbb{R}^{m} \to \mathbb{R}$ satisfying $$|D_1 f(x,y)| \le g(y)$$ for all $(x,y) \in \mathbb{R}^{m+1}$. Show that $F$ is differentiable on $\mathbb{R}$

I keep running into circular arguments when trying to prove this and would like some help.

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It can be shown that $D_1 f(x,y)$ is absolutely integrable over $\mathbb{R}^m$ for fixed $x \in \mathbb{R}$, so $\forall \epsilon > 0$, there exists an approximating sequence $A_n$ (compact/contented sets) to $\mathbb{R}^m$, and $N_1 \in \mathbb{N}$ such that $n>N_1$ implies:

$$\left\lvert \int_{\mathbb{R^m}} D_1 f(x,y)dy - \int_{A_n} D_1 f(x,y)dy \right\rvert < \frac{\epsilon}{4}$$

$A_n$ is compact/contented, so $D_1f(x,y)$ is uniformly continuous on $A_n$, and so $\exists \eta > 0$ such that $\delta < \eta$ implies $$|D_1 f(x,y) - D_1 f(x + \delta,y)| < \frac{\epsilon}{2 v(A_n)}$$

Let $\delta < \eta$. By MVT $\exists \tau$ on the line connecting $(x,y)$ to $(x + \delta,y)$ such that:

$$h_{x+\delta}(y)-h_x(y) = D_1 f(\tau,y) \delta$$ Hence, $$\frac{F(x + \delta)-F(x)}{\delta} \stackrel{?}{=} \int_{A_n} D_1 f(\tau,y)dy$$

Furthermore, $\exists N_2$ such that $n > N_2$ implies: $$\left\lvert \int_{\mathbb{R^m}} D_1 f(\tau,y)dy - \int_{A_n} D_1 f(\tau,y)dy \right\rvert < \frac{\epsilon}{4}$$

Let $N = \max \{N_1, N_1 \}$ and $n > N$:

$$\left\lvert \frac{F(x+\delta)-F(x)}{\delta} - \int_{\mathbb{R}^m} D_1 f(x,y)dy \right\rvert < \left\lvert \frac{F(x+\delta)-F(x)}{\delta} - \int_{A_n} D_1 f(x,y)dy \right\rvert + \frac{\epsilon}{4}$$ $$\le \left\lvert \frac{F(x+\delta)-F(x)}{\delta} - \int_{A_n} D_1 f(\tau,y)dy \right\rvert + \left\lvert \int_{A_n} D_1 f(x,y)dy - \int_{A_n} D_1 f(\tau,y)dy \right\rvert + \frac{\epsilon}4$$ $$= \left\lvert \int_{\mathbb{R}^m} D_1f(\tau,y)dy - \int_{A_n} D_1 f(\tau,y)dy \right\rvert + \left\lvert \int_{A_n} D_1 f(x,y)dy - \int_{A_n} D_1 f(\tau,y)dy \right\rvert + \frac{\epsilon}4$$ $$\le \left\lvert \int_{A_n} D_1 f(x,y)dy - \int_{A_n} D_1 f(\tau,y)dy \right\rvert + \frac{\epsilon}2$$ $$\le \int_{A_n} |D_1 f(x,y) - D_1 f(\tau, y)| dy + \frac{\epsilon}2$$

$|D_1 f(x,y) - D_1 f(\tau, y)| < \frac{\epsilon}{2 v(A_n)}$, but this $A_n$ may not be what we are currently integrating over, so this does nothing for us. I also can't change the location of the uniform continuity statement, because $\tau$ depends on $\delta$ which depends on $\eta$.

Answer 1

We can salvage your approach and prove as you intended.

First, you are correct that for sufficiently large $n > N_1$ we find

$$\tag{1}\left\lvert \int_{\mathbb{R^m}} D_1 f(x,y)\,dy - \int_{A_n} D_1 f(x,y)\,dy \right\rvert < \frac{\epsilon}{4}, $$

but we can make a stronger statement that this is true for all and not just fixed $x$. This follows because $|D_1f(x,y)| \leqslant g(y)$ and the convergence $\int_{A_n} \to \int _\mathbb{R^m}$ is uniform by the Weierstrass test.

Define $F_n(x) = \int_{A_n} f(x,y)\,dy $ and let $n > N_1$ be fixed.

Similar to your attempt, we can apply the triangle inequality and (1) to obtain

$$\tag{2}\left\lvert \frac{F(x+\delta)-F(x)}{\delta} - \int_{\mathbb{R}^m} D_1 f(x,y)\,dy \right\rvert \\ \leqslant \left\lvert \frac{F(x+\delta)-F(x)}{\delta} - \frac{F_n(x+\delta)-F_n(x)}{\delta}\right\rvert+ \left\lvert \frac{F_n(x+\delta)-F_n(x)}{\delta} - \int_{A_n} D_1 f(x,y)\,dy \right\rvert + \frac{\epsilon}{4} $$

We can use your argument based on the mean value theorem to show that for some $\tau_y$ between $x$ and $x + \delta$ we have

$$\frac{F_n(x+\delta)-F_n(x)}{\delta} = \int_{A_n} D_1 f(\tau_y,y)\,dy, $$

and

$$\left\lvert \frac{F_n(x+\delta)-F_n(x)}{\delta} - \int_{A_n} D_1 f(x,y)\,dy \right\rvert \leqslant \int_{A_n} |D_1 f(\tau_y,y) - D_1f(x,y)|\,dy .$$

Using uniform continuity of $D_1f$ on the compact set $A_n$, there exists $\eta > 0$ such that when $\delta < \eta$ we have $|D_1f(\tau_y,y) - D_1f(x,y)| < \epsilon/(4v(A_n)) $ and

$$\left\lvert \frac{F_n(x+\delta)-F_n(x)}{\delta} - \int_{A_n} D_1 f(x,y)\,dy\right\rvert < \frac{\epsilon}{4}.$$

It just remains to show that for sufficiently large $n$, the first term on the RHS of (2) is smaller than $\epsilon/2$.

For $m > n$ we apply the mean value theorem to find $z$ between $x$ and $x + \delta$ such that

$$\left|\frac{F_m(x+\delta)-F_m(x)}{\delta} - \frac{F_n(x+\delta)-F_n(x)}{\delta}\right| = \left|\int_{A_m} D_1f(z,y) \, dy - \int_{A_n} D_1f(z,y) \, dy \right| $$

By uniform convergence of $\int_{A_n}$, there exists $N_2$ such that for all $m > n > N_2$ we have

$$\left|\frac{F_m(x+\delta)-F_m(x)}{\delta} - \frac{F_n(x+\delta)-F_n(x)}{\delta}\right| < \frac{\epsilon}{2}.$$

Taking the limit of both sides, as $m \to \infty$ we get

$$\left|\frac{F_(x+\delta)-F_(x)}{\delta} - \frac{F_n(x+\delta)-F_n(x)}{\delta}\right| \leqslant \frac{\epsilon}{2}.$$

Therefore, we can find $\eta>0$ such that if $\delta < \eta$ and we have

$$\left\lvert \frac{F(x+\delta)-F(x)}{\delta} - \int_{\mathbb{R}^m} D_1 f(x,y)\,dy \right\rvert < \epsilon.$$

Note that $\eta$ may depend of the fixed $n > \max(N_1,N_2)$ used in the estimate, but this is enough to reach the conclusion that $F$ is differentiable with $F'(x) =\int_{\mathbb{R}^m}D_1f(x,y) \, dy.$

Answer 2

Fix $x\in \mathbb R,$ and fix a real sequence $h_n \to 0$ through nonzero values. Define

$$ d_n(y) = \frac{f(x+h_n,y)-f(x,y)}{h_n}.$$

Then $d_n(y) \to D_1f(x,y)$ for every $y\in \mathbb R^m.$

By the mean value theorem, for each $y\in \mathbb R^m,$ there exists $c(y,n)$ between $x$ and $x+h_n$ such that

$$d_n(y) = D_1 f(c(y,n),y).$$

From our hypothesis on $D_1f,$ it follows that $|d_n(y)| \le g(y)\in L^1(\mathbb R^m).$ Thus by the dominated convergence theorem,

$$\tag 1 \int_{\mathbb R^m} d_n(y)\, dy \to \int_{\mathbb R^m} D_1f(x,y)\, dy.$$

Now $(F(x+h_n) - F(x))/h_n$ equals the left side of $(1),$ so $(F(x+h_n) - F(x))/h_n$ $\to \int_{\mathbb R^m} D_1f(x,y)\, dy.$ Because $h_n$ was an arbitrary sequence $\to 0,$ we have $F'(x) = \int_{\mathbb R^m} D_1f(x,y)\, dy.$ Finally, since $x$ was an arbitrary real number, we have $F$ differentiable on $\mathbb R$ as desired.