# $\frac{\partial}{\partial x_i} \int_{\mathbb{R^d}} f(x_0,y)\, dy = \int_{\mathbb{R^d}} \frac{\partial}{\partial x_i} f(x_0,y)\, dy$

Let $\Omega$ be an open set in $\mathbb{R^d}$ and $f:\Omega \times \mathbb{R^d} \to \mathbb{R}$ be an integrable function. Suppouse that $f$ satisties the following conditions:

1) For all $x_0\in \Omega$, $f(x_0,y)\in L(\mathbb{R^d})$ [Lebesgue integrable in $\mathbb{R^d}$]

2) Given $x_0$, there exists $\varepsilon>0$ and $g\in L(\mathbb{R^d})$ such that $B_{\varepsilon}(x_0)\subset \Omega$ and

$$\left| \frac{\partial}{\partial x_i}f(\gamma,y) \right| \leq g(y)$$

for all $\gamma \in B_{\varepsilon}(x_0),\, 1\leq i\leq d$.

I have to prove that

$$\frac{\partial}{\partial x_i} \int_{\mathbb{R^d}} f(x_0,y)\, dy = \int_{\mathbb{R^d}} \frac{\partial}{\partial x_i} f(x_0,y)\, dy$$ and conclude that if $f$ is infinitely differentiable then its integral is as well.

My attempt son far:

Let $x_n$ be a sequence sucha that $x_n\to x_0$. Let $\varepsilon>0$, and assume $x_n \in B_{\varepsilon}(x_0)$ for all $n\in \mathbb{N}$. Consider the sequence

$$F_{n,i}(y) = \frac{\partial}{\partial x_i} f(x_n,y)$$ By hypothesis, for a fixed $i$, there exists a function $g$ such that

$$|F_{n,i}(y)|\leq |g(y)|$$ I have been trying to understand how to use dominated convergence but can't find out how. Any suggestions? And Also for the concluding part...

• Hint --- dominated convergence is about when you can "commute limits and integrals". Can you write $\frac{\partial}{\partial x_i}$ as a limit? – Mark Nov 7 '17 at 2:43

## 1 Answer

Hint: $\dfrac{1}{|h|}|f(x_{0}+he_{i},y)-f(x_{0},y)|=\left|\dfrac{\partial f}{\partial x_{i}}(\eta_{h},y)\right|\leq g(y)$ for sufficiently small $h$, here $\eta_{h}$ is chosen by Mean Value Theorem. Lebesgue Dominated Convergence Theorem will take place by this observation and note that the difference quotient converges to the partial derivative.

• Thanks! this was helpful! – michaelaba Nov 7 '17 at 3:08