Proving that an integral is differentiable

Question

Given an open interval $I\subset\mathbb{R}$, $ i_0\in I$, and $f:I\times\mathbb{R} \rightarrow\mathbb{R}$ such that

1. For all $i\in\mathbb{R}:y\mapsto f(i,y)$ is integrable

2. For all $y \in \mathbb{R}$ the partial derivative $\frac{\partial f}{\partial i}(i_0,y)$ exists and

3. T.e. a neighbourhood $V$ of $i_0$ and an integrable function $h:\mathbb{R}\rightarrow\mathbb{R}$ s.t. for all $i \in V\cap I, i\neq i_0$ and for all $y\in\mathbb{R}$ we have $\frac{f(i,y)-f(i_0,y)}{i-i_0}\le h(y)$

I want to prove that $$F:I\rightarrow\mathbb{R}, i\mapsto\int_{\mathbb{R}}f(i,y)dy$$ is differentiable at $i_0$ and that $$F'(i_0)=\int_{\mathbb{R}}\frac{\partial f}{\partial i}(i_0,y)dy$$

I know that for $F$ to be differentiable I have to prove that $$\lim_{i\rightarrow i_0}\frac{F(i)-F(i_0)}{i-i_0}$$ exists. If this turns out to be the expression from above, I'm done I think. So at the outset I have $$\frac{F(i)-F(i_0)}{i-i_0}=\frac{\int_{\mathbb{R}}f(i,y)dy-\int_{\mathbb{R}}f(i_0,y)dy}{i-i_0}=\frac{\int_{\mathbb{R}}(f(i,y)-f(i_0,y))dy}{i-i_0}$$

What do I do next?

Answer 1

Fix an $i \in I$ and define the function

$$u(y,h) := \begin{cases} \frac{f(y,i+h) - f(y,i)}{h}, & \text{if } h \ne 0 \\\\ \frac{\partial f}{\partial i}(y,i), & \text{if } h = 0 \end{cases}$$

From our conditions, this function is well defined for almost all $y \in \mathbb{R}$ and since the partial derivative exists, $i \mapsto u(y,i)$ is continuous on $I$ for almost all $y \in I$. Also, $i \mapsto f(y,i)$ is continuously differentiable, so the mean value theorem implies the existence of a $c \in I$ with $\lvert c - i \rvert < \lvert h \rvert$ such that

\begin{align} u(x,h) &= \frac{f(y,i+h) - f(y,i)}{h} \\ &= \frac{\partial f}{\partial c} (y,c) \end{align}

Hence, by assumption

\begin{align} \lvert u(x,h) \rvert &= \bigg \lvert \frac{\partial f}{\partial c} (y,c) \bigg \rvert \\ &\le g(y) \end{align}

(I used $g(y)$ not $h(y)$ to differentiate the function from the step size $h$) for almost all $y \in \mathbb{R}$ and all $h \in \mathbb{R}$ with $i + h \in I$. So the requisite assumptions are satisfied and so

\begin{align} F'(i) &= \lim_{h \to 0} \frac{F(i+h) - F(i)}{h} \\ &= \lim_{h \to 0} \int_{\mathbb{R}} u(x,h) dy \\ &=\int_{\mathbb{R}} u(y,0) dy \\ &= \int_{\mathbb{R}} \frac{\partial f}{\partial i} (y,i) dy \end{align}

Answer 2

Fix an $i \in \mathbb{R}$ and define the function

$$u(y,h) := \begin{cases} \frac{f(y,i+h) - f(y,i)}{h}, & \text{if } h > \ne 0 \\\\ \frac{\partial f}{\partial i}(y,i), & \text{if } h = 0 > \end{cases}$$

From our conditions, this function is well defined for almost all $y > \in \mathbb{R}$ and since the partial derivative exists, $i \mapsto > u(y,i)$ is continuous on $I$ for almost all $y \in I$. Also, $i > \mapsto f(y,i)$ is continuously differentiable, so the mean value theorem implies the existence of a $c \in I$ with $\lvert c - i \rvert > < \lvert h \rvert$ such that

-> the opening post was f(i,y), does it mean $y \in R$, doesn't it? and the variable h is equivalent to which one I saw in your post they are all f(y,i), but in the opening post was f(i,y)

\begin{align} u(x,h) &= \frac{f(y,i+h) - f(y,i)}{h} \\ &= \frac{\partial f}{\partial c} (y,c) \end{align}

Hence, by assumption

\begin{align} \lvert u(x,h) \rvert &= \bigg \lvert \frac{\partial f}{\partial c} (y,c) \bigg \rvert \\ &\le g(y) \end{align}

(I used $g(y)$ not $h(y)$ to differentiate the function from the step size $h$) for almost all $y \in \mathbb{R}$ and all $h \in \mathbb{R}$ with $i + h \in I$. So the requisite assumptions are satisfied and so

\begin{align} F'(i) &= \lim_{h \to 0} \frac{F(i+h) - F(i)}{h} \\ &= \lim_{h \to 0} \int_{\mathbb{R}} u(x,h) dy \\ &=\int_{\mathbb{R}} u(y,0) dy \\ &= \int_{\mathbb{R}} \frac{\partial f}{\partial i} (y,i) dy \end{align}

and what happens to i_0?