Differentiability of Functions Defined by Lebesgue Integrals
Differentiability of Functions Defined by Lebesgue Integrals
Let $I$ and $J$ be intervals in $\mathbb{R}$ and let $f : I \times J \to \mathbb{R}$. On the Continuity of Functions Defined by Lebesgue Integrals page we saw conditions which made the function $F : J \to \mathbb{R}$ defined below continuous at each $y \in J$:
(1)\begin{align} \quad F(y) = \int_I f_y(x) \: dx = \int_I f(x, y) \: dx \end{align}
We will now look at conditions which makes the function $F$ differentiable.
| Theorem 1: Let $I$ and $J$ be intervals and let $f : I \times J \to \mathbb{R}$. Suppose that: 1) For every fixed $y \in J$, the function $f_y :I \to \mathbb{R}$ defined by $f_y(x) = f(x, y)$ is a measurable function on $I$. 2) There exists a $y_0 \in J$ such that $f_{y_0}$ is Lebesgue integrable on $I$. 3) For every point $(x, y) \in \mathrm{int} (I \times J)$, the partial derivative $\displaystyle{\frac{\partial}{\partial y} f(x, y)}$ exists. 4) There exists a nonnegative function $g$ that is Lebesgue integrable on $I$ and such that $\displaystyle{\biggr \lvert \frac{\partial}{\partial y} f(x, y) \biggr \rvert \leq g(x)}$ for all $\displaystyle{(x, y) \in \mathrm{int} (I \times J)}$. Then we can conclude that: a) For every $y \in J$, $f_y$ is Lebesgue integrable on $I$, i.e., $\displaystyle{\int_I f_y(x) \: dx = \int_I f(x, y) \: dx}$ exists. b) If $F : J \to \mathbb{R}$ is defined for all $y \in J$ by $\displaystyle{F(y) = \int_I f_y(x) \: dx = \int_I f(x, y) \: dx}$ then $F$ is differentiable at every point $(x, y) \in \mathrm{int} (I \times J)$ and $\displaystyle{F'(y) = \int_I \frac{\partial}{\partial y} f(x, y) \: dx}$. |
- Proof of a) Fix $x \in I$. Then $f(x, y)$ is a single-variable function in terms of $y$. From (3), since the partial derivative $\displaystyle{\frac{\partial}{\partial y} f(x, y)}$ exists for all $(x, y) \in \int (I \times J)$ (i.e., for all $y \in \mathrm{int} (J)$ which is an open interval), we have by the mean value theorem that there exists a $\xi$ between $y$ and $y_0$ such that:
\begin{align} \quad f(x, y) - f(x, y_0) = \frac{\partial}{\partial y} f(x, \xi) [y - y_0] \\ \quad f(x, y) = f(x, y_0) + \frac{\partial}{\partial y} f(x, \xi) [y - y_0] \end{align}
- Taking the absolute value of both sides and using the triangle inequality gives us:
\begin{align} \quad \mid f(x, y) \mid \leq \mid f(x, y_0) + \frac{\partial}{\partial y} f(x, \xi) [y - y_0] \mid \\ \quad \mid f(x, y) \mid \leq \mid f(x, y_0) \mid + \biggr \lvert \frac{\partial}{\partial y} f(x, \xi) \biggr \rvert \mid y - y_0 \mid \end{align}
- Now from (4), since $(x, \xi) \in \mathrm{int} (I \times J)$ we have that $\biggr \lvert \frac{\partial}{\partial y} f(x, \xi) \biggr \rvert < g(x)$ where $g$ is a Lebesgue integrable function on $I$, and so:
\begin{align} \quad \mid f(x, y) \mid & \leq \mid f(x, y_0) \mid + g(x) \mid y - y_0 \mid \\ \quad \mid f_y(x) \mid & \leq f_{y_0}(x) + g(x) \mid y - y_0 \mid \end{align}
- Notice that $f_{y_0}(x)$ and $g(x)$ are Lebesgue integrable on $I$ and $\mid y - y_0 \mid$ is a constant. So the sum is a Lebesgue integrable function. Moreover, from (1) we have that $f_y$ is a measurable function. So, by the Criterion for a Measurable Function to be Lebesgue Integrable we see that $f_y$ is Lebesgue integrable for all $y \in J$, i.e., the following integral exists for all $y \in J$:
\begin{align} \quad \int_I f_y(x) \: dx = \int_I f(x, y) \: dx \quad \blacksquare \end{align}
- Proof of b) Fix $y \in J$ and let $(y_n)_{n=1}^{\infty}$ be a sequence of real numbers in $J$ that converges to $y$, i.e., $\displaystyle{\lim_{n \to \infty} y_n = y}$ and such that $y_n \neq y$ for all $n \in \mathbb{N}$. Define a new sequence of functions $(h_n(x))_{n=1}^{\infty}$ for each $n \in \mathbb{N}$ by:
\begin{align} \quad h_n(x) = \frac{f(x, y_n) - f(x, y)}{y_n - y} \end{align}
- We have already established from earlier that $f_{y_n} = f(x, y_n)$ and $f_y = f(x, y)$ are Lebesgue integrable on $y$, and since $y_n - y$ is a constant, we have the Linearity of Lebesgue Integrals that each $h_n$ is Lebesgue integrable on $I$. Moreover, $(h_n(x))_{n=1}^{\infty}$ converges to $\frac{\partial}{\partial} f(x, y)$ almost everywhere on $I$ since as $n \to \infty$, $y_n \to y$ and furthermore, by the mean value theorem we have that $h_n(x) = \frac{\partial}{\partial y} f(x, \xi_n)$ for some $\xi_n$ between $y$ and $y_n$, for each $n \in \mathbb{N}$. But $(4)$ we have that for all $n \in \mathbb{N}$ that:
\begin{align} \quad \mid h_n(x) \mid = \biggr \lvert \frac{\partial}{\partial y} f(x, \xi_n) \biggr \rvert \leq g(x) \end{align}
- So, by the Lebesgue's Dominated Convergence Theorem we have that the limit function $\displaystyle{\frac{\partial f}{\partial}}$ is Lebesgue integrable and moreover:
\begin{align} \quad \int_I \frac{\partial}{\partial y} f(x, y) \: dx = \int_I \lim_{n \to \infty} \underbrace{\frac{f(x, y_n) - f(x, y)}{y_n - y}}_{h_n(x)} \: dx = \lim_{n \to \infty} \int_I \underbrace{\frac{f(x, y_n) - f(x, y)}{y_n - y}}_{h_n(x)} \: dx \quad (*) \end{align}
- Now consider the quotient:
\begin{align} \quad \frac{F(y_n) - F(y)}{y_n - y} = \frac{\int_I f_{y_n}(x) \: dx - \int_I f_y(x) \: dx}{y_n - y} = \int_I \frac{f(x, y_n) - f(x, y)}{y_n - y} \: dx \end{align}
- Taking the limit as $n \to \infty$ and using $(*)$ gives us that:
\begin{align} \quad F'(y) = \lim_{n \to \infty} \frac{F(y_n) - F(y)}{y_n - y} = \lim_{n \to \infty} \int_I \frac{f(x, y_n) - f(x, y)}{y_n - y} \: dx \overset{\mathrm{by} \: (*)} = \int_I \lim_{n \to \infty} \frac{f(x, y_n) - f(x, y)}{y_n - y} \: dx = \int_I \frac{\partial}{\partial y} f(x, y) \: dx \quad \blacksquare \end{align}