Differentiation under the integral · Selected prerequisite proofs

L9.1.1: Compact one-parameter integral derivative with uniform-continuity/MVT proof and explicit integral-error bound.

Suppose

f \colon [a,b] \times [c,d] \to \R

is a continuous function, such that

\frac{\partial f}{\partial y}

exists for all

(x,y) \in [a,b] \times [c,d]

and is continuous. Define

g \colon [c,d] \to \R

\begin{equation*} g(y) \coloneqq \int_a^b f(x,y) \,dx . \end{equation*}

Then

g

is continuously differentiable and

\begin{equation*} g'(y) = \int_a^b \frac{\partial f}{\partial y}(x,y) \,dx . \end{equation*}

Fix

y \in [c,d]

and let

\epsilon > 0

be given. As

\frac{\partial f}{\partial y}

is continuous on

[a,b] \times [c,d]

it is uniformly continuous. In particular, there exists

\delta > 0

such that whenever

y_1 \in [c,d]

with

\sabs{y_1-y} < \delta

and all

x \in [a,b]\text{,}

we have

\begin{equation*} \abs{\frac{\partial f}{\partial y}(x,y_1)-\frac{\partial f}{\partial y}(x,y)} < \epsilon . \end{equation*}

Suppose

h

is such that

y+h \in [c,d]

and

\sabs{h} < \delta\text{.}

Fix

x

for a moment and apply the mean value theorem to find a

y_1

between

y

and

y+h

such that

\begin{equation*} \frac{f(x,y+h)-f(x,y)}{h} = \frac{\partial f}{\partial y}(x,y_1) . \end{equation*}

\sabs{y_1-y} \leq \sabs{h} < \delta\text{,}

\begin{equation*} \abs{ \frac{f(x,y+h)-f(x,y)}{h} - \frac{\partial f}{\partial y}(x,y) } = \abs{ \frac{\partial f}{\partial y}(x,y_1) - \frac{\partial f}{\partial y}(x,y) } < \epsilon . \end{equation*}

The argument worked for every

x \in [a,b]

(different

y_1

may have been used). Thus, as a function of

x

\begin{equation*} x \mapsto \frac{f(x,y+h)-f(x,y)}{h} \qquad \text{converges uniformly to} \qquad x \mapsto \frac{\partial f}{\partial y}(x,y) \qquad \text{as } h \to 0 . \end{equation*}

We defined uniform convergence for sequences although the idea is the same. You may replace

h

with a sequence of nonzero numbers

\{ h_n \}_{n=1}^\infty

converging to

0

such that

y+h_n \in [c,d]

and let

n \to \infty\text{.}

Consider the difference quotient of

g\text{,}

\begin{equation*} \frac{g(y+h)-g(y)}{h} = \frac{\int_a^b f(x,y+h) \,dx - \int_a^b f(x,y) \,dx }{h} = \int_a^b \frac{f(x,y+h)-f(x,y)}{h} \,dx . \end{equation*}

Uniform convergence implies the limit can be taken underneath the integral. So

\begin{equation*} \lim_{h\to 0} \frac{g(y+h)-g(y)}{h} = \int_a^b \lim_{h\to 0} \frac{f(x,y+h)-f(x,y)}{h} \,dx = \int_a^b \frac{\partial f}{\partial y}(x,y) \,dx . \end{equation*}

Then

g'

is continuous on

[c,d]

by Proposition 7.5.12 mentioned above.