Problem 3-32.
Let $f:[a,b]\times [c,d]\to\mathbb{R}$ be continuous and suppose $D_2f$ is continuous. Define $F(y)=\int_a^b f(x,y) dx$. Prove Leibnitz's rule: $F'(y)=\int_a^b D_2f(x,y) dx$. Hint: $F(y)=\int_a^b f(x,y)dx=\int_a^b\left(\int_c^y D_2f(x,y)dy+f(x,c)\right)dx$. (The proof will show that continuity of $D_2f$ may be replaced by considerably weaker hypotheses.)
I solved this problem as follows.
Since $D_2f$ is continuous on $[a,b]\times [c,d]$, $D_2f$ is uniformly continuous on $[a,b]\times [c,d]$.
So, for arbitrary positive real number $\varepsilon$, there is a positive real number $\delta$ such that $(x_1,y_1)\in [a,b]\times [c,d]$ and $(x_2,y_2)\in [a,b]\times [c,d]$ and $|(x_1,y_1)-(x_2,y_2)|<\delta\implies |D_2f(x_1,y_1)-D_2f(x_2,y_2)|<\frac{\varepsilon}{b-a}$.
By Fubini's theorem, $F(y)=\int_a^b f(x,y)dx=\int_a^b\left(\int_c^y D_2f(x,y)dy+f(x,c)\right)dx=\int_c^y\left(\int_a^b D_2f(x,y)dx\right)dy+C$, where $C$ is some real number.
Let $(x,y)\in [a,b]\times [c,d]$.
Let $z\in [c,d]$ and $|z-y|<\delta$.
Then $|(x,z)-(x,y)|=|z-y|<\delta$.
So, $|D_2f(x,z)-D_2f(x,y)|<\frac{\varepsilon}{b-a}$.
So, $\left|\int_a^b D_2f(x,z)dx-\int_a^b D_2f(x,y)dx\right|\leq\int_a^b \left|D_2f(x,z)-D_2f(x,y)\right|dx<\int_a^b \frac{\varepsilon}{b-a}dx=\varepsilon$.
So, $G$ such that $G(y):=\int_a^b D_2f(x,y)dx$ is continuous on $[c,d]$.
So, $F(y)=\int_c^y G(y)dy+C$ is differentiable at $y\in [c,d]$ and $F'(y)=G(y)=\int_a^b D_2f(x,y)dx$.
The author wrote "The proof will show that continuity of $D_2f$ may be replaced by considerably weaker hypotheses.".
What are "considerably weaker hypotheses"?
My attempt:
Is there a function $f:[a,b]\times [c,d]\to\mathbb{R}$ which satisfies the following conditions?
- $D_2f$ exists on $[a,b]\times [c,d]$.
- $D_2f$ is integrable on $[a,b]\times [c,d]$.
- $D_2f$ is not continuous on $[a,b]\times [c,d]$.
- For any $\varepsilon>0$ and any $y\in [c,d]$, there is $\delta>0$ such that $|D_2f(x,z)-D_2f(x,y)|<\varepsilon$ if $x\in [a,b]$ and $z\in [c,d]$ and $|y-z|<\delta$.
- For any $y\in [c,d]$, $[a,b]\ni x\mapsto f(x,y)\in\mathbb{R}$ is integrable on $[a,b]$.
- For any $y\in [c,d]$, $[a,b]\ni x\mapsto\int_c^y D_2f(x,y)dy\in\mathbb{R}$ is integrable on $[a,b]$.
- For any $y\in [c,d]$, $[a,b]\ni x\mapsto D_2f(x,y)\in\mathbb{R}$ is integrable on $[a,b]$.
Suppose such $f$ exists.
By 5, $F(y)=\int_a^b f(x,y)dx$ exists for any $y\in [c,d]$.
By 4, $\int_a^b f(x,y)dx=\int_a^b\left(\int_c^y D_2f(x,y)dy+f(x,c)\right)dx$.
By 6 and 5, $\int_a^b\left(\int_c^y D_2f(x,y)dy+f(x,c)\right)dx=\int_a^b\left(\int_c^y D_2f(x,y)dy\right)dx+C$, where $C$ is some real number.
By 2 and 4 and 7 and Fubini's theorem, $\int_a^b\left(\int_c^y D_2f(x,y)dy\right)dx+C=\int_c^y\left(\int_a^b D_2f(x,y)dx\right)dy+C$.
By 4, $G$ such that $G(y):=\int_a^b D_2f(x,y)dx$ is continuous on $[c,d]$.
So, $F(y)=\int_c^y G(y)dy+C$ is differentiable at $y\in [c,d]$ and $F'(y)=G(y)=\int_a^b D_2f(x,y)dx$.
But by 3, $D_2f$ is not continuous on $[a,b]\times [c,d]$.