Prove that $\Phi(y, u, v) := \int_{u}^{v} f(x, y) dx$ is a $C^1$ function.(differentiation under the integral sign)

Question

I am reading "A Course in Analysis vol.3" by Kazuo Matsuzaka.

The author wrote the following fact without a proof.
I think it is easy to show that $D_2 \phi(y, u, v) = f(v, y)$ and $D_3 \phi(y, u, v) = -f(u, y)$ are continuous.
But I cannot show that $D_1 \phi(y, u, v) = \int_{u}^{v} D_2 f(x, y) dx$ is continuous.
Please tell me the proof.

Let $f(x, y)$ be a continuous function from $S = [a, b] \times I \subset \mathbb{R}^2$ to $\mathbb{R}$, where $I$ is an interval in $\mathbb{R}$.
Suppose that $D_2 f(x, y)$ exists in $S$ and is continuous in $S$.
Let $\Phi(y, u, v) := \int_{u}^{v} f(x, y) dx$ be a function from $I \times [a, b] \times [a, b]$ to $\mathbb{R}$.

Then, $\Phi(y, u, v)$ is a $C^1$ function.

Answer 1

Disclaimer:

Each time you see a number at the side of an mathematical expression there is something to red at the bottom of my answer!

As you already pointed out, we have that: $$D_u \Phi(y,u,v)=-f(u,y)\text{ and }D_v\Phi(y,u,v)=f(v,y)$$ And thus both $D_u\Phi$ and $D_v\Phi$ exist and are continuous.

For $D_y\phi$ consider using the definition of partial derivatives: $$D_y\Phi(y,u,v)=\lim_{t\rightarrow0}\frac{\Phi(y+t,u,v)-\Phi(y,u,v)}{t}$$ So: \begin{align} D_y\Phi(y,u,v) &=\lim_{t\rightarrow0}\frac{1}{t}\left(\int_u^v f(x,y+t)dx-\int_u^v f(x,y)dx\right)\\ &=\lim_{t\rightarrow0}\left(\int_u^v \frac{f(x,y+t)-f(x,y)}{t} dx\right) \end{align} So since $D_y f$ exists and is continuous we have: $$D_y\Phi(y,u,v)=\int_u^v D_y f(x,y) dx$$ Let us also prove that swapping the integral and the limit is valid: Since $D_y f(x,y)$ exists and is continuous, the integral of $D_y f(x,y)$ exists and by the definition of the partial derivative we have for each $\varepsilon>0$ a $\delta>0$ such that for all $x$ with $\vert x\vert<\delta$: $$\left\vert D_y f(x,y)-\frac{f(x,y+t)-f(x,y)}{t}\right\vert<\frac{\varepsilon}{2\vert v-u\vert}$$ So for all $x$ with $\vert x\vert<\delta$: \begin{align} &\left\vert \int_u^v D_y f(x,y) dx-\int_u^v \frac{f(x,y+t)-f(x,y)}{t} dx\right\vert\\ =&\left\vert \int_u^v D_y f(x,y)-\frac{f(x,y+t)-f(x,y)}{t} dx\right\vert\\ \le & \left\vert\int_u^v \left\vert D_yf(x,y)-\frac{f(x,y+t)-f(x,y)}{t}\right\vert dx\right\vert\;\;\;\;\;\;(1)\\ \le & \left\vert\int_u^v \frac{\varepsilon}{2\vert v-u\vert} dx\right\vert\\ =&\frac{\varepsilon}{2}\\ <&\varepsilon \end{align} So in particular: $$\int_u^v\frac{f(x,y+t)-f(x,y)}{t} dx\longrightarrow \int_u^v D_yf(x,y) dx$$

Notice that $$F:I^2\rightarrow\mathbb{R},(x,y)\mapsto\int_a^x D_y(t,y)\,dt$$ exists and is a antiderivative of $D_y f$ with respect to $x$ as $D_y f$ exists and is continuous. So we have: $$D_y\Phi(y,u,v)=\int_u^vD_y f(t,y)\,dt=F(v,y)-F(u,y)$$ This tells us that $D_y\Phi$ is continuous if $F$ is continuous!

Let's test for continuity of $F$ via the sequence test:

Fix $a:= \inf I$, take any pair $(x,y)\in I^2$ and any sequence $(x_n,y_n)\rightarrow (x,y)$. Since $D_y f$ is continuous and thus continuous in $y$ we have for any $\varepsilon>0$ a $n_0$ such that for all $n\ge n_0$: $$\vert D_y f(x,y_n)-D_y f(x,y)\vert<\frac{\varepsilon}{2(x-a)}$$ But since $x_n\rightarrow x$ we also have a $m_0$, such that for any $n\ge m_0$: $$\vert x_n-x\vert<\frac{\varepsilon}{2\,\sup_{t\in I} D_yf(t,y_n)}$$

So for every $n\ge\max\{n_0,m_0\}$: \begin{align} &\left\vert F(x_n,y_n)-F(x,y)\right\vert\\ =&\left\vert \int_a^{x_n} D_y f(t,y_n)\,dt-\int_a^x D_y f(t,y)\,dt\right\vert\\ =&\left\vert\int_a^x D_yf(t,y_n)\,dt+\int_x^{x_n} D_yf(t,y_n)\,dt-\int_a^xD_yf(t,y)\,dt\right\vert\\ =&\left\vert \int_a^x D_yf(t,y_n)-D_yf(t,y)\,dt+\int_x^{x_n} D_yf(t,y)\,dt\right\vert\\ \le&\int_a^x\left\vert D_yf(t,y_n)-D_yf(t,y)\right\vert\, dt+\vert x_n-x\vert\cdot\sup_{x\leftrightarrow x_n} D_y f(t,y_n)\;\;\;\,(2)\\ <&\int_a^x\frac{\varepsilon}{2(x-a)}\,dt+\frac{\varepsilon}{2\sup_{t\in I}D_yf(t,y_n)}\cdot\sup_{x\leftrightarrow x_n} D_y f(t,y_n)\\ <&\frac{\varepsilon}{2}+\frac{\varepsilon}{2}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(3)\\ =&\varepsilon \end{align} And thus: $$F(x_n,y_n)\longrightarrow F(x,y)$$ and $F$ (thus $D_y\Phi$) is continuous!

Remarks:

(1): The extra pair of $\vert\cdot\vert$ outside of the integral in are needed, as it may be possible that $\int_u^v \dots \,dt$ is negative even though the integrand is positive. This would be the case if $v<u$.

(2): The notation $x\leftrightarrow x_n$ stands for either $[x,x_n]$ or $[x_n,x]$ as only one set is well defined (we don't know whether $x<x_n$ or $x_n<x$).

(3): Since $x\leftrightarrow x_n\subseteq I$ we have that $\sup_{x\leftrightarrow x_n}/\sup_{t\in I}\le 1$.

Answer 2

Well, $\Phi$ will be $C^1$ as soon as it has continuous partial derivatives in every direction, as you noted. I'll just treat $D_1$, as you seem to have a grasp of the other partial derivatives.

So let $y$ be an interior point of $I$ and note that there exists some $r>0$ such that $[a,b]\times [y-r,y+r]\subseteq I\times [a,b]$. Now, on this set, $D_2f$ is uniformly continuous, so for any $\varepsilon>0$, there exists some $r>\delta>0$ such that $||(x,y)-(x',y')||<\delta$ implies $|D_2f(x,y)-D_2f(x',y')|<\varepsilon.$ Thus, for $y'\in [y,y+\delta]$ and any $u,v\in [a,b]$, we see that

\begin{align} &\left|\int_u^v f(x,y')\textrm{d}x-\int_u^v f(x,y)\textrm{d}x-(y'-y)\int_u^v D_2f(x,y)\textrm{d}x\right|\\ =& \left|\int_u^v \left(\int_y^{y'} D_2f(x,t)\textrm{d}t+f(x,y) \right)\textrm{d}x-\int_u^v f(x,y)\textrm{d}x-(y'-y)\int_u^v D_2f(x,y)\textrm{d}x\right|\\ =& \left| \int_u^v \left(\int_y^{y'} (D_2f(x,t)-D_2f(x,y))\textrm{d}t\right)\textrm{d}x\right|\\ \leq& \left|\int_{u}^v \int_y^{y'} |D_2 f(x,t)-D_2f(x,y)|\textrm{d}t\textrm{d}x \right| \\ \leq& (b-a) (y'-y) \varepsilon, \end{align} which is less than $\varepsilon$ for $y'-y<\min \{\delta,\frac{1}{b-a}\}$. The calculation is analogous for $y'\in [y-\delta,y],$ but all the inner integrals change orders.

Accordingly, $\Phi$ has a continuous $y$ derivative on the interior of $I\times [a,b]^2$, but the argument also immediately generalises to boundary points. Combining this with your own observations implies that $\Phi\in C^1(I\times [a,b]^2)$.