# Using dominated convergence to prove partial derivative and integral can be interchanged

Hi guys doing a self study here, and came across this problem. I know the same question with slightly different hypotheses has been asked before but I was a bit confused on the answers given and not sure how to completely adjust those arguments to solve my problem and also want to clear up and confirm some things.

It says the following: Assume $f(x,y)$ and $\frac{\partial f}{\partial x}$ are continuous functions on $\mathbb R x \mathbb R$ and let $[a,b]$be a finite interval. Show that $\int _{[a,b]} \frac{\partial f(x,y)}{\partial x} dy=\frac{ \partial f(x,y)}{\partial x} \int_{[a,b]} f(x,y) dy$

First of all, does f(x,y) continuous in $R^2$ imply it is continuous at a fixed x as a function of y?!If so does this imply f is uniformly continuous on [a,b] as a function of y? If so then f is bounded by its supremum on the finite interval and hence its integral exists? Do we turn this problem into a more tractable form by changing the derivative into a limit of a sequence of functions$n( f(x+\frac{1}{n},y)-f(x,y))$? But then how do we find a bound for this sequece by an integrable function? basically I would appreciate it if somebody explained to me one by one why the hypothesis are the way they are. Thanks in advance guys!

First of all, does $f(x,y)$ continuous in $\mathbb{R}^2$ imply it is continuous at a fixed $x$ as a function of $y$?

Yes, that's correct and holds for any continuous function $f: \mathbb{R}^2 \to \mathbb{R}$. For a fixed point $(x_0,y_0)$ continuity at $(x_0,y_0)$ means that for all $\epsilon>0$ there exists $\delta>0$ such that $$|f(x,y)-f(x_0,y_0)| \leq \epsilon \qquad \text{for all} \, \, |(x_0,y_0)-(x,y)| \leq \delta. \tag{1}$$ Now for fixed $x_0,y_0$ and $\epsilon>0$ choose $\delta>0$ as above, then we have $$|f(x_0,y)-f(x_0,y_0)| \leq \epsilon \qquad \text{for all} \, \, |y-y_0| \leq \delta$$ which shows that $f(x_0,\cdot)$ is continuous at $y=y_0$. Since $x_0,y_0$ are arbitrary, this finishes the proof.

If so does this imply $f$ is uniformly continuous on $[a,b]$ as a function of $y$?

Yeah, any continuous function which is defined on a compact interval is uniformly continuous, see this question; however, this is not needed for the proof of the assertion.

If so then $f$ is bounded by its supremum on the finite interval and hence its integral exists?

Correctly. (For this we don't need absolute continuity; continuity is enough.) Note that the continuity also implies measurability of $f(x,\cdot)$ for each $x$.

Do we turn this problem into a more tractable form by changing the derivative into a limit of sequence of functions $n(f(x+1/n,y)-f(x,y))$.

Yes, exactly. Set

$$F(x) := \int_{[a,b]} f(x,y) \, dy.$$

We have to show that $$\frac{\partial}{\partial x} F(x) = \int_{[a,b]} \partial_x f(x,y) \, dy. \tag{2}$$

First of all, note that $F$ is well-defined because $f(x,\cdot)$ is continuous (see above) and, moreover,

$$\frac{F(x+1/n)-F(x)}{1/n} = n \int_{[a,b]} (f(x+1/n,y)-f(x,y)) \, dy.$$

For fixed $x$ we define a sequence of auxiliary functions by $$u_n(y):= n (f(x+1/n,y)-f(x,y)).$$

Since $f(\cdot,y)$ is, by assumption, differentiable, we have

$$u_n(y) \xrightarrow[]{n \to \infty} \partial_x f(x,y).$$

On the other hand, the mean value theorem shows

$$|u_n(y)| \leq \sup_{\lambda \in [0,1]} |\partial_x f(x+\lambda/n,y)| \leq \sup_{y \in [a,b]} \sup_{|u-x| \leq 1} |\partial_x f(u,y)|.$$

Since $\partial_x f$ is continuous, the right-hand side is a (finite) constant and therefore integrable on the finite interval $[a,b]$. This means that we have found an integrable dominating function for $u_n$. Applying the dominated convergence theorem, we find that

$$\frac{F(x+1/n)-F(x)}{1/n} = \int_{[a,b]} u_n(y) \, dy \xrightarrow[]{n \to \infty} \int_{[a,b]} \partial_x f(x,y) \, dy.$$

This proves $(2)$.

• Thank you for some reason I had not seen this answer!! – user172377 Mar 15 '17 at 22:27
• @Socchi You are welcome. – saz Mar 16 '17 at 6:20
• Hi, could you elaborate a bit on the mean value part? Why can't we just say $\frac{f(x+1/n,y)-f(x)}{1/n} = \frac{\partial f(c,y)} {\partial x}$ for some c in [a,b] $\leq \sup_{x,y} {\frac{\partial f(x,y)}{ \partial x}}$ which is continuous on a closed interval hence integrable. Thanks. – user172377 Apr 11 '17 at 6:21
• @Socchi Well, essentially that's what I did; I just formulated the mean value theorem in a slightly different way. – saz Apr 11 '17 at 7:07
• Thanks, so mine is correct too? – user172377 Apr 11 '17 at 7:17