Taking out limit from integral of limit.

Question

I have a function $f(x, y)$ defined on a set $[a,b] \times [c,d]$ such that $\partial f/\partial y$ is continuous. I want to show that $g(y) = \int_a^bf(x,y)\, dx$ is differentiable with $g'(y) = \int_a^b\frac{\partial f}{\partial y}(x,y)\, dx$. By hypothesis, I know that $\int_a^b\frac{\partial f}{\partial y}(x,y)\, dx$ exists and equals $\int_a^b\lim_{h \rightarrow 0}\frac{f(x, y) - f(x, y + h)}{h}\, dx$. At this point, can I move the limit outside of the integral (and if so please explain why)? Otherwise how do I proceed with proof?

Answer 1

Applying the mean value theorem, there exists $\xi_y \in (y, y+h)$ such that

$$\left|\int_a^b \frac{f(x,y+h) - f(x,y)}{h} \, dx - \int_a^b \partial_yf(x,y) \, dx\right|\\ = \left|\int_a^b \partial_y f(x, \xi_y) \, dx - \int_a^b \partial_yf(x,y) \, dx\right|\\ \leqslant \int_a^b |\partial_yf(x,\xi_y) - \partial_yf(x,y)| \, dx.$$

Since $f_y$ is continuous on $[a,b] \times [c,d]$ it is uniformly continuous and for any $\epsilon > 0$ there exists $\delta >0$ such that if $|h| < \delta$ then $ |\partial_yf(x,\xi_y) - \partial_yf(x,y)| < \epsilon/(b-a)$ and

$$\int_a^b |\partial_yf(x,\xi_y) - \partial_yf(x,y)| \, dx < \epsilon.$$

Thus,

$$\lim_{h \rightarrow 0} \int_a^b \frac{f(x,y+h) - f(x,y)}{h} \, dx = \int_a^b \partial_y f(x,y) \, dx \\ = \int_a^b \lim_{h \rightarrow 0} \frac{f(x,y+h) - f(x,y)}{h} \, dx . $$