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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?

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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,

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

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    I think formally, this shows that $\lim_{h \rightarrow 0}\left|\int_a^b \frac{f(x,y+h) - f(x,y)}{h} \, dx - \int_a^b \partial_yf(x,y) \, dx\right| = 0$, but this is not quite what we want? We need the limit to be inside the absolute value – 2017-02-25
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    @ilmth: This shows by the $\epsilon-\delta$ definition that $\lim_{h \rightarrow 0} \int_a^b \frac{f(x,y+h) - f(x,y)}{h} \, dx = \int_a^b \partial_y f(x,y) \, dx$ which is equivalent to what you write. But also we have $\partial_y f(x,y) = \lim_{h \rightarrow 0} \frac{f(x,y+h) - f(x,y)}{h}$. Hence we have shown that the limit and integral can be switched. – 2017-02-25
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    Not really, since the limit is nowhere to be seen in the proof. Otherwise, could you make the proof work by starting with $\left|\lim_{h \rightarrow 0}\int_a^b \frac{f(x,y+h) - f(x,y)}{h} \, dx - \int_a^b \partial_yf(x,y) \, dx\right|$ and showing that this is less than $\epsilon$? – 2017-02-25
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    I think you missed my point. You want to show $\lim_{h \to 0} \int g(x,h) \, dx = \int \lim_{h \to 0} g(x,h) \, dx$ where in this case $g(x,h) = (f(x,y+h) - f(x,y))/h$ and we know $\lim_{h \to 0} g(x,h) = \partial_y f(x,y)$ since you assumed the partial derivative exists (and is continuous). One way you show that the limit of the integral equals the integral of the limit is to show if $|h|$ is sufficiently small then $|\int g(x,h) \,dx - L| < \epsilon$ where $L$ is the integral of the limit. This is easier under your conditions with compact domain and continuous derivative. – 2017-02-25
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    Otherwise you can have to look for monotone or dominated convergence. There you usually take the course of showing $\liminf \int g = \limsup \int g = L$. You can look that up in a real analysis book -- look for Fatou's lemma and monotone or dominated convergence theorem. However we don't need that heavy machinery here under your more restricted conditions. – 2017-02-25
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    Nevermind, I see what you mean, you were using epsilon delta definition *of the limit*, thanks! – 2017-02-25
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    That's great! You're welcome. – 2017-02-26