I am trying to find "minimal" conditions for Leibniz Rule: $$\dfrac{d}{dx}\left(\displaystyle\int_{a}^{b}f(x,y)\;dy\right)=\displaystyle\int_{a}^{b}f_x(x,y)\;dy.$$

Let $$F(x)=\displaystyle\int_{a}^{b}f(x,y)\;dy,$$ which is a single variable function.

Let us take the derivative of $F(x)$ using the definition of derivative: $$\dfrac{d}{dx}F(x)=\lim\limits_{h\to0}\dfrac{\displaystyle\int_{a}^{b}f(x+h,y)dy-\displaystyle\int_{a}^{b}f(x,y)dy}{h}$$



To rest of prove, I want to continue with definition of limit:

Given $\epsilon>0$. I need to find $\delta >0$ such that if $$|h|<\delta,$$ then $$\left|\displaystyle\int_{a}^{b}\dfrac{f(x+h,y)-f(x,y)}{h}dy-\displaystyle\int_{a}^{b}f_x(x,y)dy\right|$$$$=\left|\displaystyle\int_{a}^{b}\left[\dfrac{f(x+h,y)-f(x,y)}{h}-f_x(x,y) \right]dy\right|<\epsilon.$$

I know that $$\left|\displaystyle\int_{a}^{b}\left[\dfrac{f(x+h,y)-f(x,y)}{h}-f_x(x,y) \right]dy\right|\le \displaystyle\int_{a}^{b}\left|\dfrac{f(x+h,y)-f(x,y)}{h}-f_x(x,y) \right|dy.$$

Therefore, I want to continue with this inequality.

If I will find a $\delta>0$ for all $y\in[a,b]$ such that $$\left|\dfrac{f(x+h,y)-f(x,y)}{h}-f_x(x,y) \right|<\frac{\epsilon}{b-a}$$ whenever $|h|<\delta$, then the proof will complete.

First Attempt: (Uniformly Continuous) If $f_x(x,y)$ is continuous on $C\times[a,b]$ where $x\in C$ is a closed interval having more than one element, then it will be uniformly continuous and so we can find such $\delta>0$.

Second Attempt: If $f_x(x,y)$ is piece-wise continuous on $y$, we can separate the integral and make the pieces continuous and then apply the First Attempt to the pieces.

(I wrote that proof by myself. Am I correct with this proof; and are the Attempts correct?)

How can I make the conditions weaker? (Please don't say dominated convergence theorem directly, I am trying to understand proofs).

Thanks for help in advance.


Towards the weakest requirements, provided that $f_x(x,y)$ exists, for every $(x,y)$, and $f_x(x,\cdot)$ is Riemann integrable over $[a,b]$.

Observe that $$ \frac{1}{h}\bigg(\int_a^b f(x+h,y)\,dy-\int_a^b f(x,y)\,dy\,\bigg)-\int_a^b f_x(x,y)\,dy\\ =\frac{1}{h}\int_a^b\int_x^{x+h} \big(\,f_x(t,y)-f_x(x,y)\big)\,dt\,dy, $$ and hence $$ \bigg|\,\frac{1}{h}\bigg(\int_a^b f(x+h,y)\,dy-\int_a^b f(x,y)\,dy\,\bigg)-\int_a^b f_x(x,y)\,dy\,\bigg| \\ \le \frac{1}{h}\int_a^b\int_x^{x+h} \big|\,f_x(t,y)-f_x(x,y)\big|\,dt\,dy \\= \frac{1}{h}\int_x^{x+h} \int_a^b \big|\,f_x(t,y)-f_x(x,y)\big|\,dt\,dy= \frac{1}{h}\int_x^{x+h}\|\,f_x(t,\cdot)-f_x(x,\cdot)\|_{L^1[a,b]}\,dt $$ where $\,\|g\|_{L^1[a,b]}=\int_a^b |g(x)|\,dx$.

So, if $$ \lim_{h\to 0}\frac{1}{h}\int_x^{x+h}\|\,f_x(t,\cdot)-f_x(x,\cdot)\|_{L^1[a,b]}\,dt=0, \tag{1} $$ then $$ \frac{d}{dx}\int_a^b f(x,y)\,dy=\int_a^b f_x(x,y)\,dy. $$ Note that, Condition $(1)$ is significantly weaker than the condition $$ \lim_{h\to 0}\|\,f_x(x+h,\cdot)-f_x(x,\cdot)\|_{L^1[a,b]}=0, $$ which in turn, is significantly weaker than the uniform continuity of $f_x$.

  • $\begingroup$ There are some $dy$ missing. $\endgroup$ – celtschk Dec 17 '16 at 15:39
  • $\begingroup$ Thanks for your answer. Can I ask the "all" where "all $(x,y)$"? Like $U\times[a,b]$ where $x\in U$ open. $\endgroup$ – student forever Dec 17 '16 at 19:45
  • $\begingroup$ @studentforever All $(x,y)$, means ideally, for all $(x,y)\in U\times [a,b]$, where $U$ is a open set, where $x$ lies. Apparently, $x$ does not have to be scalar. $\endgroup$ – Yiorgos S. Smyrlis Dec 17 '16 at 20:02

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