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The following question is from Real Analysis by Royden (4th Edition). More precisely, it is problem 36 of chapter 4. Further, the problem follows right after the section over The Lebesgue Dominated Convergence Theorem, hence, that is what should be used. Below is the problem followed by my attempted solution which I have a question about.


Problem:

Let $f$ be a real-valued function of two variables $\left(x,y\right)$ that is defined on the following unit square $Q:=\{\left(x,y\right): 0\leq x\leq 1, 0\leq y\leq 1\}$ and is a measurable function of $x$ for each fixed value of $y$. For each $\left(x,y\right)\in Q$ let the partial derivative a $\partial f/\partial y$ exist. Suppose there is a function $g$ that is integrable over $[0,1]$ and such that,

$$\left\vert\frac{\partial f}{\partial y}\left(x,y\right)\right\vert \leq g(x) \text{ for all } \left(x,y\right)\in Q.\tag{*}$$

Prove that

$$\frac{d}{dy}\left[\int_0^1 f(x,y)dx\right] = \int_0^1\frac{\partial f}{\partial y}(x,y)dx, \text{ for all } y\in[0,1].$$


(My) Proof:

Fix $y\in[0,1]$ and consider the sequence of functions $f_n\colon[0,1]\to\mathbb{R}$ defined by,

$$f_n(x):=\frac{f(x,y+1/n) - f(x,y)}{1/n}.$$

Then, $\{f_n\}$ is a sequence of measurable functions ($f$ is a measurable function of $x$ for each fixed value of $y$) dominated by $g(x)$ that converge pointwise almost everywhere to $\partial f/\partial y$ on $[0,1]$ (for each $(x,y)\in Q$ the partial derivative $\partial f/\partial y$ exists). Therefore, by The Lebesgue Dominated Convergence Theorem,

$$\lim_{n\to\infty}\int_0^1f_n=\int_0^1f$$ $$\lim_{n\to\infty}\int_0^1\frac{f(x,y+1/n) - f(x,y)}{1/n}dx = \int_0^1\frac{\partial f}{\partial y}(x,y)dx.\tag{1}$$

Notice,

$$\lim_{n\to\infty}\int_0^1\frac{f(x,y+1/n) - f(x,y)}{1/n}dx = \lim_{h\to 0}\frac{\int_0^1f(x,y+h)dx-\int_0^1f(x,y)dx}{h}= \frac{d}{dy}\left[\int_0^1f(x,y)dx\right].\tag{2}$$

Thus, combining (1) and (2),

$$\frac{d}{dy}\left[\int_0^1f(x,y)dx\right] = \int_0^1\frac{\partial f}{\partial y}(x,y)dx,$$

which was to be shown.


My main concern with my proof is that, I am not sure how the sequence $\{f_n\}$ are dominated by $g(x)$. I know by (*) that the limit of the sequence $\{f_n\}$ is bounded by $g(x)$. Also, if there are any mistakes in the proof, please let me know!

Edit:

Does the boundedness of the $\{f_n\}$ come from the fact that every convergent sequence is a bounded sequence?

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  • $\begingroup$ to avoid confusing, you shouldn't write $\lim_{n\to\infty}\int_0^1f_n=\int_0^1f$ as $f$ is the initial function. $\endgroup$ – Leon Jan 4 at 9:40
  • $\begingroup$ Is $f_n$ only "almost everywhere" converging? If so, can someone explain why? $\endgroup$ – Rem Jan 4 at 10:07
  • $\begingroup$ @Rem Let fixed $y$. For every $x$, we have $\lim_n f_n(x)=\partial_y f(x,y)$. $\endgroup$ – Leon Jan 4 at 10:11
  • $\begingroup$ @Leon So we make no assumption that it is only "almost everywhere" pointwise convergent, right? I understand that we don't need more than that for the proof though. $\endgroup$ – Rem Jan 4 at 10:19
  • $\begingroup$ Yes, i think so. $\endgroup$ – Leon Jan 4 at 11:32
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Consider $y \in [0,1]$ and any sequence $y_n \in [0,1]$ such that $y_n \to y$. By the mean value theorem, there exists $\xi_{x,n}$ between $y$ and $y_n$ such that

$$|f_n(x)| = \left|\frac{f(x,y_n) - f(x,y)}{y_n - y} \right| = \left|\frac{\partial f}{\partial y}(x,\xi_{x,n})\right| \leqslant g(x)$$

We have

$$\lim_{n \to \infty}\frac{f(x,y_n) - f(x,y)}{y_n - y} = \frac{\partial f}{\partial y}(x,y),$$

and by the dominated convergence theorem

$$\lim_{n \to \infty}\int_0^1\frac{f(x,y_n) - f(x,y)}{y_n - y}\, dx = \int_0^1\lim_{n \to \infty}\frac{f(x,y_n) - f(x,y)}{y_n - y}\, dx = \int_0^1 \frac{\partial f}{\partial y}(x,y) \, dx$$

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  • $\begingroup$ This makes sense, but is it wrong to say "the boundedness of the $\{f_n\}$ come from the fact that every convergent sequence is a bounded sequence"? $\endgroup$ – user853982 Jan 3 at 22:51
  • $\begingroup$ The principle you seem to be invoking is that a sequence $(a_n)$ in $\mathbb{R}$ that is convergent is bounded -- meaning that there exists $B>0$ such that $|a_n| \leqslant B$ for all $n$. Here we can say that convergence of $f_n(x) \to f(x)$ implies that $f_n(x)$ is bounded in the pointwise sense. $\endgroup$ – RRL Jan 3 at 22:59
  • $\begingroup$ ... but what you really have is $f_n(x,y)$ depending on both variables. So this all gets taken care of nicely with MVT. $\endgroup$ – RRL Jan 3 at 23:02
  • $\begingroup$ So, if we can say that the convergence of $f_n \to f$ implies that $f_n(x)$ is bounded in the pointwise sense, why does this not work to conclude The Lebesgue Dominated Convergence Theorem? But yes, I ultimately agree the MVT works very well here. $\endgroup$ – user853982 Jan 3 at 23:04
  • $\begingroup$ If all I know is that $f_n(x,y)$ converges pointwise with respect to $x,y$ as $n \to \infty$, then I can conclude that there exists $B(x,y)>0$ such that $|f_n(x,y)| \leqslant B(x,y)$ for all $n \in \mathbb{N}$ with $x,y$ fixed. I can't say more about $B$ at this point -- for example, that there is a uniform bound for all $y$. $\endgroup$ – RRL Jan 3 at 23:08
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Another way to prove that $f_n$ is bounded by $g$ is \begin{align} |f_n(x)| = & \bigg \vert \frac{ f(x, y+1/n)-f(x,y)}{1/n}\bigg \vert\\ = & \bigg \vert \frac{ \int_y^{y+1/n} \partial_y f(x,y) dy }{1/n}\bigg \vert\\ \leq & \frac{ \int_y^{y+1/n} |\partial_y f(x,y)| dy }{1/n}\\ \leq & g(x). \end{align}

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  • $\begingroup$ This is a cool way, thanks for presenting. The last inequality is bugging me though. Mind explaining it? I know that $|\partial_{y}f(x,y)| \leq g(x)$. $\endgroup$ – user853982 Jan 4 at 14:47
  • $\begingroup$ Yes, thank you! the last one is because $\int_y^{y+1/n} dy=(y+1/n)-y=1/n$. $\endgroup$ – Leon Jan 4 at 16:42
  • $\begingroup$ Ahhh, I see. The $g(x)$ is not a function of $y$. $\endgroup$ – user853982 Jan 4 at 16:54

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