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Let $f:[0,1] \times [0,1] \to \mathbb R$ be a function such that:

(a) for each $x \in [0,1]$, the function $y \to f(x,y)$ is Lebesgue integrable on $[0,1]$.

(b) $\dfrac{\partial f}{\partial x}(x,y)$ is a bounded function of $(x,y)$.

Show that for each $x$, the function $y \to \dfrac{\partial f}{\partial x}(x,y)$ is measurable and $\dfrac{d}{dx} \int_0^1 f(x,y)dy=\int_0^1 \dfrac{\partial f}{\partial x}(x,y)dy$.

I am a bit stuck on the exercise, to prove that $g_x(y)=\dfrac{\partial f}{\partial x}(x,y)$ I did the following:

If for each $x$, $f_x(y)=f(x,y)$ is integrable, by definition, it is measurable. So given $x$ in [0,1), the function $f_{x+\frac{1}{n}}$ is measurable for each $n>\dfrac{1}{x+1}$ and the function $f_{1-\frac{1}{n}}$ is measurable for $x=1$, but then for $0 \leq x <1$ we have $$\dfrac{\partial f}{\partial x}(x,y)=\lim_{n \to \infty} \dfrac{f(x+\frac{1}{n},y)-f(x,y)}{\frac{1}{n}}$$

and for $x=1$, $$\dfrac{\partial f}{\partial x}(x,y)=\lim_{n \to \infty} \dfrac{f(x-\frac{1}{n},y)-f(x,y)}{-\frac{1}{n}}$$

Since each function is a pointwise limit of measurable functions, then for each $0 \leq x \leq 1$, $\dfrac{\partial f}{\partial x}(x,y)$ is a measurable function.

I am not sure if my reasoning is correct and I don't know what to do for the last part of the exercise. Any help would be greatly appreciated. Thanks in advance

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    $\begingroup$ You've done well on this, but to finish, you need to use the hypothesis that a certain partial derivative is bounded. Hint: MVT $\endgroup$
    – zhw.
    Commented Sep 22, 2015 at 3:36
  • $\begingroup$ Given $x \in [0,1]$, $\dfrac{\partial f}{\partial x}(x,y) \leq M$ for all $y \in [0,1]$ and the constant function $M$ is integrable on $[0,1]$. If for each $0 \leq x <1$, I define $f_n(y)=\dfrac{f(x+\frac{1}{n},y)-f(x,y)}{\frac{1}{n}}$ , then $\lim_{n \to \infty} f_n(y)=\dfrac{\partial f}{\partial x}(x,y)$. By the domintated convergence theorem, $\int_0^1 \dfrac{\partial f}{\partial x}(x,y)dy=\lim_{n \to \infty} \int_0^1 \dfrac{f(x+\frac{1}{n},y)-f(x,y)}{\frac{1}{n}}dy= \lim_{n \to \infty} \dfrac{1}{n} (\int_0^1f(x+\frac{1}{n},y)dy-\int_0^1 f(x,y)dy)=\dfrac{d}{dx}\int_0^1f(x,y)dy$. $\endgroup$
    – user16924
    Commented Sep 22, 2015 at 15:38
  • $\begingroup$ Thanks for your suggestion, is my solution correct? It didn't occur to me to use MVT, but if my answer is wrong or you had another idea using the mean value, I would like to know your alternative solution. $\endgroup$
    – user16924
    Commented Sep 22, 2015 at 15:41
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    $\begingroup$ You invoked the dominated convergence theorem, but what is your dominating function? This is the part where one would normally use the mean value theorem. $\endgroup$ Commented Sep 24, 2015 at 3:45
  • $\begingroup$ You're right, my argument wasn't correct. However, for each $x$, $f_x(y)=f(x,y)$ is integrable on $[0,1]$, so $|f_n(y)|=|\dfrac{f(x+\frac{1}{n},y)-f(x,y)}{\frac{1}{n}}|\leq\dfrac{1}{n}(|f(x+\frac{1}{n},y)|+|f(x,y)|) \in L^1([0,1])$, since $f_{x+\frac{1}{n}}(y)$ and $f_x(y)$ are integrable, then the function of the right member of the inequality is integrable $\endgroup$
    – user16924
    Commented Sep 24, 2015 at 3:54

2 Answers 2

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The measurability of $y\to\frac{\partial f(x,y)}{\partial x}$ is OK. For the differentiation under the integral sign - I follow Theorem 2.75 here.

Take an arbitrary $x_0\in (0,1)$ and denote $\Phi(x)=\int\limits_{0}^{1}{f(x,y)dy}\,$. Let $\{x_n\}_{n=1}^{\infty}\subset [0,1]$ is an arbitrary sequence tending to $x_0$: $$\left(\frac{d}{dx}\int\limits_{0}^{1}{f(x,y)dy}\right)_{|x=x_0}=\frac{d\Phi}{d x}(x_0)=\lim\limits_{n\to\infty}{\frac{\Phi(x_n)-\Phi(x_0)}{x_n-x_0}}=\\ \lim\limits_{n\to\infty}{\int\limits_{0}^{1}{\underbrace{\frac{f(x_n,y)-f(x_0,y)}{x_n-x_0}}_{g_n(y)\to\frac{\partial f}{\partial x}(x_0,y)\,\,a.e\,\,y\in [0,1]}dy}}$$ So we have a pointwise a.e (in $y$) convergence of the sequence of functions $g_n(y)$ under the last integral to the function $\frac{\partial f}{\partial x}(x_0,y)$. Now to apply Lebesgue Dominated Convergence theorem we only need this sequence to be bounded by some summable function in $y$. We use the MVT: $$\left | \frac{f(x_n,y)-f(x_0,y)}{x_n-x_0}\right |=\left | \frac{\partial f}{\partial x}(\xi_n,y)\right|\leq M$$ where $\xi_n$ is between $x_n$ and $x_0$, and $M=const$ is some upper bound for $\frac{\partial f}{\partial x}(x,y)$ in $[0,1]\times[0,1]$. Here $M$ is summable because $[0,1]$ has finite measure. Now we apply LDC theorem to the sequence $$g_n(y)=\frac{f(x_n,y)-f(x_0,y)}{x_n-x_0}\to \frac{\partial f}{\partial x}(x_0,y)$$ and take the limit under the integral sign: $$\left(\frac{d}{dx}\int\limits_{0}^{1}{f(x,y)dy}\right)_{|x=x_0}=\lim\limits_{n\to\infty}{\int\limits_{0}^{1}{g_n(y)dy}}=\int\limits_{0}^{1}{\lim\limits_{n\to\infty}{g_n(y)}dy}=\int\limits_{0}^{1}{\frac{\partial f}{\partial x}(x_0,y)dy}$$ Finally, since $x_0\in(0,1)$ was arbitrary, the result follows.

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Measurability of $\partial_x f(x, \cdot)$ is correct. So I will only say something about $\frac{d}{dx} \int_0^1 f(x,\cdot) = \int_0^1 \frac\partial{\partial x} f(x, \cdot)$.

Your approach is correct, just little bit incomplete. You can't use a sequence as dominating "function" if its $L_1$ norm is unbounded.

Assume $\frac\partial{\partial x} f$ is bounded by $M$. Fix some $x\in[0,1]$. Let $0\ne h_n\to 0$ and define $f_n(y) = \frac{f(x+ h_n, y) - f(x,y)}{h_n}$ for $n\in\mathbb N$ and $y\in[0,1]$. By mean value theorem we have $$ |f_n(y)| = \left|\frac\partial{\partial x} f(\xi_n(y), y) \right| \le M $$ for a suitable $\xi_n(y)$ in the open interval span by $x$ and $x+h_n$ which depends on $n$ and $y$. That's $f_n$ is dominated by $M$. Since the integral domain has finite measure, the constant function $M$ is integrable and the dominated convergence theorem applies: \begin{align} \lim_{n\to\infty} \frac{1}{h_n} \left( \int_0^1 f(x+h_n, \cdot) - \int_0^1 f(x, \cdot) \right) &= \lim_{n\to\infty} \int_0^1 f_n \\ &= \int_0^1 \lim_{n\to\infty} f_n = \int_0^1 \frac\partial{\partial x} f(x, \cdot). \end{align} Since $h_n$ was arbitrary, it follows that $x\mapsto \int_0^1 f(x, \cdot)$ is differentiable at $x$ with $$ \frac{d}{dx} \int_0^1 f(x, \cdot) = \int_0^1 \frac\partial{\partial x} f(x, \cdot). $$

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