Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The problem is the following:

Let $a,b,c,d \in \mathbb R$ be given such that $a<b$ and $c<d$. Suppose $f: [a,b]\times [c,d] \to \mathbb R$ is a function such that $\partial_1 f: [a,b]\times [c,d]\to\mathbb R$ exists and is (Lebesgue-)integrable.

Show that for $t\in [a,b]$ we have

$$\frac{\mathrm d}{\mathrm dt}\int_c^d f(t,x) \, \mathrm dx = \int_c^d \partial_1 f(t,x)\, \mathrm dx$$

I have tried the following: By the fundamental theorem of calculus, we have $f(t',x) - f(t,x) = \int_{t}^{t'} \partial_1f(s,x)\, \mathrm ds$. It follows that

$$ \begin{align} \int_c^d f(t',x)-f(t,x) \, \mathrm dx &= \int_c^d \int_{t}^{t'} \partial_1f(s,x)\, \mathrm ds \, \mathrm dx \\ &= \int_{t}^{t'} \int_c^d \partial_1f(s,x)\, \mathrm dx \, \mathrm ds \end{align} $$

so that for almost all $t$, we have

$$ \begin{align} \frac{\mathrm d}{\mathrm dt}\int_c^d f(t,x) \, \mathrm dx &= \lim_{t'\to t} \frac{1}{t'-t}\int_c^d f(t',x)-f(t,x) \, \mathrm dx \\ &= \lim_{t'\to t} \frac{1}{t'-t}\int_{t}^{t'} \int_c^d \partial_1f(s,x)\, \mathrm dx \, \mathrm ds \\ &= \int_c^d \partial_1f(t,x)\, \mathrm dx \end{align} $$

since $t \mapsto \int_c^d \partial_1f(t,x)\, \mathrm dx$ is an $L^1$ function (hence almost every point $t$ is a Lebesgue-point).

Unfortunately equality for almost every $t$ is not good enough in this case.

I have been thinking about this for quite a while now, but I cannot figure out the reason for why every point $t\in [a,b]$ has to be a Lebesgue point of

$$t\mapsto \int_c^d \partial_1f(t,x)\, \mathrm dx$$

I have also been thinking about possible counterexamples, but couldn't really come up with one. (At least not if one defines $\frac{\mathrm d}{\mathrm dt} \infty = 0$ for the constant function $\infty$...)

Some help would be appreciated, thanks!

share|improve this question
add comment

2 Answers 2

up vote 4 down vote accepted

You need $\partial_1f(t,x)$ be dominated by an integrable function, i.e., you need $|\partial_1f(t,x)|\le g(x)$ for every $x,t$ for you to take the limit $t'\to t$ inside the integral for every $t$.

Just $\partial_1f(t,x)$ being integrable over $[c,d]$ for every fixed $t\in [a,b]$ isn't sufficient.

Edit:

Let me clarify things here. The problem is that you can not straightaway write $$\frac{\mathrm d}{\mathrm dt}\int_c^d f(t,x) \, \mathrm dx = \lim_{t'\to t} \frac{1}{t'-t}\int_c^d f(t',x)-f(t,x) \, \mathrm dx $$ unless you have already shown that $\int_c^d f(t,x) \, \mathrm dx$ is differentiable w.r.to $t$.

So, let $t_n$ be a sequence converging to $t$ and let $$g_n(x)=\frac{f(t_n,x)-f(t,x)}{t_n-t}$$ Then $\lim g_n(x)=\partial_1 f(t,x)$ (given).

Then by mean value theorem $$|g_n(x)|\le \sup_{t\in [a,b]}|\partial_1 f(t,x)|\le g(x)$$ so DCT applies to $g_n(x)$.

Therefore $$\lim_{t_n\to t} \frac{1}{t_n-t}\int_c^d [f(t_n,x)-f(t,x)]\mathrm dx$$ exists and is equal to $$ \int_c^d \lim_{t_n\to t}\frac{f(t_n,x)-f(t,x)}{t_n-t}\mathrm dx=\int_c^d \partial_1f(t,x) \, \mathrm dx$$

I have just reproduced here what I once learnt from Folland's Real Analysis book.

share|improve this answer
    
O.k., now I understand what you are saying. And I agree that the additional assumption of $\partial_1f(t,x)$ being uniformly bounded in $t$ by some $g(x) \in L^1[c,d]$ would be sufficient. Nevertheless, what I have written is true - as I stated - for almost all $t$. If you could provide a counterexample to prove: "Just ∂1f(t,x) being integrable over [c,d] for every fixed t∈[a,b] isn't sufficient." then that would be great! –  Sam Dec 28 '11 at 10:19
    
How can one write $\frac{\mathrm d}{\mathrm dt}\int_c^d f(t,x) \, \mathrm dx= \lim_{t'\to t} \frac{1}{t'-t}\int_c^d f(t',x)-f(t,x) \, \mathrm dx$ before knowing whether the limit exists. –  Ashok Dec 28 '11 at 10:36
    
It's existence is justified for almost all $t$ by the next line after that and the fact that $$t\to \int_c^d \partial_1f(t,x) \, dx$$ is an $L^1$ function (so almost every $t$ is a Lebesgue-point ). –  Sam Dec 28 '11 at 11:27
    
ah I see. Good point. –  Ashok Dec 28 '11 at 11:34
add comment

I have found a non-trivial counterexample:

Let $I = [0,1]$. I will construct a counterexample $f \in L^1(I\times I)$. Showing that additional assumptions along the lines of

  • For every $t_0\in I$ there exists $g\in L^1(I)$ and a neighbourhood $U$ of $t_0$, s.t. for all $t\in U$ we have: $$|\partial_1f(t,x)| \le g(x) \qquad\text{almost everywhere}$$

cannot be dispensed with (if we want the claimed equality to hold for all $t$ rather than almost all $t$).

We define $f: I\times I\to \mathbb R_{\ge 0}$ by

$$f(t,x) = \sum_{n=1}^\infty \chi_{I_n}(x) \beta_n(t)$$

where

  • $I_n = (2^{-n}, 2^{-n+1})$ and $\chi_{I_n}$ denotes the characteristic function of $I_n$,
  • $\beta_n:I \to \mathbb R$ is differentiable with
    • $0\le \beta_n(t) \le \dfrac{2^n}{n(n+1)}$, $0\le \beta_n'(t)$ for all $t\in I$,
    • $\beta_n(t) = \begin{cases} 0 & 0\le t \le \dfrac{1}{2n} \\ \dfrac{2^n}{n(n+1)} & \dfrac 1n \le t \le 1 \end{cases}$

i.e. every $\beta_n$ is a monotonic differentiable function which is constant near $t=0$.

For a fixed $x$ at most one summand in the definition of $f(t,x)$ is not equal to zero, so there are no problems with convergence.

Thus we obtain $\partial_1f(t,x) = \sum_{n=1}^\infty \chi_{I_n}(x) \beta_n'(t)$ and - using nonnegativity of all summands

$$ \begin{align} \int_0^1\int_0^1 |\partial_1f(t,x)|\, dt\, dx &= \int_0^1\int_0^1 \left(\sum_{n = 1}^\infty \chi_{I_n}(x)\beta_n'(t)\right) \, dt\, dx\\ &= \sum_{n=1}^\infty |I_n| (\beta_n(1) - \beta_n(0)) \\ &= \sum_{n=1}^\infty 2^{-n} \frac{2^n}{n(n+1)} \\ &= \sum_{n=1}^\infty \left(\frac1n - \frac1{n+1}\right) \\ &= 1 < \infty \end{align} $$ Similarly a quick calculation shows $$ \int_0^1\int_0^1 |f(t,x)|\, dt\, dx \le 1 < \infty $$ Therefore $f, \partial_1f \in L^1(I\times I)$. But now we observe $$ \begin{align} \int_0^1 \frac{f(1/m, x)-f(0,x)}{1/m} \, dx &= m \int_0^1 \left(\sum_{n=1}^\infty \chi_{I_n}(x)[\beta_n(1/m) - \beta_n(0)]\right)\, dx \\ &= m\sum_{n=1}^\infty |I_n|\beta_n(1/m) \\ &\ge m \sum_{n=m}^\infty 2^{-n} \frac{2^n}{n(n+1)} \\ &= m\cdot \frac{1}{m} = 1 \end{align} $$ for all $m\in \mathbb N$. In particular, with $\partial_1f(0,x) = \sum_n \chi_{I_n}(x) \underbrace{\beta_n'(0)}_{=0\; \forall n} = 0$ we see:

$$\limsup_{t\to 0} \int_0^1 \frac{f(t, x)-f(0,x)}{t} \, dx \ge 1 \ne 0 = \int_0^1 \partial_1f(0,x) \, dx$$

Showing that $f$ is a counterexample as claimed.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.