1
$\begingroup$

I'm trying to show the following equality using a change of integration order, but I'm unable to complete it:

$$ \int_y^1\int_0^t f(x)\,\mathrm dx\,\mathrm dt \stackrel{?}{=}\int_0^1 k(y,x)f(x)\,\mathrm dx$$

for some $k \in C([0,1]^2)$.

where $f \in C[0,1]$.

Here's what I've done so far (I might have a mistake in the change of order limits, but the idea was to capture the fact that $t$ depends either on $x$ or $y$, depending on the which is larger):

$$ \int_y^1\int_0^t f(x)\,\mathrm dx \,\mathrm dt = \int_0^1 \int_x^1 f(t)\chi_{[y,1]}(t)\,\mathrm dt\,\mathrm dx$$

(where $\chi_{[y,1]}$ is the indicator function).

EDIT: A closer option is splitting by whether $x<y$ or not, something as such:

$$ k(x,y) := \begin{cases} \int_x^1 f(t)dt &\quad\text{if } 0\leq y \leq x\\ \int_y^1 f(t)dt &\quad\text{if } x< y \leq 1 \\ \end{cases} $$

But I'm still missing the "$f(x)$" term on its own, as in the RHS of the equality.

This is as close as I got to showing there's some $k$ that fits the bill on the left-hand-side in the equality, but I'm not sure how to proceed.

Any ideas?

$\endgroup$
2
  • $\begingroup$ Do you want single $k$ that works for any $f$, or can $k$ depend on $f$? $\endgroup$
    – mihaild
    Commented Jun 15, 2022 at 17:54
  • $\begingroup$ @mihalid It has to be independent of $f$ $\endgroup$
    – Anon
    Commented Jun 15, 2022 at 17:55

1 Answer 1

1
$\begingroup$

Let $$f_n(x) = \begin{cases} 0,\ x < p - \frac{1}{n}\\ 1,\ x > p + \frac{1}{n}\\ \frac{x - p + \frac{1}{n}}{\frac{2}{n}} \end{cases}$$ Obviously, $f_n \to \chi_{[0, p]}$, and for both left and right side limit can be moved under integral. So, our property should hold for $f(x) = \chi){[0, p]}$ too.

Right side becomes $\int_0^p k(y, x)\, dx$.

If $y > p$, left side becomes $p - py$. If $y \leq p$, left side becomes $(1 - p)\cdot p + \int_y^p t\, dt = p - p^2 / 2 - y^2 /2$.

So, for any $p$, we have

$$\int_0^p k(y, x)\, dx = \begin{cases} p - p^2 / 2 - y^2 / 2,\ y \leq p\\ p - py,\ y > p \end{cases}$$

Differentiating both sides by $p$, we get $$k(y, x) = \begin{cases} 1 - x, y \leq x\\ 1 - y, y > x \end{cases}$$

Or, combining, $k(y, x) = 1 - \max(x, y)$ (also, you can get the same $k$ substituting $f(x, y) = 1$ in your last result).

The same reasoning shows that this $k$ works:

  1. It works for $\chi_{[0, p]}$ (by derivation)
  2. If the equality is true for $f$ and $g$ then it's true for $f + g$ (easy to check)
  3. If $f_n$ are all bounded by constant $M$, and $f_n \to f$ a.e., and the equality works for each $f_n$, then it works for $f$ (any theorem about convergence should work)
  4. From 1 and 2, it works for any indicator of any segment $\chi_{[q, p]}$.
  5. Any continuous function can be approximated by linear combination of indicators of segments.
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .