# Change order of integration to show an equality

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 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?

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

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.