# Why can't I interchange integration and differentiation here?

I think my questions relates to this other: "counterexamples to differentiation under integral sign"

In fact, it provides a counterexample

Consider $f(x,y)=y^3e^{-y^2x}$ and define $F(y) =\int_0^{\infty}f(x,y)dx$

We have that $F'(0)\not = \int_0^{\infty} \frac{\partial f}{\partial y}(x,0)dx$

(We calculate $F'(0)$ essentially using Monotone convergence theorem we can show that, for $y\in \mathbb{R}\setminus\{0\}$, $F(y)=y$ moreover $F(0)=0$ so $F'(0)=1$)

Now, I want to understand which hypothesis of Theorem 2 at this page does not hold. Obviously the third hypothesis does not hold, but I want to consider the case in which we replace it by a weacker condition:

"For each $b \in \mathbb{R}$, there exists an open interval $b\in J$ and an integrable function $g(x)$ over $(0, \infty)$ ,such that $| \frac{\partial f}{\partial y}(x,y)| \leq g(x)$ for every $y\in J$ and $\forall x$"

Now, the first hypothesis certainly holds as $\forall y, \ x\rightarrow f(x,y)$ is integrable $(0,\infty)$ by comparison with $e^{-kx}$ for appropriate positive value of $k$

Moreover $\frac{\partial f}{\partial y}(x,y)$ exists everywhere...

So is the last hypothesis to be problematic but I can't see how as I can bound $y$ in $J$ and then just use some linear combination of $e^{-kx}$ and $xe^{-lx}$ for suitable $k,l$ as they are both integrable over $(0,\infty)$

Thank you very much!

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How are you going to pick k and l in such a way that the limit y->0 is covered? This is not a research level question. –  Michael Renardy Dec 15 '12 at 1:44
Moritzplatz, note that $(F(y)-F(0))/y=\int_0^\infty y^2e^{-y^2x}dx$. It's the integral of $e^{-x}$ after rescaling. Rescaling is in fact the easiest way to make a counterxample to the dominated convergence theorem (then $F(y)$ has been defined consequently). If you make a picture of the graphs of $x\mapsto y^2e^{-y^2x}$ and compute $\sup_y y^2e^{-y^2x}$ it will be clear why the integrands are not dominated in $L^1$. –  Pietro Majer Dec 15 '12 at 8:43
Thanks pietro, I think I've got closer to understand the problem. Is the fact that for any $x$ big enough i can find an $y_0$ close enough to zero such that $y_0^2e^{-y_0^2x}=1/x$ and then use the fact that $1/x$ is not integrable on $(k,\infty)$ for any $k$ ? thank you very much! –  Moritzplatz Dec 15 '12 at 12:49
Precisely: there is no $g\in L^1(\mathbb{R}_+)$ such that $y^2e^{-y^2}\le g(x)$ for all $x$ and $y$, because computing the supremum you get $1/x\le g(x)$. Note that even extracting a sequence $y_n\to0$, you would always get a supremum $\sup_ n y^2 e^{-y^2 _n x}$ not integrable on $\mathbb{R}_+$. You may verify that, by a direct computation, but the simplest reason is: otherwise the dominated convergence theorem would give convergence in $L^1$ to the pointwise limit, which is $0$, whereas these functions all have $L^1$ norm equal to $1$. –  Pietro Majer Dec 15 '12 at 13:45
thank you very much, you have been very helpfull and I think I understand this problem a lot better now! –  Moritzplatz Dec 15 '12 at 13:56
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You need a convenient locally convex space $E$ of functions in $x$ such that $g\mapsto \int_0^\infty g(x)\,dx$ is a bounded linear functional, and such that $y\mapsto f(\quad,y)$ is a differentiable curve in $E$, at least of Hoelder class $C^{1,\alpha}$. Then the interchange follows by the chain rule. The space $\mathcal S(-1,\infty)$ of smooth functions which decrease rapidly on the right suggests itself. This is a Frechet space with seminorms $sup \{(1+|x|^2)^m\partial_x^{k} g(x): -1<x<\infty\}$ for fixed $m,k$

Now we look at the paper: Peter W. Michor and David Mumford: A zoo of diffeomorphism groups on $\mathbb ℝ^n$. arXiv:1211.5704. (pdf)

By repeating the proof of 3.5 there (which uses the very strong theorem 2.4 of Frölicher and Kriegl) we see that:

The space $C^\infty(\mathbb R,\mathcal S(-1,\infty))$ of smooth curves in $\mathcal S(-1,\infty)$ consists of all functions $c\in C^\infty(\mathbb R\times(-1,\infty),\mathbb R)$ satisfying the following property:

$\bullet$ For all $k,m\in \mathbb N_{\ge0}$ and $\alpha\in \mathbb N_{\ge0}^n$, the expression $(1+|x|^2)^m\partial_t^{k}\partial^\alpha_xc(t,x)$ is uniformly bounded in $x\in (-1,\infty)$, locally in $t\in \mathbb R$.

Obviously, your function $f$ satisfies this.

EDIT: The "obviously" is wrong. $(1+x^2) y^3 e^{-y^2x}$ is not uniformly bounded for $(x,y)\in (-1,\infty)\times[-1,1]$. To see this insert the curve $x(t) = 1/(1+t^2)$, $y(t)=t$ to get $$(1+x^2) y^3 e^{-y^2x}= \Big(1+\frac{1}{(1+t^2)^2}\Big) t^3 e^{-\frac{t^2}{1+t^2}}, \quad t\in[-1.1]$$ which is unbounded.

Let me try again: $g\mapsto \int_0^\infty g(x)dx$ is a bounded linear functional on $L^1(-1,\infty)$, but $y\mapsto f(\quad,y)$ is not $Lip^1$ into (otherwise you could interchange).

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Nice ! +1 for moving the goalposts. –  David Roberts Dec 15 '12 at 9:26
Thank you very much! I will save this answer and study the concepts you use (like Hoelder classes and Frechet spaces) so that I will hopefully be able to understand it. Anyway, are you saying that I actually can exchange integration and differentiation? How does this work with what I got at zero? Thank you again and sorry for being pedantic. –  Moritzplatz Dec 15 '12 at 10:42
@Moritzplatz: You are right, I did not read your question carefully enough. I shall edit my answer. –  Peter Michor Dec 15 '12 at 12:45