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Let $f : \mathbb{R}^2 \to \mathbb{R}$ be a function of two variables. Now, consider the function $g(x) = \int f(x, y)\; dy$. It is known that $g$ has exponential growth, that is, for large values of $|x|$, it behaves like a function of the type $ae^{b|x|}$, where $a, b>0$. I am wondering whether we can say that there exists $y_0$ such that $f(x, y_0)$ also grows exponentially in $x$, or can $f(x, y)$ have subexponential growth for all $y$? To make it very simple: can a family of slowly growing functions be "averaged" to find a function of faster growth?

I suspect that this might not be strictly true, but unable to come up with a counterexample. Any help would be appreciated.

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2 Answers 2

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Since you integrate over an unbounded interval, your function can even be globally bounded. For instance

$$f(x,y)=\begin{cases}1&\text{if $0\le y\le e^{\lvert x\rvert}$,}\\ 0&\text{otherwise,}\end{cases}$$

is such an example.

In contrast, it cannot happen that $\lvert f(x,y)\rvert\le a(y)b(x)$ where $a$ is integrable and $b$ has subexponential growth; if you integrate only over a bounded interval, this can be interpreted as a very weak form of “$f$ must grow exponentially w.r.t. $x$”.

However, even if you integrate over the bounded interval $[0,1]$, it is still possible that for every $y$ there holds $f(x,y)\ne0$ only if there is some natural number $n$ with $y\in(\frac1{n+1},\frac1n)$ and $x\in(n,n+1)$. In particular, for every $y\in(0,1]$ there holds $f(x,y)=0$ for $x>1+\frac1y$ and thus $$\lim_{\lvert x\rvert\to\infty}\,f(x,y)=0\quad\text{for every $y$.}$$ In fact, if $y\in(\frac1{n+1},\frac1n)$ and $x\in(n,n+1)$ just put $f(x,y):=n(n+1)2^n$ (and $f(x,y):=0$ otherwise). You can even make $f$ smooth on the rectangle $[\frac1{n+1},\frac1n]\times[n,n+1]$ by letting it smoothly go down to $0$ near the boundary of that rectangle. Then $f$ is even smooth. To see the smoothness at the points $(x_0,0)$ (with $x_0\ge0$), note that $f(x,y)=0$ if $y<\frac1{x_0+3}$ and $x<x_0+1$.

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  • $\begingroup$ Very nice answer! Just out of curiosity, suppose one posits that the integration is happening on a fixed bounded interval $I$ and further that $f(x, y)$ is jointly continuous in $x, y$. Are there still counterexamples of the above sort? $\endgroup$
    – piper1967
    Commented Jun 8 at 17:59
  • $\begingroup$ I realized that you do not need the typewriter sequence and can have an even stronger result. I modified the answer to sketch even a smooth function with the claimed property. $\endgroup$ Commented Jun 9 at 2:26
  • $\begingroup$ @MartinVäth Do you know if the opposite is possible? Integrating "faster" functions to find "slower" functions? It seems that this should not be possible, and one should be able to write a proof via the definition of Lebesgue integration (supremum of integrals of simple functions). Am I missing something? $\endgroup$
    – SMS
    Commented Jun 9 at 14:07
  • $\begingroup$ @SMS: It is probably not easy to formulate a result as you mean, but probably Fatou's lemma would be the essential tool in the corresponding proof. To settle the idea, consider a fixed sequence $x_n\to\infty$, and apply Fatou's lemma with $f_n(y)=f(x_n,y)/e^{x_n}$. If, for instance, for every $y$ there holds $\liminf_{n\to\infty}f_n(y)=a(y)>0$, then Fatou's lemma implies $\int f_n(y)dy\ge c=\int a(y)dy>0$ for all large $n$ which means $\int f(x_n,y)dy\ge ce^{x_n}$ for all large $n$. $\endgroup$ Commented Jun 9 at 16:30
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Not a complete answer but maybe this helps:

I don't know how explicitely you want to show exponential growth, but you could attempt showing that $f(x,y_0)$ grows faster than any polynomial. Assume $f(x,y)$ is bounded by some function $p(x,y)$ which is a polynomial in $x$ (it looks like a polynomial in $x$ if we view $y$ as a constant). The function $p$ would look like $p(x,y)=a_0(y)+a_1(y)x+\dots+a_n(y)x^n$, where the $a_i:\mathbb{R}\rightarrow \mathbb{R}$ are functions of $y$. Then we have $$ g(x)= \int f(x,y)dy\leq \int p(x,y) dy=\int a_0(y)dy+x\int a_1(y)dy+\dots+x^n\int a_n(y)dy$$ The right side looks again like a polynomial (provided the $a_i$ were integrable). Then $g(x)$ would also be bounded by a polynomial (a contradiction).

If you can show that such integrable $a_i$ exist you would be done.

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