# Necessary condition for square integrable functions?

I'm studying Quantum Mechanics and I came across this which I don't quite understand: For a vector space of functions $f(x)$ to be square integrable (i.e $\int{|f(x)|^2dx < \infty)}$, the necessary condition is $\lim_{|x| \to \infty}f(x) \to 0$. Can someone help me understand why that must be the case? I sort of understand that it must not blow up at the infinities but can a (continuous or piece-wise continuous) function have a non-zero finite value at the infinities and still be square integrable?

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Go back to the definition of an integral as the area under a curve. What does it look like if $f(x)$ tends to a nonzero limit, keeping in mind that the integrand is $|f(x)|^2>0$? –  Michael Brown Aug 28 '13 at 13:50
If you allow very badly behaved functions (things which don't even possess limits) you can have, for example, $f(x)=\sin(|x|)^{|x|}$ which you think might work since the width of the peaks decrease as $x\to\infty$, but in fact the integral of $f(x)^2$ still diverges, although slowly. Needless to say this is a horrible function that never comes up in physics. :) –  Michael Brown Aug 28 '13 at 14:14
Thanks Michael. That makes sense intuitively. I should have given it a little bit more thought. Is there a formal proof for it? I couldn't find one online. –  imundi Aug 29 '13 at 14:40
I believe this piecewise function is square integrable and is non-zero as $\vert x\vert\to 0$: $f(x) = \sqrt{\vert\sin(x)\vert}$ if $n\pi < x < (n/2)\pi$ and $f(x) = i\sqrt{\vert\sin(x)\vert}$ if $(n/2)\pi < x < (n+1)\pi$. –  OSE Aug 29 '13 at 21:49
Would math.stackexchange.com be a better home for this question? –  Qmechanic Aug 30 '13 at 3:20

Apart from not being sufficient to prove convergence of the integral $$\int |f(x)|^2\text dx<\infty,$$ having the vanishing limig $\lim_{x\rightarrow\infty}f(x)=0$ is only necessary for the convergence within a suitable class of "nice" functions.

Consider, for example, the function $$f(x)=\sum_{n=1}^\infty\chi_{\left[n,n+\frac1{n^2}\right]}(x)=\left\{\begin{array}&1\text{ if } n\leq x\leq n+1/n^2 \text{ for some }n=1,2,3,\ldots,\\0\text{ otherwise.}\end{array}\right.$$ (Here $\chi_A$ is the characteristic function of a set $A\subseteq\mathbb R$.) This function has a convergent $L^2$ integral but does not have a well-defined limit at infinity. While this function is not continuous, but using suitable bump functions you can make a similar $C^\infty$ function with the same properties. This is the kind of function you're allowing when you do not impose vanishing limits at infinity - i.e., pretty ugly.

More to the point, say your wavefunction obeys a stationary Schrödinger equation with energy $E$ for some potential $V$ such that $\lim_{x\rightarrow} V(x)>E$ (i.e. a bound state). Then you know that, at infinity, $f''(x)$ has the same sign as $f$, which we can assume to be positive. If $f'(x)$ is ever zero in that region, then you know that it will be positive for all $x$ after that and $f(x)$ will monotonically increase, in which case the $L^2$ integral has no chance of convergence. In this particular setting, then you can restrict yourself to monotonically decreasing functions, and those are nice enough that the vanishing limit at infinity is necessary for $L^2$ convergence.

(To be followed by a more rigorous argument if I find the time.)

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Your ugly function is the type of thing I was aiming at in my comment above, but didn't quite get. :) Good example. I think you mean the inequality to go the other way, i.e. $E<V(x)$, but nice argument. –  Michael Brown Aug 30 '13 at 4:21

Emilio Pisanty and Eckhard Giere have already given discontinuous, piecewise constant counterexamples in their answers. Here we provide for-the-fun-of-it a smooth infinitely-many-times-differentiable counterexample $f\in C^{\infty}(\mathbb{R})$ of a square integrable function $f:\mathbb{R} \to [0,1]$ that does not satisfy $\lim_{|x|\to \infty}f(x)=0$. Our counterexample is

$$\tag{1} f(x)~:=~ e^{- g(x)} ~\in ~]0,1], \qquad g(x)~:=~x^4 \sin^2 x~\in ~[0,\infty[.$$

Intuitive idea: If we imagine $x$ as a time variable, then the function $f$ returns periodically to its maximum value

$$\tag{2} f(x) =1 \quad\Leftrightarrow\quad g(x) =0 \quad\Leftrightarrow\quad \frac{x}{\pi}\in \mathbb{Z} ,$$

but spends most if its time close to the $x$-axis in order to be square integrable.

Proof: We leave a detailed rigorous epsilon-delta mathematical proof to the reader, but a sketched heuristic proof goes like this. For each very large integer $|n|\gg 1$, define a shifted variable

$$\tag{3} y~:=~x-\pi n.$$

For the fixed integer $n\in\mathbb{Z}$, always assume from now on that the $y$-variable belongs to the interval

$$\tag{4} |y|~\leq~ \frac{\pi}{2}.$$

For $|y|\ll\frac{\pi}{2}$ very small, we may approximate $g(x) \approx (\pi n)^4y^2$, so that in the interval (4), we have

$$\tag{5} g(x)~\lesssim~ \pi^4 |n| \quad \Leftrightarrow\quad |y| ~\lesssim~ |n|^{-\frac{3}{2}}.$$

Thus we may form a square integrable majorant function $h\geq f$ (outside a compact region on the $x$-axis) by defining

$$\tag{6} h(x)~:=~\left\{\begin{array}{lcl} 1 &{\rm for}& |y| ~\lesssim~ |n|^{-\frac{3}{2}}, \cr e^{-\pi^4 |n|}&{\rm for}& |n|^{-\frac{3}{2}}~\lesssim~ |y| ~\leq~ \frac{\pi}{2}, \end{array} \right. \qquad |n|\gg 1.$$

The function $h\in {\cal L}^2(\mathbb{R})$ is square integrable on the whole $x$-axis, since

$$\tag{7} \sum_{n\neq 0} |n|^{-\frac{3}{2}} ~<~ \infty$$

and

$$\tag{8} \pi \sum_{n\in\mathbb{Z}}e^{-2\pi^4 |n|}~<~\infty$$

are convergent series.

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Some simple example that illustrates that the condition $$\lim_{|x|\to \infty} f(x) = 0 \quad (1)$$ is not necessary. If the condition were necessary $f\in L^2$ would imply that the limit in (1) holds.

Take in dimension 1 the function $$f(x) = \sum_{n=2}^{\infty} \chi_{I_n}(x)$$ where $\chi_{I_n}$ is the characteristic function of the interval $I_n = [n-\frac{1}{n^2}, n+\frac{1}{n^2}]$ then the integral evaluates to $$\int |f(x)|^2 dx = \sum_{n=2}^{\infty} |I_n| = \sum_{n=2}^{\infty} \frac{2}{n^2} < \infty\ .$$ But the function does not converge to zero for $|x|\to \infty$.

Sorry: Forgot to center the intervals around n. Now corrected.

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Am I missing something here? The widest interval is the one for $n=1$, in which case $f(x) = 0,\,\forall |x| > 1$. –  WetSavannaAnimal aka Rod Vance Aug 29 '13 at 21:41
@WetSavannaAnimal aka Rod Vance You are right I didn't center the intervals around $n$ - corrected. Just saw that a correct answer has already be given. –  Eckhard Giere Aug 30 '13 at 4:21

Let $g(x)$ be a positive continuous function, such that $$\lim_{x\to\infty}g(\pm x)=0.$$ Suppose its $L^2$ integral converges. Now consider a function $f(x)=g(x)+C,\; C>0$. It'll have $$\lim_{x\to\infty}f(\pm x)=C.$$ Now its $L^2$ integral will look like this: $$\int_\mathbb{R}|f(x)|^2 dx=\int_\mathbb{R} g^2(x)dx+C\int_\mathbb{R} g(x)dx+C^2\int_\mathbb{R} dx$$ First integral converges, second is positive and may converge, but the third one obviously diverges. This means that $L^2$ integral of $f(x)$ also diverges. This should be easy to generalize for non-positive $g(x)$.

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Ok, so this shows that a function that is bounded below by $C>0$ and that converges to $C$ as $x\to\infty$ can not be in $L^2$. But you realise that this is a very special case, yes? –  UwF Aug 30 '13 at 8:00