## Quasi-bleg: Why are there bump functions?

When I was learning analysis (beyond, say, first-year calculus), one of the facts that most surprised me was the fact that there are functions that were smooth (i.e., infinitely differentiable) and yet compactly supported. Of course, I didn’t think about it with that phrasing; there’s a pretty simple geometric interpretation of smoothness for most functions one encounters in calculus (actually, one rarely sees differentiable functions that aren’t smooth!) Specifically, if a function isn’t smooth at $x$, then there’s some sort of a “kink” at that point, or at least “around” that point.

Is this justified? Well, not totally, but let’s give a couple of examples to at least show why it’s a good first approximation. We’ll start by modifying the canonical grade-school-calculus function that’s continuous but not differentiable: the absolute value function, $sgn(x) x$. Taking the antiderivative a couple of times and ignoring constant factors, we get $sgn(x) x^3$, which looks like this:

Although it’s hard to put into words, it’s sort of intuitively clear that there’s something not quite right about this graph — it’s sort of got the wrong shape to be symmetric about zero, it flattens too unexpectedly, just something. As we go to higher and higher classes of functions, this feeling gets weaker and more nebulous, but this is easily dismissed as a fault of our feeble, finite minds.

Now let’s look at a very different non-smooth function, specifically $x^2 sin(1/x)$:

(Legally required aside which I’d most likely include anyway: The above plots are courtesy of Wolfram Alpha, which has proven itself rather unexpectedly quite useful over the past few weeks. If you for whatever reason don’t have access to a system like Mathematica or Maple, or just don’t want to wait for it to start up for a simple, quick calculation, it’s probably the best general resource out there.)

Here, it’s a little clearer what’s the problem is: as $x$ goes to 0, the function is certainly going to 0 as well, but it’s too “spiky” to be smooth. One way to make this a little more formal is to note that the function changes sign infinitely many times in any neighborhood of 0 — and since it’s continuous, it therefore has infinitely many zeros in that neighborhood.

This is a big clue as to where this intuition about what a smooth function “should” be is coming from: it’s coming from what we know about polynomials! The real problem with the first graph is that it certainly doesn’t look like $x^3$, but it doesn’t look quite like $x^4$ either; it’s masquerading as a polynomial but isn’t one. And $sin(1/x)$ and its cousins behave like no polynomial in existence, having so many zeros in one place.

Of course, there are other smooth functions that don’t behave like polynomials — stuff like $e^x$ (which grows way too fast to be polynomial in one direction, and refuses to grow at all in the other) and $sin(x)$ (which has infinitely many zeros and is bounded). But here’s the key: they do behave like polynomials, locally! They have convergent Taylor series expansions (that agree with the function at least locally) that let them fool us into thinking that they’re polynomials at any given point, so they don’t upset our intuition.

There’s a name for this kind of function; they’re called (real) analytic. And, indeed, analytic functions have nice properties that agree with our intuition about smooth functions; for instance, the only compactly supported analytic function is the zero function.

So the question isn’t so much “Why are there bump functions?” as “Why are there non-analytic smooth functions?” And, unfortunately, I don’t have a very good answer to that. There are simple constructions of smooth functions that aren’t analytic, which are easy to check, but there’s still not an intuitive real-analytic reason as to why it’s there.

The obvious way to get a smooth function that isn’t analytic is to show that the function has a Taylor series around some point which obviously doesn’t describe the function at a neighborhood of the point. A good candidate Taylor series would be the identically-zero series; and, in fact, this is what the above-linked construction uses. But there are some conceptual problems. First, for the function not to agree with the Taylor series on a neighborhood of x, the derivatives should fall off quickly as they approach x from the right, say. But if the n-th derivative falls off quickly, then the (n+1)-th derivative would be expected to be rather large! The sequence of derivatives at a point close to x doesn’t seem to act like a “usual” sequence of functions.

So this is sort of a bleg: Readers, do you have a good explanation as to why these functions exist, that would be comprehensible from the viewpoint of a bright first- or maybe second-year calculus student?

In the meantime, I want to give a bit of an argument as to why these functions (at least bump functions) aren’t intuitively obvious. The reason is that they clear up some obstructions to the theory of distributions, which can be quite counterintuitive. Perhaps the central problem in analysis is approximation: we have a space of very nice functions which can be dealt with quite simply; the problem is how to deal with less-nice functions. The answer is often to show that the less-nice function can be approximated, in some sense, by a sequence of the nice functions. (Perhaps the quintessential example is Weierstrass’ approximation theorem, which is at least the one most often taught to undergraduates.)

So say we want to approximate some non-smooth function by a sequence of smooth functions. As I understand it (and I don’t know very much about real analysis at all, so I may be completely off), bump functions are used to construct objects called mollifiers, which let us do this very well — we can even approximate “infinitely sharp functions” like the Dirac delta distribution. (Mathworld has a nice .gif of this.) Furthermore, the notion of convergence is very strong. This seems like an unexpectedly strong result when you’re used to approximation by polynomials.

In addition, I’m not sure how intuitive of an idea this is, but Wikipedia tells me that the dual of the space of bump functions is the space of distributions. This sort of vaguely makes sense to me — these unexpected smooth functions being dual to these unexpected sharp functions — but I certainly can’t start to justify it with any degree of rigor. If anyone wants to take a crack at doing so, they’re more than welcome :).

One last thing: Qiaochu pointed out to me during a short conversation about this topic that it’s not difficult to see that there are non-analytic smooth functions via complex analysis. (This despite — maybe because of — the fact that, in the complex domain, all smooth functions are analytic!) The conceptual reason is that an essential singularity anywhere in the complex plane prevents the function from having an analytic continuation, but in general a one-dimensional slice can’t “see” the two-dimensional singularity. So we can take a slice of the complex function that doesn’t include the singularity — this is smooth if the original function was meromorphic on the slice, but may not be real analytic. (So, why are there essential singularities? Well, one conceptual way to construct one relies on the fact that $1/z$ maps every neighborhood of the origin to a very non-compact set, so if we compose this with a function, like $e^z$, that behaves very differently as we approach infinity in different directions, we’ll get something that has really pathological behavior around the origin. Not so surprisingly, then, one of the standard examples of a smooth function that’s not analytic is related to $e^{1/x}$). This is quite a nice explanation, but isn’t really what I want. So I’ll be a blegger and a chooser: does anyone have an explanation that isn’t so complex?

Keepin’ it real,

Harrison

### 5 Responses to “Quasi-bleg: Why are there bump functions?”

1. Akhil Mathew Says:

I’m pretty sure you can use the Baire category theorem to show that there are C-infinity functions which are nowhere analytic, and moreover they form a residual set in the space of C-infinity functions, with (I think) the topology defined by the metric: $d(f,g) := \sum_n \frac{1}{2^n} \sup \left( \frac{ |f^{(n)} - g^{(n)}| }{ 1 + |f^{(n)} - g^{(n)}| } \right)$.

• Harrison Says:

Unfortunately I don’t know nearly enough analysis to be able to tell if that works. (Someone who does know analysis?) If it does, it’s still not entirely satisfactory, since the Baire category theorem is totally out of reach of your generic very bright high-schooler or reasonably bright undergraduate. (I’m actually not even convinced that its statement should be obvious, even for $\mathbb{R}$, but there might be a different formulation that is.)

There’s also the fact that the Baire category theorem is serious overkill for this problem, but I’m okay with that :).

• Akhil Mathew Says:

Well, you did say you wanted to avoid complex analysis, hence the choice of Baire’s theorem. I recall reading it in _Topology,_ by Dugundji, but I’d have to think for some time before I could actually reconstruct it.

I think another observation that may be helpful is that in general, the derivatives will grow too fast for the Taylor series to converge (that is more or less the idea behind the Baire category proof), but this doesn’t happen for complex functions because of the Cauchy formulas.

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3. colinwytan Says:

Regarding your belief that bump functions (unexpected smooth functions) give rise to distributions (unexpected sharp functions), let me try to be the devil’s advocate.

First I give the heuristic definition that distributions are those “functions/fictions” (perhaps not always smooth) that you can do analysis with. And what kind of analysis, you might ask? Let’s say, integration and differentiation. Basic calculus.

Given that bump functions can approximate the Dirac delta finction, using your analogy, distributions should approximate functions. Tada! By approximate, what I mean is that distributions may not have notion of eat-a-value-spit-a-value, but in the “limit” they should have pointwise values just like functions, who eat values and spit values. And what is this “in the limit” you might ask? This means, that to the extent that we will to do calculus, say local operations like integration and differentiation, we can do it. But ask for pointwise information, and you will get fiction (what you call the “counterintuitive” properties of the Dirac).

I’m still pondering about your question as to why there are bump functions, but perhaps let me just say that your question of “why are there non-analytic smooth functions” reduces to “why there are bump functions”. This is because bump functions “generate” all smooth functions. Generate means that take a smooth function, multiply it with bump functions of larger and larger radius, and these will approximate your original smooth function.

As far as I can see, the question you are pondering amounts to asking why bump functions can exhibit non-local behaviour. That is, why infinitely differentiability, unlike analyticity, to cause the local vanishing of a function to effect a global vanishing.