Distributional Calculus Part 4: Properties of Distributions

So, in the previous post, we found that distributions gave an alternate way to characterize functions; that is, by mapping from a set of test functions instead of from a compact set $X$. Test functions turn out to be completely central to how operations are performed! In fact, I'll spoil the entire content of this post by saying any operation on $f$ can be 'moved' to apply on the set of test functions instead.

But before that, let's list some basic properties which are more evocative of elementary real analysis than anything else. For a distribution $\langle f, \phi\rangle$:

• Linearity, i.e. $f(a\phi_1+\phi_2)$ for any real constant $a$ and test functions $\phi_1,\ \phi_2$;
• There exists a sequence of test functions $\{\phi_n\}$ such that $\phi_n \to f$

All of these properties are necessary, but we'll be making the most use out of the second one. Recall the super-useful integral characterization of a distribution $$T_f(\phi)=\int_{\mathbb{R}}f(x)\phi(x)\,dx?$$ That can only be expressed if $f$ is a function with no weird generalized properties. Yet now, if we consider $f$ as the limit of a sequence of test functions, $\phi_n$ is a classically defined function for all $n$ and it is now possible to write  $$T_f(\phi)=\lim_{n\to\infty}\int_{\mathbb{R}}\phi_n(x)\phi(x)\;dx$$ for any generalized function $f$.* Great! Now we can look at any and all distributions the easy way.

The real magic starts when we attempt to translate the distribution. Recall that any function can be translated $y$ units by taking $f(x-y)$ instead of $f(x)$; the same thing can be done for generalized functions by considering $\lim_{n\to\infty}\langle \phi_n(x-y),\phi(x)\rangle$. (Let's define the translation function tau as $\tau_y\phi(x)=\phi(x-y)$.) Using some simple $u$-substitution magic, \begin{align}\langle\tau_yT_f,\phi\rangle &=\lim_{n\to\infty}\langle \phi_n(x-y),\phi(x)\rangle\\&=\lim_{n\to\infty}\int_{\mathbb{R}}\phi_n(x-y)\phi(x)\;dx;\qquad u = x-y\\&=\lim_{n\to\infty}\int_{\mathbb{R}}\phi_n(u)\phi(u+y)\;du\\&=\langle T_f,\tau_{-y}\phi\rangle.\end{align} We have essentially found that any distribution can be translated by applying the opposite translation to every test function in $\mathcal{D}$. To reiterate:$$\langle\tau_yT_f,\phi\rangle =\langle T_f,\tau_{-y}\phi\rangle.$$ Hooray!

Differentiating a distribution works in much the same way as translation in that the operation gets pawned off onto the test function but with an extra minus sign. However, it does involve an extra technique: integration by parts. I assume that nobody who is reading this is unfamiliar with the practice, but, for the sake of cute mnemonics, a friend of my fiancé's refers to $$\int u\;dv = uv - \int v \;du$$as "sudv uv svidoo."

Let's take a moment to appreciate how adorable that is.

The actual fancy differentiation trick can be proved in essentially one integration-by-parts step:\begin{align}\left\langle \frac{d}{dx}T_f,\phi\right\rangle &= \lim_{n\to\infty}\int_\mathbb{R}\left(\frac{d}{dx}\phi_n(x)\right)\phi(x)\;dx\\&= \lim_{n\to\infty}-\int_\mathbb{R}\phi_n(x)\left(\frac{d}{dx}\phi(x)\right)\;dx \\ &=\left\langle T_f, -\frac{d}{dx}\phi(x)\right\rangle\end{align}(brownie points if you've already figured out what happened to the $uv$ term). This identity is essential for a crazy number of distributional calculus proofs.

For example, we can directly use this identity to prove the Dirac delta function is the distributional derivative of the Heaviside function in two seconds. Let $T_H= \int_0^\infty \phi(x)\,dx$ represent the Heaviside distribution. Now, from the above identity, we conclude $$\left\langle \frac{d}{dx}T_H(x),\phi\right\rangle=\left\langle T_H(x),\frac{d}{dx}\phi\right\rangle=-\int_0^\infty \phi(x)\,dx=\phi(0)-\phi(\infty)=\phi(0),$$ that is, because $\phi$ is zero at infinity. Yet $\phi(0)=\langle \delta, \phi\rangle$ by definition! We're done here.

As super awesome as that is, there should be some material on how all this pertains to weak solutions of DEs up on Thursday. Woooo! This is basically my definition of a party!



* The MCT happened here. Shhh.

Convexity and You: Unpacking the Definition

Real Analysis is notorious for taking easy-to-understand concepts and repackaging them in a thick theoretical barrier. Take the epsilon-delta definition of continuity---it's impossible to prove anything with the information "the function, uh, doesn't have any holes," but it's impossible to develop a mental picture given only the theoretical perspective. For this reason, one of the biggest barriers to learning any type of analysis is properly connecting the intuitive idea and the theoretical representation.

We'll focus here on one of the less transparent definitions: convex functions. Convex functions can be understood intuitively as "the area above the function is a shape that doesn't go inwards on itself"... and theoretically as

Given convex set $X$, a function $f:X\to\mathbb{R}$ is convex if for all $x_1,\ x_2\in X$ and $t \in [0,1]$, $f(tx_1+(1-t)x_2)\leq tf(x_1)+(1-t)f(x_2)$.

What.

This is the part where, during an analysis course, you are expected to nod your head at the alphabet vomit (at least this time it's the Roman alphabet, not the Greek, that tossed its cookies). Let's make some sense out of what information is being conveyed.

First of all, to understand the definition of convex functions, you must know what convex sets are. A set is convex if any two points (call them $x_1$ and $x_2$) can be connected by a straight line that is contained in the set. If the set is not convex (i.e. "goes inwards" visually), then there will be at least two points whose connecting line goes outside the set.

Now the domain of $f$ is a convex set $X$, which should explain what the points $x_1$ and $x_2$ are doing in the definition: they correspond to the two arbitrary points that we want to try and connect with a line. This brings us to the purpose of defining $t \in [0,1]$. Consider the function $y(t)=tx_1+(1-t)x_2$. Since $y(0)=x_2$, $y(1)=x_1$ and $y$ itself is a linear functional, this function represents a straight line segment starting at $x_2$ and ending at $x_1$. Thus the purpose of $t$ is to create the parametrized line segment joining points $x_1$ and $x_2$.

We are given that $X$ is a convex set, so it is certainly true that the line $tx_1+(1-t)x_2$ is completely contained in $X$, the domain of $f$. This makes it completely legit to consider $f(tx_1+(1-t)x_2)$ as the image of this line. The image of a straight line in the domain won't necessarily be a straight line itself, but will instead be a path along the function starting at $f(x_2)$ and ending at $f(x_1)$. Hence the expression $f(tx_1+(1-t)x_2)$ is asking us to consider the section of $f(x)$ that connects* $f(x_1)$ and $f(x_2)$.

This brings us to the last part of the inequality
$$f(tx_1+(1-t)x_2)\leq tf(x_1)+(1-t)f(x_2).$$
Just as before, the second expression $tf(x_1)+(1-t)f(x_2)$ is representing a parametrized line segment, joining the points $f(x_2)$ and $f(x_1)$. We are now comparing two paths between $f(x_1)$ and $f(x_2)$: one is a straight line, and the other a path on the function. The inequality places a lower bound on where the straight line can be. If the straight line is above the path on $f$ everywhere---that is, if it satisfies the above inequality---it is contained in the area above $f(x)$ (the epigraph of $f$).

That's exactly the definition of a convex set, but applied to the space above $f$... cool.

Here's a picture for $X = \mathbb{R}$:

That's what the definition is communicating. I hope that was insightful for someone!




*(does not refer to connectedness in the mathematical sense)

Grading Stories: "Cheese Weight" and Thusforthwith

One thing I love about the internet is being able to share stories and moments from everyday life. Here are a couple about something I'm sure other academics will be able to relate to: grading stories.

Cheese Weight

My alma mater enforced mathematical writing guidelines and the use of $\LaTeX$ very strongly. Yet some people, notably non-majors, chose to ignore those guidelines completely and complain when points were taken off for writing. Some people handed in scratch work done in pen on graph paper in consistently gigantic writing. Some people *coughEigenpetercough* printed out the questions in $\LaTeX$... two problems to one page, in landscape form... then did them out by hand in tiny writing. Some people *coughalsoEigenpetercough* did the homework in $\LaTeX$ but omitted large amounts of information to fit every proof-based problem on one side of one page.

Then there are the people with just plain bad handwriting. While grading with a friend, I encountered a homework that exemplified this while grading a core class; apparently, one of the people in the class was secretly a chicken tied to a Ouija board. Here's how it went down.

Me: Hey, do you have any idea what these two words are?

Friend: ..........

Me: It looks like it says "cheese weight".

Friend: It does, but that doesn't have anything to do with the problem.

Me: Can you tell from context?

Friend: .... no.... (to another person) Hey, do you know what this says?

Someone else: ..... looks like "cheese weight"?

Friend: How about you?

Yet another person: I have no idea.

Me: Well, "cheese weight" it is then.

And that's how someone got their work back with "what's a cheese weight?" written as a comment.

Runner-up for best handwriting-related mishap goes to the person who tried to write "I used Professor X's code," but botched the last two letters in "code" in a way that evoked, erm, Little Professor X.

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Thusforthwith!

As a fan of both analysis and silly things, I can't help but enjoy when they're combined. This story is about a friend who perfected this combination.

My friend, at the time, was taking the same real analysis course I was grading, so I mentioned to him how funny it was when people used archaic connecting words: "thusly", "wither" and the like. From there we started trying to come up with the most ridiculous word. Thenceforth! Thuswith! Whencehence!

So of course every homework I got from this friend had at least one made-up connecting word (despite being typed up quite nicely). This continued without incident, until one day:

Me: This is hilarious! I'm worried about you slipping up and doing it on the test, though.

Him: Why not?

Me: Well... the professor might notice, and you might get docked some points...

Him: Hmm...

Which obviously culminated in him PUTTING FAKE WORDS ON THE ANALYSIS TEST.

And guess what?

THE PROFESSOR DIDN'T NOTICE.

Jesus.

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Readers, do you have any grading stories? Let me know if anyone tries to pull off using fake connecting words---not everyone may be as lucky!

Adventures in Linear Algebra with the Prismatoy

As the nature of the first few posts here should somewhat suggest, my fiancé and I spend a whole lot of time talking to each other about math. He needs a nom. Let's call him Eigenpeter.

The latest installment of "Peter finds an interesting idea, spends 1 hour worth of whiteboard lecture on representation theory to his algebra-phobic lover and makes a Mathematica toy in 15 minutes" is brought to you by Prismatoy, a cube that can be collapsed into a parallelpiped:

Basically, we wandered into a puzzle store where he picked one of these up and didn't put it down. (We did pay before leaving!)

I like this because, when restricted to any one of the 6 faces, it gives a visualization of the linear transformation
$$\left[\begin{array}{cc}1 & \cos\theta\\ 0 & \sin\theta\end{array}\right]$$
(up to transformations, scaling and unitary operations) with $0<\theta\leq \frac{\tau}{4}$* being the acute angle in the final configuration. You could derive this quickly at home by imagining one face as a unit square, then exploiting some basic trig to find that the transformation maps (0,1) to ($\cos\theta$,$\sin\theta$), (1,1) to ($1+\cos\theta$, $\sin\theta$), and leaves the bottom side of the square unchanged. The above then follows from knowing how the transformation acts on the standard basis vectors for $\mathbb{R}^2$.

Peter noted that the volume of this structure is given by the area of the base times the height, which, in this case, is the determinant of the linear transformation that takes it from cube form to its parallelpiped shape. To demonstrate this, we name the three vectors along the given three sides $\vec{a}$, $\vec{b}$ and $\vec{c}$:

The area of the base is given by $\lvert\vec{a} \times \vec{b}\rvert$---one can see this because
$$\lvert\vec{a}\times\vec{b}\rvert=\vec{a}\vec{b}\sin\theta,$$
which corresponds exactly to the area in the first picture, except the vectors are no longer of unit length. Now recall that $\vec{a}\times\vec{b}$ is a vector perpendicular to both $\vec{a}$ and $\vec{b}$. Taking the dot product $(\vec{a} \times \vec{b})\cdot \vec{c}$ only takes into account the component of $\vec{c}$ that is parallel to $\vec{a}\times\vec{b}$---in other words, perpendicular to both $\vec{a}$ and $\vec{b}$---in other words, the height of the parallelpiped! Hence taking the magnitude of this quantity gives us base times height, which is volume.

But wait, there's more! The quantity $\left\lvert(\vec{a} \times \vec{b})\cdot \vec{c}\right\rvert$ can be written as
$$(\vec{a} \times \vec{b})\cdot \vec{c}=\sum_{i=1}^3\left(\sum_{j=1}^3\sum_{k=1}^3 \epsilon_{ijk}a_jb_k\right)c_i$$
where (in case the reader hasn't seen it before) the Levi-Civita symbol $\epsilon_{ijk}$ essentially acts as the 'opposite' of the Kroenecker $\delta$ function, i.e.
$$\epsilon_{ijk}=\left\{\begin{array}{ll}1 & i=1, j=2, k=3;\ i=3, j=1, k=2;\ i=2, j=3, k=1\\ 0 & i=k=j\\ -1 & \textrm{else}\end{array}\right..$$
Now imagine taking the determinant of the matrix
$$\left[\begin{array}{ccc} \lvert & \lvert & \lvert\\ \vec{c} & \vec{a} & \vec{b}\\ \lvert & \lvert & \lvert\end{array}\right].$$
I won't put the algebra all out here, but calculating the determinant according to the definition and rearranging it will give the previous nested sum. This technique can also be used to prove that
$$(\vec{a} \times \vec{b})\cdot \vec{c}=(\vec{b} \times \vec{c})\cdot \vec{a}=(\vec{c} \times \vec{a})\cdot \vec{b}.$$
Yay! It is now evident that
$$V_{ppiped} = \left\lvert(\vec{a} \times \vec{b})\cdot \vec{c}\right\rvert = \textrm{det}[\vec{c}\  \vec{a}\ \vec{b}].$$
More generally, the determinant of a matrix is a factor indicating what change in volume (or area, or the appropriate dimensional quality) it produces.

EPILOGUE: He spent most of Saturday trying to sketch the manifold of possible shape configurations of this object, then trying to determine whether the set of operations on the toy in SL(3) was a group, then made several demonstrations of these and similar phenomena in Mathematica. However, he did not do the dishes. I cut the lecture for brevity.

*YES, TAU.

5 New High-Level Math Jokes

We here at the Analysisters enjoy puns, and are happy to contribute to the already-gigantic list of math jokes every once in a while. Here are some we've come up with over the years!

1. A topologist walks into $\bar A$. It's closed.

2. (3 variants) Q: If chocolate is a Hilbert space and peanut butter is its dual, why can every element in peanut butter be written as an inner product $\langle y,x \rangle$, where $y$, $x$ are chocolates and $x$ is uniquely fixed?
        A: Reese's Representation Theorem.

      Q: Why is peanut butter the adjoint of chocolate? Why is chocolate the adjoint of peanut butter?
      A: Reese's Representation Theorem.
      (Thanks, Paul!)

      Q: What's a mathematician's favorite candy?
      A: Riesz's Pieces.


3. What did the analyst have for dinner?
Limsup.

4. Your mama isn't Lebesgue integrable because she doesn't vanish at infinity!

5. Eight mathematicians walk into a diner. The first one says they're not hungry, and orders nothing. The next three order a beer, a hamburger, and french fries, respectively. The next three order a burger and fries, fries and a beer, and a burger and a beer, respectively. The last one orders a burger, fries, and a beer.

"I'm sorry, I can't fill your order," she says.

"Why is that?"

"This is only a partial order."

BONUS: Not necessarily in joke format, but I refuse to refer to
$$F_n = \frac{1}{\sqrt{5}}(\phi^n - \bar{\phi}^{n})$$
as anything except "that formula that shoots water up your butt".