Study Music. Click to play or pause. After it starts, press the Space Bar to play or pause. If enabled, it will resume across pages.

A Counterexample That Teaches Analysis and Partial Differential Equations Better Than a Lecture

Written by

admin

in

In analysis and PDE, “regularity” is the quiet promise in the background. You set up a weak formulation because the data are rough or the domain is irregular, but you still hope the solution you obtain is not merely an abstract object in a function space. You want to know whether it is bounded, continuous, differentiable, or even classical. Many theorems say: under the right hypotheses, yes. The point of a counterexample is to show you precisely what “the right hypotheses” are buying you.

A single borderline phenomenon does more teaching than a dozen slogans:

Premium Gaming TV
65-Inch OLED Gaming Pick

LG 65-Inch Class OLED evo AI 4K C5 Series Smart TV (OLED65C5PUA, 2025)

LG • OLED65C5PUA • OLED TV
LG 65-Inch Class OLED evo AI 4K C5 Series Smart TV (OLED65C5PUA, 2025)
A strong fit for buyers who want OLED image quality plus gaming-focused refresh and HDMI 2.1 support

A premium gaming-and-entertainment TV option for console pages, living-room gaming roundups, and OLED recommendation articles.

$1396.99
Price checked: 2026-03-23 18:31. Product prices and availability are accurate as of the date/time indicated and are subject to change. Any price and availability information displayed on Amazon at the time of purchase will apply to the purchase of this product.
  • 65-inch 4K OLED display
  • Up to 144Hz refresh support
  • Dolby Vision and Dolby Atmos
  • Four HDMI 2.1 inputs
  • G-Sync, FreeSync, and VRR support
(paid link)
View LG OLED on Amazon
Check the live Amazon listing for the latest price, stock, shipping, and size selection.

Why it stands out

  • Great gaming feature set
  • Strong OLED picture quality
  • Works well in premium console or PC-over-TV setups

Things to know

  • Premium purchase
  • Large-screen price moves often
See Amazon for current availability
As an Amazon Associate I earn from qualifying purchases.
  • In low dimensions, a little integrability buys you a lot of regularity.
  • At the critical exponent, the same estimate stops upgrading the solution.
  • Just below the threshold, the failure is not cosmetic; it is structural.

The cleanest place to see all of this at once is the Sobolev embedding at the two-dimensional threshold.

The promise you want (and the place it breaks)

Let $\Omega \subset \mathbb{R}^2$ be a bounded domain. If $u \in W^{1,p}(\Omega)$ with $p>2$, the Sobolev embedding theorem gives that $u$ has a Hölder-continuous representative. In particular, $u$ is bounded. This is exactly the kind of statement PDE analysts love because it turns an energy estimate into a pointwise conclusion.

At $p=2$, the formal scaling turns “just enough derivatives” into a knife-edge case. There is still a powerful substitute, the Moser–Trudinger inequality, but the naive hope “$W^{1,2}$ implies boundedness” is false. Not false in a contrived way, but false in the only way that matters: there exist functions with uniformly controlled $W^{1,2}$ norm whose peaks grow without bound.

That is the counterexample.

The counterexample as a sequence, not a single function

One can construct a single unbounded function in $W^{1,2}$, but the sequence version teaches more. It shows that the failure is not an isolated pathology; it is stable under the very estimates we routinely use.

Work in the unit disk $B_1(0) \subset \mathbb{R}^2$. Define, for integers $k\ge 2$, a radial function $u_k : B_1(0) \to \mathbb{R}$ by

$$ u_k(x) = \begin{cases} \sqrt{\log k}, & |x| \le \frac{1}{k},\$$4pt] \displaystyle \frac{\log\!\big(\frac{1}{|x|}\big)}{\sqrt{\log k}}, & \frac{1}{k} < |x| \le 1. \end{cases} $$

This is continuous, radial, and its maximum is $\|u_k\|_{L^\infty(B_1)} = \sqrt{\log k}$, which tends \to infinity with $k$. The only question is whether the $W^{1,2}$ norm stays controlled.

Because $u_k$ is constant on $|x|\le 1/k$, its gradient vanishes there. On the annulus $1/k < |x| \le 1$, write $r=|x|$. Then

$$ u_k(r) = \frac{\log(1/r)}{\sqrt{\log k}}, \qquad u_k'(r) = -\frac{1}{r \sqrt{\log k}}. $$

So $|\nabla u_k(x)| = |u_k'(r)| = \frac{1}{r\sqrt{\log k}}$ almost everywhere on the annulus.

Compute the Dirichlet energy:

$$ \int_{B_1} |\nabla u_k|^2\,dx = \int_{1/k}^1 \left(\frac{1}{r^2\log k}\right)\, (2\pi r)\,dr = \frac{2\pi}{\log k}\int_{1/k}^1 \frac{1}{r}\,dr = \frac{2\pi}{\log k}\, \log k = 2\pi. $$

The energy is constant, independent of $k$. In particular, the $W^{1,2}$ seminorm is uniformly bounded.

The $L^2$ norm can also be bounded uniformly. A rough estimate suffices: on $|x|\le 1/k$, $u_k^2 = \log k$ but the area is $\pi/k^2$, so that piece contributes $\pi (\log k)/k^2 \to 0$. On $1/k

$$ \int_{1/k}^1 \frac{\log(1/r)^2}{\log k}\, (2\pi r)\,dr \le \frac{2\pi}{\log k} \int_0^1 \log(1/r)^2\, r\,dr $$

and the last integral is finite (it is a standard calculus exercise). So $\|u_k\|_{L^2}$ stays bounded while $\|u_k\|_{L^\infty}$ blows up.

This is the lesson in one line:

  • Uniform control of $\int |\nabla u_k|^2$ does not prevent arbitrarily tall spikes in two dimensions.

Why this is not a gimmick: scaling is the culprit

The sequence is not random; it is tuned \to the scaling of the $W^{1,2}$ seminorm in $\mathbb{R}^2$. If you rescale a function by concentrating it near a point, the gradient energy behaves differently depending on the dimension.

For intuition, suppose you try \to create a spike of height $A$ supported on a ball of radius $\varepsilon$. A typical gradient size is about $A/\varepsilon$. The energy scales like

$$ \int_{B_\varepsilon} |\nabla u|^2 \sim \left(\frac{A^2}{\varepsilon^2}\right)\varepsilon^2 = A^2 $$

in two dimensions: the $\varepsilon$ cancels. That cancellation is the criticality. In higher dimensions $n\ge 3$, the energy would scale like $A^2 \varepsilon^{n-2}$, and shrinking the support would reduce the energy, making spikes cheap; in one dimension, spikes are expensive. The two-dimensional case is exactly where the “spike cost” becomes independent of scale.

The $u_k$ above is a refined version of this idea: it concentrates logarithmically rather than by a simple cutoff, precisely because the critical scale is so delicate.

How this interacts with PDE

The counterexample sits in function spaces, but PDE is where the stakes are. Here is a common pattern:

  • You solve an elliptic equation in weak form: find $u \in W^{1,2}_0(\Omega)$ such that
$$ \int_\Omega \nabla u \cdot \nabla \varphi\,dx = \int_\Omega f \varphi\,dx \quad \text{for all } \varphi \in C_c^\infty(\Omega). $$

This is the weak formulation of $-\Delta u = f$ with zero boundary data.

  • By Lax–Milgram, you get existence and uniqueness provided $f \in H^{-1}(\Omega)$ (or $f \in L^2$ if you like).
  • From the variational structure, you get an energy estimate $\int |\nabla u|^2 \le C\|f\|_{H^{-1}}^2$.

At this point, a newcomer often expects boundedness or continuity “because solutions of Poisson’s equation are nice.” But the counterexample tells you what you must check: the energy estimate alone does not force boundedness in dimension two. Whether $u$ is bounded depends on stronger information about $f$, the domain, and which regularity theorem you can legitimately invoke.

In fact, there are two distinct messages hidden here:

  • Even when $u$ is harmonic on an annulus (as $u_k$ essentially is away from the origin), it can have large peaks if you allow singular behavior at a point.
  • To rule out these peaks for weak solutions, you need hypotheses that exclude concentration of the \right type.

What replaces the false embedding

The failure of $W^{1,2}\hookrightarrow L^\infty$ in $\mathbb{R}^2$ does not mean “no control is possible.” Instead, the correct statement is “the control becomes exponential.”

A representative form of the Moser–Trudinger inequality is:

  • If $u \in W^{1,2}_0(\Omega)$ with $\int_\Omega |\nabla u|^2\,dx \le 1$, then there exists $C$ (depending on $\Omega$) such that
$$ \int_\Omega \exp\!\big(4\pi u^2\big)\,dx \le C. $$

The constant $4\pi$ is not decoration; it is sharp, and the sequence $u_k$ above is designed to sit near that sharpness. The point is that $u_k$ shows a bounded energy class where the natural integrability gain is “exponential in $u^2$,” not “boundedness of $u$.”

That is exactly how borderline analysis feels: you do not lose everything, but the theorem changes its shape.

The practical PDE takeaway: look for the upgrade step

When proving regularity for a PDE, you almost always follow a chain of upgrades:

  • Start with a weak solution in a Sobolev space.
  • Use the equation to show higher integrability or higher derivatives are controlled.
  • Apply an embedding to turn that into continuity or boundedness.

The counterexample tells you where the chain can stall. If your only estimate is $\|\nabla u\|_{L^2}\le C$, and the dimension is two, then “apply Sobolev embedding to get $u\in L^\infty$” is an illegal step. The right upgrade might be:

  • get $u \in W^{1,p}$ for some $p>2$, then embed \to $C^\alpha$, or
  • get a De Giorgi–Nash–Moser type estimate (if the operator and data permit), or
  • accept the exponential integrability conclusion when the problem lives at the critical exponent.

This is why analysts love to record the exponent explicitly. “$p>2$” is not a technicality; it is the border between boundedness and the possibility of concentration spikes.

A compact “what this counterexample teaches” table

| Hope you might have | What is actually true in $\mathbb{R}^2$ | What to use instead |

|—|—|—|

| Energy control $\int|\nabla u|^2$ forces boundedness | False: bounded energy allows arbitrarily large peaks | Exponential integrability (Moser–Trudinger), or higher $p$ estimates |

| Weak solutions automatically become classical | Not without an upgrade theorem matching your data and operator | Caccioppoli + bootstrapping, Calderón–Zygmund, De Giorgi–Nash–Moser, depending on structure |

| “Critical” is a mild inconvenience | Critical means scaling cancels and concentration can survive the estimates | Track scaling; prove the missing gain explicitly |

The deeper moral: a counterexample is a map of the border

In PDE, the difference between a theorem and a false statement is often a single exponent, a single integrability hypothesis, or a single structural condition (uniform ellipticity, divergence form, bounded coefficients). The counterexample above does not merely say “boundedness fails.” It says where and why:

  • It fails exactly at the scaling where the energy estimate stops penalizing concentration.
  • It fails in a way that survives the standard a priori bounds.
  • It suggests the correct replacement theorem by pointing to the sharp regime.

Once you have internalized this, your reading of PDE papers changes. Every time you see a regularity conclusion, you ask: where did the gain come from, and how does it beat concentration? If the proof has no genuine gain step, the conclusion is not believable. If it does, you can often predict the sharpness and the likely counterexamples that sit at the boundary.

That is why this one sequence teaches analysis and PDE better than a lecture: it turns “regularity” from a wish into a quantified, checkable upgrade mechanism.

Books by Drew Higgins

Explore this field
Analysis and Partial Differential Equations
Library Analysis and Partial Differential Equations
Mathematics
Complex Analysis
Functional Analysis
Partial Differential Equations
Real Analysis
Algebra
Category Theory
Combinatorics
Dynamical Systems
Geometry

Comments

Leave a Reply

Your email address will not be published. Required fields are marked *

More posts