Discontinuous Linear Functions?!

For your example to work you need to restrict the domain of your functions to some compact e.g. because the uniform norm requires the functions to be bounded.

Also note this example works because you're not using the "usual" topology on $C^{1} ([0, 1])$ which also includes the uniform norm of the derivative and makes the space complete. It is much more difficult if the domain is complete!

[-]CuriouslyNuclear3mo60

For your example to work you need to restrict the domain of your functions to some compact e.g. because the uniform norm requires the functions to be bounded.

Hmm, is this really a substantive problem? Call it an "extended norm" instead of a norm and everything in the post works, right? My reasoning: An extended norm yields an extended metric space, which still generates a topology — it's just that points which are infinitely far apart are in different connected components. Since you get a perfectly valid topology, it makes perfect sense for the post to talk about continuity. Or at least I think so; am I missing something?

[-]Jeffrey Liang3mo40

Perhaps! I'm not familiar with extended norms. But when you say "let's put the uniform norm on " warning bells start going off in my head 😅

[-]Zack_M_Davis3mo40

Thanks for commenting—and for your patience. I've changed the domain to be an arbitrary closed interval and credited you at the bottom.

[-]Gurkenglas3mo72

for any ε, we can take

You mean take $N := ⌈ \frac{1}{ε} ⌉$ .

sqrt(x) is continuous at 0 with an infinitely steep slope.

[-]Zack_M_Davis3mo20

Thanks for commenting—and for your patience. I've corrected the k/N slip-up, deleted the misleading clause about "a function not having any 'jumps' impl[ying] that it can't have an 'infinitely steep' slope", and thanked you at the bottom.

[-]Gurkenglas3mo20

you haven't deleted it.

[-]Zack_M_Davis3mo20

Now I have. (Sorry, apparently I got confused while trying to make parallel updates to the original and this mirrorpost; they're not using the same source file due to differences between platform markup engines.)

[-]Gurkenglas3mo120

a small change in the input can't correspond to an arbitrarily large change in the output

The sign function doesn't have an arbitrarily large change in the output. Do you maybe mean that an infinitesimal change in the input can only produce an infinitesimal change in the output? I don't see how that fails, but maybe just because I don't have a definition for it at hand.

[-]Alex Gibson3mo30

Specifically for linear maps these are the same thing, but yeah I guess that kinda defeats the purpose of the post.

[-]CuriouslyNuclear3mo*50

This is a good post. I remember being so confused in a real analysis class when my professor started talking about how important it is that we restrict our attention to continuous linear functions (what on Earth was a discontinuous linear function supposed to be?). This post explains what's going on better than my professor or textbook did.

I agree with one of the other commenters that this part is not technically phrased accurately:

One way to think about continuity is that a function not having any "jumps" implies that it can't have an "infinitely steep" slope

Because eg the derivative of at $x = 0$ is $+ \infty$ despite the fact that it's continuous there.

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45 Discontinuous Linear Functions?!

by Zack_M_Davis

6th Jun 2025

3 min read

45

This is a linkpost for http://zackmdavis.net/blog/2025/06/discontinuous-linear-functions/

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Discontinuous Linear Functions?!

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[-]Jeffrey Liang3mo123

Okay I took the nerd bait and signed up for LW to say:

For your example to work you need to restrict the domain of your functions to some compact e.g. because the uniform norm requires the functions to be bounded.

[-]CuriouslyNuclear3mo60

For your example to work you need to restrict the domain of your functions to some compact e.g. because the uniform norm requires the functions to be bounded.

[-]Jeffrey Liang3mo40

Perhaps! I'm not familiar with extended norms. But when you say "let's put the uniform norm on " warning bells start going off in my head 😅

[-]Zack_M_Davis3mo40

Thanks for commenting—and for your patience. I've changed the domain to be an arbitrary closed interval and credited you at the bottom.

[-]Gurkenglas3mo72

for any ε, we can take

You mean take $N := ⌈ \frac{1}{ε} ⌉$ .

sqrt(x) is continuous at 0 with an infinitely steep slope.

[-]Zack_M_Davis3mo20

[-]Gurkenglas3mo20

you haven't deleted it.

[-]Zack_M_Davis3mo20

[-]Gurkenglas3mo120

a small change in the input can't correspond to an arbitrarily large change in the output

[-]Alex Gibson3mo30

Specifically for linear maps these are the same thing, but yeah I guess that kinda defeats the purpose of the post.

[-]CuriouslyNuclear3mo*50

I agree with one of the other commenters that this part is not technically phrased accurately:

One way to think about continuity is that a function not having any "jumps" implies that it can't have an "infinitely steep" slope

Because eg the derivative of at $x = 0$ is $+ \infty$ despite the fact that it's continuous there.

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We know what linear functions are. A function f is linear iff it satisfies additivity and homogeneity $f (a x) = a f (x)$ .

We know what continuity is. A function f is continuous iff for all ε there exists a δ such that if $| x - x_{0} |$ < δ, then $| f (x) - f (x_{0}) |$ < ε.

An equivalent way to think about continuity is the sequence criterion: f is continuous iff a sequence $(x_{k})$ converging to $x$ implies that $(f (x_{k}))$ converges to $f (x)$ . That is to say, if for all ε there exists an N such that if k ≥ N, then $| x_{k} - x |$ < ε, then for all ε, there also exists an M such that if k ≥ M, then $| f (x_{k}) - f (x) |$ < ε.

Sometimes people talk about discontinuous linear functions. You might think: that's crazy. I've seen many linear functions in my time, and they were definitely all continuous. f(x): ℝ → ℝ := ax is continuous for any a ∈ ℝ. T(x⃗): ℝ² → ℝ² := $(\begin{matrix} a & b c & d \end{matrix}) \to x$ is continuous no matter what the entries in the matrix are. Stop being crazy!!

Actually, it's not crazy. It's just that all the discontinuous linear functions live in infinite-dimensional spaces.

Take, say, the space $C^{1} ([a, b])$ of continuously differentiable functions from a closed interval [a,b] to ℝ, with the uniform norm. (The uniform norm means that the "size" of a function for the purposes of continuity is the least upper bound of its absolute value.) If you think of a vector in the n-dimensional $R^{n}$ as a function from {1...n} to ℝ, then you can see why a function from a continuous (not even countable) domain would be infinite-dimensional.

Consider the sequence of functions $(f_{k}) = (\frac{sin k x}{k})_{k = 1}^{\infty}$ in $C^{1} ([a, b])$ . The sequence converges to the zero function: for any ε, we can take $N := ⌈ \frac{1}{ε} ⌉$ and then $\frac{sin k x}{k} \leq \frac{1}{⌈ \frac{1}{ε} ⌉} \leq \frac{1}{\frac{1}{ε}} = ε$ .

Now consider that the sequence of derivatives is $(\frac{k cos k x}{k})_{k = 1}^{\infty} = (cos k x)_{k = 1}^{\infty}$ , which doesn't converge. But the function D: $C^1([a,b]) \rightarrow $C^{0} ([a, b])$ that maps a function to its derivative is linear. (We have additivity because the derivative of a sum is the sum of the derivatives, and we have homogeneity because you can "pull out" a constant factor from the derivative.)

By exhibiting a function D and a sequence $(f_{k})$ for which $(f_{k})$ converges but $(D (f_{k}))$ doesn't, we have shown that the derivative mapping D is a discontinuous linear function, because the sequence criterion for continuity is not satisfied. If you know the definitions and can work with the definitions, it's not crazy to believe in such a thing!

The infinite-dimensionality is key to grasping the ultimate sanity of what would initially have appeared crazy. One way to think about continuity is that a small change in the input can't correspond to an arbitrarily large change in the output.

Consider a linear transformation $T$ on a finite-dimensional vector space; for simplicity of illustration, suppose it's diagonalizable with eigenbasis ${{\to u}_{j}}$ and eigenvalues ${λ_{j}}$ . Then for input $\to x = \sum_{j} c_{j} {\to u}_{j}$ , we have $T (\to x) = \sum_{j} c_{j} λ_{j} {\to u}_{j}$ : the eigencoördinates of the input get multiplied by the eigenvalues, so the amount that the transformation "stretches" the input is bounded by ${max}_{j} | λ_{j} |$ . The linearity buys us the "no arbitrarily large change in the output" property which is continuity.

In infinite dimensions, linearity doesn't buy that. Consider the function $T (x_{1}, x_{2}, x_{3}, . . .) = (x_{1}, 2 x_{2}, 3 x_{3}, . . .)$ on sequences with finitely many nonzero elements, under the uniform norm. The effect of the transformation on any given dimension is linear and bounded, but there's always another dimension that's getting stretched more. A small change in the input can result in an arbitrarily large change in the output, by making the change sufficiently far in the sequence (where the input is getting stretched more and more).

(Thanks to Jeffrey Liang and Gurkenglas for corrections to the original version of this post.)