We know what linear functions are. A function f is linear iff it satisfies additivity f(x + y) = f(x) + f(y) and homogeneity f(ax) = af(x).

We know what continuity is. A function f is continuous iff for all ε there exists a δ such that if |x − x₀| < δ, then |f(x) − f(x₀)| < 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 := $\begin{pmatrix} a & b \\ c & d \end{pmatrix} \boldsymbol{\vec{v}}$ 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 C¹(ℝ) of continuously differentiable functions ℝ → 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 ℝⁿ 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) = $(\frac{\sin kx}{k})_{k=1}^{\infty}$ in C¹(ℝ). The sequence converges to the zero function: for any ε, we can take $k := \lceil \frac{1}{\varepsilon} \rceil$ and then $\frac{\sin kx}{k} \le \frac{1}{\lceil \frac{1}{\varepsilon} \rceil} \le \frac{1}{\frac{1}{\varepsilon}} = \varepsilon$.

Now consider that the sequence of derivatives is $(\frac{k \cos kx}{k})_{k=1}^{\infty} = (\cos kx)_{k=1}^{\infty}$, which doesn’t converge. But the function D: C¹(ℝ) → C⁰(ℝ) 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 function not having any “jumps” implies that it can’t have an “infinitely steep” slope: 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 {u⃗_j} and eigenvalues {λ_j}. Then for input x⃗ = Σ_j c_ju⃗_j, we have T(x⃗) = Σ_j c_jλ_ju⃗_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(x₁, x₂, x₃, …) = (x₁, 2x₂, 3x₃, ...) 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).