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Class Inclusions C⁰ ⊃ D¹ ⊃ C¹

We can now summarize the hierarchy of function classes: $C^0 \supset D^1 \supset C^1$ . $C^0$ (continuous): $f$ has no jumps. $D^1$ (differentiable): $f$ is $C^0$ and $f'(x)$ exists at every point (but $f'$ might not be continuous). $C^1$ (continuously differentiable): $f$ is $D^1$ and $f'$ is itself continuous.

Each inclusion is strict: there exist $C^0$ functions that are not $D^1$ (e.g., $|x|$ at $0$ ), and there exist $D^1$ functions that are not $C^1$ (the classic example: $f(x) = x^2 \sin(1/x)$ for $x \neq 0$ , $f(0) = 0$ , which is differentiable at $0$ but has a discontinuous derivative there).

For multivariate functions (Chapters 11–12), these same classes generalize: $C^0$ means continuous, $D^1$ means partial derivatives exist everywhere, and $C^1$ means partial derivatives are continuous. The key result: $C^1$ implies $D^1$ , and in multiple dimensions, $C^1$ implies the full chain rule holds — this is not guaranteed for merely $D^1$ functions.

Formal View

Theorem 10.4 — Strict Inclusions C⁰ ⊃ D¹ ⊃ C¹

The inclusions

C^0 \supset D^1 \supset C^1

are all strict: -

|x| \in C^0 \setminus D^1

(continuous but not differentiable at 0). -

f(x) = x^2 \sin(1/x)

(extended by

f(0)=0

)

\in D^1 \setminus C^1

(differentiable but derivative not continuous at 0). - Every polynomial

\in C^1

For optimization: $C^1$ functions have Lipschitz gradients under mild assumptions, enabling standard convergence proofs for gradient descent.

Why This Matters

Knowing which class a function belongs to determines what algorithms and theorems apply.

Gradient descent convergence proofs require $C^1$ with Lipschitz derivative — $D^1$ alone is insufficient.
Implicit function theorem applies to $C^1$ mappings — not just $D^1$ .
Numerical ODE solvers have higher-order accuracy for $C^k$ functions with larger $k$ .

Learning Resources

Classes of differentiable functions

MIT OpenCourseWare

Rigorous treatment of function regularity classes from MIT 18.01.

45 min

Differentiability vs smooth

Khan Academy

Connecting differentiability classes to practical smoothness requirements.

8 min

Quiz

Question 1

Which function is in $D^1$ but NOT in $C^1$ ?

Question 2

Every $C^1$ function is also in $D^1$ .

Common Mistakes

Thinking differentiable and continuously differentiable are the same — the example $x^2\sin(1/x)$ shows they are not.
Assuming that for smooth-looking functions, the derivative is automatically continuous — check carefully for oscillatory behavior.