$f(x) = x^2$ on $\reals$
The squaring function, as a function $f : \reals \to \reals$.
Terms
| Term | This example | Why |
|---|---|---|
| domain | $\reals$ | $f$ is declared as a function $\reals \to \reals$. |
| codomain | $\reals$ | $f$ is declared as a function $\reals \to \reals$. |
| image | $f(\reals) = [0, \infty)$ | Squares of real numbers are exactly the nonnegative numbers. |
| preimage | $f^{-1}(\setdef{4}) = \setdef{-2, 2}$ and $f^{-1}(\setdef{-1}) = \emptyset$ | The preimage $f^{-1}(\setdef{4})$ exists even though $f$ has no inverse. No real number squares to $-1$, so $f^{-1}(\setdef{-1})$ is empty. |
| range image | $f(\reals) = [0, \infty)$ | |
| range codomain | $\reals$ | |
| continuously differentiable | ✓ | $x \mapsto x^2$ on $\reals$ is continuously differentiable because its derivative $2x$ is continuous. |
| continuous | ✓ | $x \mapsto x^2$ is continuous. |
| sequentially continuous | ✓ | |
| function | ✓ | $f(x) = x^2$ is a function and a map, as is every function. “Function” is the more common word here, because the inputs and outputs are numbers. |
| map | ✓ | |
| positive definite function | ✓ | $x \mapsto x^2$ on $\reals$ is positive definite because it is positive away from $0$ and zero at $0$. |
| convex function | ✓ | The function $x \mapsto x^2$ is convex, because its second derivative is $2 \geq 0$. |
| locally Lipschitz continuous | ✓ | $x \mapsto x^2$ on $\reals$ is locally Lipschitz continuous, because on each interval $[-R, R]$ its slope $2x$ is bounded by $2R$. |
| Lipschitz continuous | ✗ | $x \mapsto x^2$ on $\reals$ is not Lipschitz continuous, because its slope $2x$ is unbounded on $\reals$. |
| uniformly continuous | ✗ | $f(x) = x^2$ is not uniformly continuous on $\reals$, because its graph gets steeper and steeper, so no single closeness of inputs works everywhere. |
| class K | ✗ | |
| operator | ✗ | $f(x) = x^2$ on $\reals$ is not an operator, because its inputs are numbers, not elements of a space of functions. |
| functional | ✗ | $f(x) = x^2$ on $\reals$ is not a functional, because its inputs are numbers, not elements of a space of functions. |
| transform | ✗ | $f(x) = x^2$ on $\reals$ is not a transform, because its inputs are numbers, not elements of a space of functions. |