Because of the nature of their range, real valued functions can be characterized using some special features.
Also consider properties of functions over ordered semigroups described elsewhere.
Topological properties
Limits, continuity, smoothness, steepness. See topology ref.
Limits
See topology ref. Left and right handed limits. \(\pm \infty\) as limits.
Limits of sums, products, quotients. Squeeze or pinching theorem: If \(f(x) \leq g(x) \leq h(x)\), and if \(\lim_{x \to a} f(x) = \lim_{x \to a} h(x) = b\), \(\lim_{x \to a} g(x)=b\).
f:F to F: Find limits
Try Substitution, factorization, rationalization.
L’Hopital rule: If \(lt_{x \to c}f(x) = lt_{x \to c}g(x) = 0, L = \lim_{x\to c}\frac{f(x)}{g(x)} = \frac{f(x)’}{g(x)’}\): from definitions or from generalized mean value thm: define f(c) = g(c) = 0, so derivatives exist around c; as \(\exists L, g’(x) \neq 0\).
Closed functional f
All sublevel sets of \(f\) are closed. Equivalently, the epigraph is closed. \why
Consider/ visualize the epigraph: If \(f\) is continuous, dom(f) is closed, then \(f\) is closed. Also, if \(f\) is continuous, dom(f) is open, \(f\) is closed iff \(f\) converges to \(\infty\) along every sequence converging to bd(dom(f)). \why
Continuity
See topology ref.
If \(f, g\) are continuous, then f+g, fg, f/g are continous.
Absolute continuity of f:Rm to Rn
More powerful/ specialized than uniform continuity.
\(\forall \eps \geq 0, \exists d \geq 0: \forall \) finite sequence of pairwise disjoint sub-intervals \((x_k, y_k)\) : \(\sum_{k} | y_k - x_k| < \delta \implies \sum_{k} | f(y_k) - f(x_k) | < \epsilon.\) This can be extended to \(f:R^{m} \to X\) for any topological space X.
Simple discontinuity
Upper and lower limits exist, but different: \(\floor{x}\). Non-simple disc: f(x): 1 if \(x \in Q\), 0 else.
Fixing discontinuities. \(\lim_{x \to 0} \frac{\sin x}{x} = 1\).
Extreme value existence, boundedness
(Weierstrass) If real valued \(f\) is continuous over compact (closed and bounded in R) \(S = [x_{1}, x_{2}]\), it attains maximum and minimum value somewhere in S.
Proof
As S compact, f(S) compact [See topology ref.]. So \(f\) closed and bounded. By LUB property of R, \(\sup \)f\( = M\); take \(d_{n}\): \(M - n^{-1}\leq f(d_{n}) \leq M\); so \(f(d_{n}) \to M\); by Bolzano Weierstrass take convergent subseq \((d_{n_{k}}) \to d\); \(d \in S\) as S closed; as \(f\) cont, so \(f(d) = M\).
If S not compact, there can be: unbounded but cont f: \(S = (0,1), f(x) = x^{-1}\); cont \(f\) without max: f(x) = \(x\) on (0,1); cont but not uniformly cont: \(S = (0,1), f(x) = x^{-1}\).
Intermediate value theorem
If continuous f(x):[a, b] \(\to\) R , \ \(u \in [f(a), f(b)], \exists c \in [a, b] : f(c) = u\): [a,b] connected, so f([a,b]) connected.