A lower bound of a subset SSS of an ordered set (X,≤)(X,\le)(X,≤) is an element ℓ∈X\ell\in Xℓ∈X such that ∀s∈S, ℓ≤s.\forall s\in S,\ \ell\le s.∀s∈S, ℓ≤s.Lower bounds formalize the idea that a set lies entirely to the “right” of some point. They are the dual notion to upper bounds and are used in defining infimum. Examples: In (R,≤)(\mathbb{R},\le)(R,≤), the set S=(0,1)S=(0,1)S=(0,1) has lower bounds ℓ=0\ell=0ℓ=0, ℓ=−1\ell=-1ℓ=−1, and in fact every ℓ≤0\ell\le 0ℓ≤0. In (R,≤)(\mathbb{R},\le)(R,≤), the set S={x∈R:x≥2}S=\{x\in\mathbb{R}: x\ge 2\}S={x∈R:x≥2} has lower bounds ℓ=2\ell=2ℓ=2 and every ℓ≤2\ell\le 2ℓ≤2. In (Z,≤)(\mathbb{Z},\le)(Z,≤), the set S={n∈Z:n≥0}S=\{n\in\mathbb{Z}: n\ge 0\}S={n∈Z:n≥0} has lower bounds ℓ=0\ell=0ℓ=0, ℓ=−5\ell=-5ℓ=−5, etc.

Let f:[a,b]→Rf:[a,b]\to\mathbb{R}f:[a,b]→R be bounded and let P:a=x0<⋯<xn=bP:a=x_0<\cdots<x_n=bP:a=x0​<⋯<xn​=b be a partition. For each subinterval, define mi:=inf⁡{f(x):x∈[xi−1,xi]}.m_i := \inf\{f(x): x\in[x_{i-1},x_i]\}.mi​:=inf{f(x):x∈[xi−1​,xi​]}. The lower sum of fff with respect to PPP is ...

Let XXX be a set and let fn:X→Rf_n:X\to\mathbb{R}fn​:X→R (or C\mathbb{C}C). Suppose: each fnf_nfn​ is continuous on XXX (when XXX is a metric space ), and there exist Mn≥0M_n\ge 0Mn​≥0 with ∣fn(x)∣≤Mn|f_n(x)|\le M_n∣fn​(x)∣≤Mn​ for all x∈Xx\in Xx∈X, and ∑Mn\sum M_n∑Mn​ converges . Corollary: ...

Let (X,≤)(X,\le)(X,≤) be an ordered set and let S⊆XS\subseteq XS⊆X. An element m∈Sm\in Sm∈S is a maximum of SSS (written m=max⁡Sm=\max Sm=maxS) if ∀s∈S, s≤m.\forall s\in S,\ s\le m.∀s∈S, s≤m.A maximum is an upper bound that actually lies in the set. If a maximum exists, it is unique and equals the supremum : max⁡S=sup⁡S\max S = \sup SmaxS=supS. ...

Let (X,d)(X,d)(X,d) be a metric space . A set M⊆XM\subseteq XM⊆X is meager (or of first category) if there exist nowhere dense sets A1,A2,⋯⊆XA_1,A_2,\dots\subseteq XA1​,A2​,⋯⊆X such that M=⋃n=1∞An. M=\bigcup_{n=1}^\infty A_n. M=⋃n=1∞​An​. ...

Let U⊆RnU\subseteq\mathbb{R}^nU⊆Rn be open and let f:U→Rmf:U\to\mathbb{R}^mf:U→Rm be of class $C^1$ . Fix a∈Ua\in Ua∈U. Mean value estimate lemma: For every ε>0\varepsilon>0ε>0 there exists δ>0\delta>0δ>0 such that if x,y∈Ux,y\in Ux,y∈U satisfy ∥x−a∥<δ\|x-a\|<\delta∥x−a∥<δ, ∥y−a∥<δ\|y-a\|<\delta∥y−a∥<δ, and the line segment [x,y]⊆U[x,y]\subseteq U[x,y]⊆U, then ∥f(x)−f(y)−Df(a)(x−y)∥≤ε ∥x−y∥. \|f(x)-f(y)-Df(a)(x-y)\|\le \varepsilon\,\|x-y\|. ∥f(x)−f(y)−Df(a)(x−y)∥≤ε∥x−y∥. In particular, ∥f(x)−f(y)∥≤(∥Df(a)∥+ε) ∥x−y∥. \|f(x)-f(y)\|\le \bigl(\|Df(a)\|+\varepsilon\bigr)\,\|x-y\|. ∥f(x)−f(y)∥≤(∥Df(a)∥+ε)∥x−y∥. ...

Mean value inequality (multivariable): Let U⊆RnU\subseteq\mathbb{R}^nU⊆Rn be open and let f:U→Rmf:U\to\mathbb{R}^mf:U→Rm be differentiable . Suppose x,y∈Ux,y\in Ux,y∈U and the line segment [x,y]={x+t(y−x):0≤t≤1} [x,y]=\{x+t(y-x):0\le t\le 1\} [x,y]={x+t(y−x):0≤t≤1} is contained in UUU. If there is a constant MMM such that ∥Df(z)∥≤Mfor all z∈[x,y] \|Df(z)\| \le M \quad \text{for all } z\in[x,y] ∥Df(z)∥≤Mfor all z∈[x,y] (where ∥Df(z)∥\|Df(z)\|∥Df(z)∥ is the operator norm ), then ∥f(y)−f(x)∥≤M∥y−x∥. \|f(y)-f(x)\|\le M\|y-x\|. ∥f(y)−f(x)∥≤M∥y−x∥. ...

Mean Value Theorem: Let f:[a,b]→Rf:[a,b]\to\mathbb{R}f:[a,b]→R be continuous on [a,b][a,b][a,b] and differentiable on (a,b)(a,b)(a,b). Then there exists c∈(a,b)c\in(a,b)c∈(a,b) such that f′(c)=f(b)−f(a)b−a. f'(c)=\frac{f(b)-f(a)}{b-a}. f′(c)=b−af(b)−f(a)​. ...

Mean Value Theorem for integrals: Let f:[a,b]→Rf:[a,b]\to\mathbb{R}f:[a,b]→R be continuous . Then there exists c∈[a,b]c\in[a,b]c∈[a,b] such that ∫abf(x) dx=f(c) (b−a). \int_a^b f(x)\,dx = f(c)\,(b-a). ∫ab​f(x)dx=f(c)(b−a). Equivalently, f(c)=1b−a∫abf(x) dx. f(c)=\frac{1}{b-a}\int_a^b f(x)\,dx. f(c)=b−a1​∫ab​f(x)dx. ...

Mertens theorem (Cauchy products): Let ∑n=0∞an\sum_{n=0}^\infty a_n∑n=0∞​an​ and ∑n=0∞bn\sum_{n=0}^\infty b_n∑n=0∞​bn​ be convergent series (real or complex). Define the Cauchy product coefficients cn=∑k=0nakbn−k.c_n=\sum_{k=0}^n a_k b_{n-k}.cn​=∑k=0n​ak​bn−k​. If at least one of the series ∑an\sum a_n∑an​ or ∑bn\sum b_n∑bn​ converges absolutely , then the Cauchy product series ∑cn\sum c_n∑cn​ converges and ∑n=0∞cn=(∑n=0∞an)(∑n=0∞bn).\sum_{n=0}^\infty c_n = \left(\sum_{n=0}^\infty a_n\right)\left(\sum_{n=0}^\infty b_n\right).∑n=0∞​cn​=(∑n=0∞​an​)(∑n=0∞​bn​). ...