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Given a choice function on all nonempty subsets of $X$, recursively choose the next point from the complement of all earlier choices. Hartogs theorem forces the recursion to exhaust $X$ before it defines an injection from $h(X)$ into $X$, producing a bijection from an ordinal to $X$.

Generalized least squares minimizes $(Y-X\beta {)}^T{\Omega }^{-1}(Y-X\beta )$ for a known covariance shape $\Omega$.

Writing $G_{2r}={\sum }_{\omega \ne 0}{\omega }^{-2r}$ for the lattice Eisenstein sums, expansion about zero givesThus the constant coefficient vanishes and, for $k\ge 1$, the coefficient of $z^{2k}$ is $(2k+1)G_{2k+2}$.

The Weierstrass zeta function satisfies ${\zeta }^{\prime }=-\wp$ and is quasi-periodic: for every period $\omega$, the difference $\zeta (z+\omega )-\zeta (z)$ is constant. It has a simple pole of residue one at each lattice point. Consequently, if ${\sum }_j{c}_j=0$, thenis an elliptic function.

It is the quasi-periodic zeta member of the Weierstrass functions.

For a function $f:[0,1]\to \mathbb{R}$ and an integer $n\ge 1$, its Bernstein polynomial isThe Bernstein basis supplies nonnegative weights whose sum is one. Equivalently, the weights are the probabilities of a binomial distribution, and the result is an expected value. At a given $x$, this averages the sampled values $f(k/n)$ with weights concentrated near $k/n=x$.^[ref]

The endpoints are exact: $B_{n}f(0)=f(0)$ and $B_{n}f(1)=f(1)$. Interior sampled points need not be interpolated. For example, binomial distribution moments give $B_n(x)=x$ and $B_n(x^2)=x^2+x(1-x)/n$. Thus every affine function is reproduced exactly, whereas $x^2$ is generally changed. Bernstein polynomial degree preservation is a statement about degree of a polynomial, not equality of the entire polynomial.

For a continuous $f$, the uniform continuity of $f$ and the variance $\mathrm{Var}(K/n)=x(1-x)/n\le 1/(4n)$ explain uniform convergence. Split the expected value of the error into $|K/n-x|<\delta$, where uniform continuity makes the change small, and its complement, whose probability is at most $1/(4n{\delta }^2)$ by Chebyshev's inequality. The bounded function $f$ bounds the remaining error. This proves convergence without assuming differentiability.

For $f\in C([0,1{]}^d)$, the tensor-product Bernstein polynomial is the expected value of $f(X_1/n,\dots ,X_d/n)$ for independent random variables $X_j\sim \mathrm{Bin}(n,x_j)$. The same uniform continuity and coordinatewise variance estimates give uniform convergence for each fixed finite $d$.

For diagonal transition probability $1-2\alpha$ and each off-diagonal probability $\alpha$,with $0\log 0=0$.

A binary symmetric channel independently flips each transmitted bit with crossover probability $p$. Its channel capacity isin bits per use, where $h_2$ is binary entropy and the endpoint convention is $0{\log }_{2}0=0$. Knowing the input leaves just the uncertainty of the flip, so $H(Y\mid X)=h_2(p)$ for every input probability distribution. Since the output is a bit, $H(Y)\le 1$, and thus $I(X;Y)\le 1-h_2(p)$. A fair input produces a fair output and reaches the bound. This is the two-symbol case of weakly symmetric channel capacity; Eytan Modiano, MIT 16.36, Channel Capacity and Coding, slides 4–5 gives this calculation.

For crossover probabilities $p=0.01,0.1,0.25$, the resulting channel capacities are approximately $0.91920686$, $0.53100441$, and $0.18872188$ bits per use. At $p=0$, the communication channel is noiseless; at $p=1/2$, the output is independent of the input and the capacity is zero. At $p=1$, every bit is inverted, a known reversible rule, and capacity is one again. More generally $C(p)=C(1-p)$, because inverting every received bit exchanges the two crossover probabilities. The existing mutual information figure shows this full symmetric curve.

An uncoded transmission illustrates why the raw bit-error probability is not itself channel capacity. At $p=0.1$, transmitting 100 uncoded bits succeeds on all of them with probability $0.9^{100}\approx 0.00002656$. The probability of at least one error is therefore approximately $0.99997344$. A suitable block code changes which received words count as recoverable evidence of a message; simply repeating uncoded transmission at rate one does not obtain the reliability promised at rates below $C$.

The closed convex hull of a weakly compact subset of a Banach space is weakly compact.

The other standard formulation says that a convex subset $C$ of a Banach-space dual is weak-star closed if its intersection with every norm-closed ball is weak-star closed.

A set is weakly sequentially compact when every sequence in it has a subsequence that converges weakly to a point of the set.

For a weak-star compact set $K\subseteq X^{\ast }$, its Szlenk derivation at scale $\epsilon >0$ isIt removes points having a relatively weak-star open neighbourhood of norm diameter at most $\epsilon$.

Transfinite iteration of this operation defines the Szlenk index; one derivation step is not itself the index.

If $X$ is separable, the weak-star topology on $B_{X^{\ast }}$ is metrizable. For a dense sequence $(x_n)$ in $B_X$, a compatible metric is obtained by summing bounded multiples of $2^{-n}|f(x_n)-g(x_n)|$.

The Galerkin method seeks an approximate solution in a finite-dimensional subspace and requires the equation's residual to be orthogonal to that subspace. Uniform energy estimates and compactness can then produce a weak solution as the subspaces become dense.

A weak formulation multiplies a differential equation by a test function, integrates, and uses integration by parts to move derivatives away from the unknown function. Natural boundary conditions appear in the resulting boundary terms.

The solution operator $(-\Delta +m^2{)}^{-1}:L^2(U)\to L^2(U)$ is compact on every bounded open set because it maps boundedly into $H_0^1(U)$ and the Rellich-Kondrashov compactness theorem for H01 embeds that space compactly into $L^2(U)$.

A norm-closed convex subset of a Banach space contains every weak limit of its sequences. Apply Mazur lemma to obtain norm-convergent convex combinations that remain in the set.

Every orthonormal sequence in a Hilbert space converges weakly to zero, by Bessel inequality, but no subsequence converges strongly because distinct terms remain distance $\sqrt{2}$ apart.

In a Hilbert space, weak convergence $x_n\rightharpoonup x$ together with $\Vert x_n\Vert \to \Vert x\Vert$ implies strong convergence $x_n\to x$.

If $x_n\rightharpoonup x$ in a Hilbert space, thenFor $x\ne 0$, take the inner product with $x$ and apply Cauchy-Schwarz before passing to the lower limit.

Every bounded sequence in a separable Hilbert space has a weakly convergent subsequence. Successive subsequences make each basis coordinate converge, and a diagonal subsequence converges in every coordinate; the coordinate criterion for weak convergence in a separable Hilbert space finishes the proof.

For a Hilbertian basis $(e_i)$, a sequence $x_n$ converges weakly if and only if $(x_n)$ is norm bounded and every coordinate sequence $⟨x_n,e_i⟩$ converges.