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Expand a real function with zero endpoint values in its Fourier sine series, $y={\sum }_{n\ge 1}{a}_n\sin (n\pi x/L)$. The Parseval identity gives $\int y^2=(L/2)\sum a_n^2$. For its weak derivative, the Fourier cosine series has coefficients $(n\pi /L)a_n$ and zero constant term, soThe derivative coefficient identity follows by integration by parts, rather than by assuming pointwise differentiation of an arbitrary series. This proves the Wirtinger inequality and its equality case. Under unit L2 norm, the only real minimizers are $y=\pm \sqrt{2/L}\sin (\pi x/L)$.

Suppose the Weierstrass necessary condition applies to a strong local minimum of this fixed-endpoint integral. It requires the tangent to $g$ at $u^{\prime }(x)$ to lie below the entire graph for almost every $x$. A line of nonzero slope is unbounded above in one direction, while $g$ is bounded; hence this supporting line has slope zero. Its contact value is then a global minimum of $g$. The resulting function minimizes the integral globally, since its integrand is almost everywhere equal to the global lower bound. Thus for a bounded differentiable slope-only integrand a strong local minimum can exist only at global energy level. Weak local minima can behave differently because their competitors have nearby derivatives.

For a weakly holomorphic Banach-valued function, choose a closed disk contained in its domain. Each continuous functional is bounded on the image of that disk by scalar continuity. The Uniform boundedness principle, applied to its evaluation functionals in the bidual, gives a common bound $M$ on the norms of the image vectors. Cauchy's formula on the disk's boundary then bounds the scalar difference between difference quotients at increments $h,k$ by a constant times $M\Vert \phi \Vert |h-k|$, uniformly for sufficiently small increments. Explicitly the difference kernel isThe Hahn-Banach theorem dual norm formula converts this into the same norm bound for the vector difference quotients. They are Cauchy, so completeness gives a norm derivative. This also explains why weak holomorphy suffices without a separately assumed norm-continuity hypothesis.

A bounded measurable one-periodic function rescaled to frequency $j$ converges weak-star in $L^{\infty }(\mathbb{R})$ to its period average. Subtract the average and take a bounded periodic primitive $F$. Its integral on $[a,b]$ after rescaling is $[F(jb)-F(ja)]/j$, which tends to zero. Sequential Banach-Alaoglu with separable predual $L^1$ then identifies every cluster point by interval integrals; uniqueness of all cluster points gives full weak-star convergence. Localizing to bounded intervals avoids falsely assuming that a bounded cluster function is globally integrable.

The same Banach space can acquire incomparable weak-star topologies from different isometric preduals. Pair ${\ell }^1$ with $c_0$ by $\sum {\alpha }_n{\beta }_n$ and with the space $c$ of convergent sequences by ${\alpha }_1{\gamma }_{\infty }+{\sum }_{n\ge 2}{\alpha }_n{\gamma }_{n-1}$. The functional $\alpha \mapsto {\alpha }_1$ is continuous only for the first topology: $e_1-e_n\to 0$ for the second but its values remain one. Conversely $\alpha \mapsto \sum {\alpha }_n$ is continuous only for the second: $e_n\to 0$ for the first but its values remain one. More generally, comparable weak-star topologies arising from isometric preduals agree on the common closed unit ball, since its identity map is a continuous bijection from a compact space to a Hausdorff space by the Banach-Alaoglu theorem.

The continuous dual space of ${\ell }^1$ is ${\ell }^{\infty }$ under the complex-linear coordinate pairing. Its subspace $c_0$ is norm closed: uniform approximation by a sequence with a small tail leaves a small tail. Nevertheless its preannihilator in ${\ell }^1$ is zero, because the coordinate unit vectors belong to $c_0$ and test every coordinate. Its double continuous annihilator is therefore all of ${\ell }^{\infty }$. It is weak-star dense directly as well: the finite truncations of any bounded sequence lie in $c_0$ and converge against every ${\ell }^1$ test by absolute summability. This demonstrates why norm closure cannot replace closure in the weak-star topology in the dual-side annihilator theorem.

On $x,t>0$, the displayed smooth nonzero function satisfies $w_t=x{w}_{xx}$. For each fixed $t>0$ it tends to zero at $x=0$ and at spatial infinity; for each fixed $x>0$ it tends to zero as $t\downarrow 0$. Thus these three pointwise limits do not determine a unique solution: a multiple of $w$ may be added to any particular solution of the same forced equation.

To verify the PDE, set $\eta =x/t$. Both $w_t$ and $x{w}_{xx}$ equal $\eta (\eta -2)e^{-\eta }/t^2$. The missing control is at the corner: $w(t,t)=1/(et)$ diverges as $t\downarrow 0$. There is no contradiction with the weak parabolic maximum principle, because a supremum on a past strip may be infinite or unattained. A bounded-solution class excludes this example. A standard uniqueness proof there applies the maximum principle to the difference minus $\delta (1+x+t)$ and lets $\delta \downarrow 0$. Conditions at infinity and at initial time must control the relevant strip, not only each fixed-variable limit.

If a locally integrable function has weak derivatives represented by a continuous vector field $g$, it has a representative $U\in C^1$ with $\nabla U=g$. The assertion is local, so work on a ball compactly contained in the domain and regularize using a mollifier. The regularizations satisfy the segment identity $u_{\epsilon }(x)-u_{\epsilon }(y)={\int }_0^1\nabla u_{\epsilon }(y+t(x-y))\cdot (x-y)dt$. Averaging over a smaller ball in $y$ fixes the additive constant through local $L^1$ convergence. The gradients converge uniformly on compact sets, so the averaged identity gives local uniform convergence to $U$. Passing to the limit on each segment then gives $U(x+h)-U(x)-g(x)\cdot h=o(|h|)$. Continuity of $g$ proves that $U$ is continuously differentiable.

The displayed spikes have $L^p$ norm one. Pairing with $g\in L^{p^{\prime }}$ and applying Holder inequality bounds the pairing by $({\int }_0^{1/j}|g{|}^{p^{\prime }}{)}^{1/p^{\prime }}$, which tends to zero. At exponent one, however, $j{1}_{(0,1/j)}$ cannot converge weakly in $L^1$: every interval away from zero forces any limit to vanish there, while the constant test function preserves integral one. Their failure of uniform integrability records exactly the obstruction in the Dunford-Pettis theorem.

The displayed nonzero rational sequence tends to zero both in the usual real absolute value and in every p-adic absolute value. In the real absolute value, $|y_n|\le 1/n!\to 0$. For a fixed prime $p$ and $n\ge p$, the denominator is congruent to one modulo $p$, so its P-adic valuation is zero. Therefore $v_p(y_n)=v_p(n!)\to \infty$, proving $|y_n{|}_p\to 0$.

This is convergence at each fixed place; it is not uniform convergence over all primes. For each particular $n$, new prime divisors of the denominator can compensate the small values at the previously fixed places, consistently with the product formula. Thus a product identity for a single rational number does not prohibit pointwise convergence at infinitely many absolute values.

For nonnegative Lebesgue measurable $f$ on $(0,1)$, the displayed Volterra power-kernel integral is defined for every $0<x<1$ as a nonnegative extended Lebesgue integral. The factor is finite and continuous on the open integration interval $(0,x)$, and an endpoint value can be assigned arbitrarily. The result may initially be infinite; for example $f(y)=1/y$ makes every value infinite. At $\beta =1$, it reduces to the ordinary Volterra operator.

For joint integration, use the integral kernel $(x-y{)}^{\beta -1}$ on $0<y<x<1$ and zero elsewhere. This piecewise prescription avoids the singular diagonal and never raises a negative number to a noninteger power. The L1 bound for a Volterra power-kernel integral gives an integrable output, finite almost everywhere, whenever $f$ is integrable.

On $L^2(-a,a)$, the adjoint operator is $T^{\ast }g(s)={\int }_s^{a}g(t)dt$ for $s>0$, and $T^{\ast }g(s)=-{\int }_{-a}^{s}g(t)dt$ for $s<0$. These formulas follow by reversing the triangular integration order separately on the two half-intervals. Consequently $(T+T^{\ast })g$ is constant on each half, with values ${\int }_0^{a}g$ and $-{\int }_{-a}^{0}g$. Applying this adjoint to a real weight transfers a weighted integral of an anchored function to an integral against its derivative.

A clinical consequence of vitamin C deficiency, with impaired collagen formation and fragile connective tissue. ^[ref]

For infinite spaces over a field of size $q$, two-variable types are determined by the kernel of the linear map sending $(a,b)$ to $ax+by$. Its ranks give the zero pair, $q+1$ one-dimensional directions and the independent-pair type. All $q+3$ types are isolated by finite formulas. Quantifier elimination makes the Lindenbaum algebra their full power-set Boolean algebra. The one-variable algebra has just the zero and nonzero atoms, hence four Elements.

The positive-radius closed Euclidean balls in ${\mathbb{R}}^d$ have VC dimension $d+1$ for $d\ge 1$. Vertices of a regular simplex are shattered: interpolate any desired signs by an affine function $a\cdot x+b$, then perturb a sphere through the vertices to the score $1+\eta b+\eta a\cdot x-\Vert x{\Vert }^2$. For the upper bound, take an affine dependence $\lambda$ among $d+2$ points and orient it so that $Q={\sum }_i{\lambda }_i\Vert x_i{\Vert }^2\ge 0$. Labelling the positive coefficients inside and the negative coefficients outside would force ${\sum }_i{\lambda }_i(r^2-\Vert x_i-c{\Vert }^2)>0$, while the affine dependence makes this sum equal $-Q\le 0$. The weaker $d+2$ upper bound from unrestricted lifted affine linear classifiers loses their fixed quadratic coefficient.