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The indicator function example reaches a squared L2 norm ratio of two. To approach four, concentrate more strongly near zero. For $0\le a<1/2$, use the measurable functionAll integrals use Lebesgue measure $dx$. Keeping both portions of the squared L2 norm givesConsequently,Every allowed member has ratio strictly below four. For any proposed constant $C<4$, a sufficiently large $a<1/2$ has ratio greater than $C$. Combined with Hardy's integral inequality, this proves that four is the smallest possible squared L2 norm constant.^[ref]

The endpoint is excluded: at $a=1/2$, ${\int }_0^1|f_a{|}^{2}dx={\int }_0^1{x}^{-1}dx=\infty$, and the inner portion of $\Vert H{f}_a{\Vert }_2^2$ also diverges. Its tail still has finite integral four. There is no finite L2 norm ratio at this endpoint, hence no equality example obtained by substituting it. The limit four is unattained by this family. Multiplying each $f_a$ by $\sqrt{1-2a}$ makes its L2 norm one without changing the ratio.

The triangular function $f(x)=(1-|x|)$ on $|x|<1$, zero elsewhere, has Fourier transform $\hat{f}(k)=[\sin (k/2)/(k/2){]}^2$ in the angular-frequency convention. Its squared L2 norm is $2/3$. Parseval's identity therefore givesThis proves the displayed value. The removable value at zero is one, and the fourth-power decay bounds the tail. The sinc-squared integral follows similarly from a rectangular function rather than a triangle.

For a divergence-free drift $\mathbf{b}$ and a function $v$ with zero boundary Sobolev trace, integration by parts gives ${\int }_{\Omega }v\mathbf{b}\cdot \nabla v=\frac{1}{2}{\int }_{\partial \Omega }{v}^2\mathbf{b}\cdot \mathbf{n}-\frac{1}{2}{\int }_{\Omega }(\nabla \cdot \mathbf{b})v^2=0$. Consequently the advection matrix in a conforming Galerkin method has zero quadratic form; its symmetric part vanishes. Adding a diffusion stiffness matrix gives a nonpositive spatial generator and a nonincreasing L2 norm in the semidiscrete advection-diffusion equation. The same identity holds with a suitable zero normal-flux boundary term instead of zero trace.

For a compact Hausdorff space $K$, $C(K)$ consists of continuous scalar-valued functions with the supremum norm. It is a Banach space and, with complex scalars, pointwise multiplication and conjugation make it a commutative unital C-star algebra.

Why does completeness hold? A Cauchy sequence in the supremum norm is Cauchy at every point, hence has a pointwise limit. The uniform Cauchy estimate passes to that limit, giving uniform convergence; the uniform limit theorem makes the limit continuous. Thus the norm and the chosen regularity of the functions cooperate to keep every limit inside $C(K)$.

Changing to the L2 norm on $C([0,1])$ loses this property. For integers $n\ge 2$, the continuous functions $h_n(x)=\min (1,\max (0,n(x-1/2)))$ approximate the step function $h(x)=0$ for $x\le 1/2$ and $h(x)=1$ for $x>1/2$. Direct integration givesThe triangle inequality makes $(h_n)$ a Cauchy sequence, but no continuous function agrees with $h$ almost everywhere: continuity would force value zero on the whole left open interval and one on the whole right open interval. Hence this normed space is incomplete. The missing limit exists in L2 space, its completion of a normed space, showing exactly what the larger space adds.

Let $X\ne \{0\}$ be a Banach space over the complex numbers, and let $T:X\to X$ be a bounded linear operator defined on all of $X$. Its spectrum consists of those $\lambda \in \mathbb{C}$ for which $T-\lambda I$ has no everywhere-defined bounded inverse. Its complement is the resolvent set of an operator $\rho (T)$. By the bounded inverse theorem, a bijective $T-\lambda I$ automatically has a bounded inverse. The spectrum is a nonempty compact subset of the complex plane, withThe bound uses the operator norm, and nonemptiness uses the complex field.^[ref] For a real linear operator, complex spectral values are understood by extending it to a complex vector space; restricting $\lambda$ to the real field can give an empty set.

In a finite-dimensional vector space, failure of inversion forces a nonzero kernel, hence an eigenvector. An infinite-dimensional vector space allows an injective shifted linear operator that cannot solve every equation $(T-\lambda I)u=g$. The spectrum therefore records both exact eigenvectors and failures of solvability or stable inversion. Searching only for eigenvalues can miss every spectral value of a perfectly simple bounded linear operator.

Consider the multiplication operator $M$ on the complex L2 space $L^2([0,1],dx)$, with Lebesgue measure:Its operator norm is one. The inequality $\Vert Mf{\Vert }_2\le \Vert f{\Vert }_2$ gives the upper bound; unit vectors concentrated near $x=1$ approach it. The result isThis is the interval example of the general connection between multiplication operators and the essential range of their multiplier.^[ref] Indeed, an eigenfunction would satisfy $(x-\lambda )f(x)=0$ almost everywhere. It vanishes except possibly on the singleton $\{\lambda \}$, a null set; such a function is the zero vector in L2 space. Concentrating at one point cannot produce a nonzero L2 function.

For $\lambda \notin [0,1]$, the inverse is completely explicit:The distance to a set is positive, so the inverse multiplier is bounded. Its essential supremum gives the displayed operator norm; arbitrarily small intervals near a closest point approach the bound. Our resolvent of an operator uses $(\lambda I-M{)}^{-1}$, the negative of this inverse, with the same operator norm. For $\lambda =-1/4$, $5/4$, and $1/2+i/10$, the inverse norms are respectively $4$, $4$, and $10$. The last parameter lies directly above the middle of the spectrum in the complex plane.

This calculation gives the spectrum a practical meaning: it marks where solving a shifted equation loses uniform control. At $\lambda =1/2+i\eta$, the largest amplification of an input error is $1/|\eta |$. Changing $\eta$ from $0.1$ to $0.01$ increases that factor from ten to one hundred. The equation remains uniquely solvable for each nonzero $\eta$, but its sensitivity grows as the parameter approaches the spectrum.

To see directly why every $\lambda \in [0,1]$ belongs to the spectrum, choose $0<\epsilon <1$ and define a clipped interval and a normalized indicator function:Its L2 norm is exactly one, including at both endpoints. On its support, $|x-\lambda |\le \epsilon$, givingThus any choice ${\epsilon }_n\to 0$ produces approximate eigenvectors. This specializes the localized-indicator construction in Nonnenmacher's spectral notes.^[ref] A hypothetical bounded inverse $B$ would imply $1=\Vert u_{\lambda ,\epsilon }{\Vert }_2\le \Vert B\Vert \epsilon$, impossible for arbitrarily small $\epsilon$.

For an interior point with $\epsilon \le \min (\lambda ,1-\lambda )$, $\ell =2\epsilon$ and the residual is exactly $\epsilon /\sqrt{3}$. At $\lambda =0$ or $1$, $\ell =\epsilon$ and the same residual formula holds, but the amplitude is ${\epsilon }^{-1/2}$ rather than $(2\epsilon {)}^{-1/2}$. At $\lambda =1/2$, taking $\epsilon =0.2,0.1,0.05$ gives amplitudes approximately $1.5811,2.2361,3.1623$ and residuals $0.11547,0.057735,0.028868$. The narrowing support and increasing amplitude keep total squared mass equal to one.

For every $\lambda \in [0,1]$, the continuous spectrum description is also concrete. Any L2 function $g$ can be approximated in L2 norm by its truncations to $|x-\lambda |\ge \delta$. Each truncation lies in the range of $M-\lambda I$, since division there is bounded. The range is therefore dense. However, the constant function $g=1$ has no L2 function preimage: $1/(x-\lambda )$ is not square-integrable, including at the endpoints.

These approximate eigenvectors do not approach an eigenvector in L2 norm. Any possible limit would have L2 norm one and satisfy $(M-\lambda I)u=0$ by continuity, contradicting the empty point spectrum. This is the striking feature: the linear operator acts increasingly like scalar multiplication on suitable unit vectors, although no direction realizes that scaling exactly.

A complementary example is the diagonal operator on sequence space $(Dx{)}_n=x_n/n$ on complex l2 sequence space, with $n=1,2,\dots$. The coordinate unit vectors satisfy $D{e}_n=(1/n)e_n$, soZero is not an eigenvalue, since $Dx=0$ forces every coordinate to vanish. Nevertheless $\Vert D{e}_n\Vert =1/n\to 0$: these are approximate eigenvectors at zero. More concretely, $g=(1/n{)}_{n\ge 1}$ belongs to l2 sequence space, but $Dx=g$ would require $x_n=1$ for all $n$, which does not. Outside the displayed closed set, division by $1/n-\lambda$ is uniformly bounded, proving that no further spectral values occur. Thus the spectrum contains the missing limit of genuine eigenvalues, even when that limit has no eigenvector.

Let $S_n$ be a simple symmetric random walk started at a deterministic value $S_0=s$, and let $T$ be a nonnegative integer-valued stopping time with $\mathbb{E}T<\infty$. The stopped second-moment identity isIn particular, $\mathbb{E}{S}_T^2=\mathbb{E}T$ requires the zero start $s=0$.

For a direct argument, write $S_{T\wedge n}-s={\sum }_{k=1}^n{\xi }_k{1}_{\{T\ge k\}}$. The indicators depend only on the past. Centered independent increments therefore give zero cross terms andFor $m>n$, the same calculation gives the squared L2 norm of the difference as $\mathbb{E}[(T\wedge m)-(T\wedge n)]$, which tends to zero. Thus the stopped positions converge in L2 space, as well as eventually pointwise to $S_T$, and both their first and second moments pass to the limit. This supplies the required justification for an unbounded stopping time; almost-sure convergence alone would not pass the second moment.

For a regular self-adjoint Sturm-Liouville problem, choose its eigenfunctions $y_n$ orthogonal in the weighted Hilbert space $L^2((a,b),wdx)$. For real or complex $f$ in that space,The expansion converges in the weighted L2 norm. Pointwise, uniform or differentiated convergence needs extra assumptions; it does not follow from square integrability alone. The conjugation fixes the complex inner-product convention.

For the Dirichlet boundary conditions on $(0,\pi )$ with $p=w=1$, $q=0$, this is the Fourier sine series: $y_n=\sin (nx)$ and $c_n=(2/\pi ){\int }_0^{\pi }f(x)\sin (nx)dx$. The ordinary sine basis is therefore one instance of a general spectral expansion rather than an unrelated Fourier trick.

On an orthogonal diffusion eigenmode, the theta method has the displayed amplification with $s=-\Delta t\lambda$. The identity $(1+as{)}^2-[1-(1-a)s{]}^2=2s+(2a-1)s^2$ proves contraction at all nonnegative $s$ exactly when $a\ge 1/2$. For $a<1/2$, contraction holds up to $s\le 2/(1-2a)$, with a positive denominator in that range. Orthogonality is what transfers these modal bounds to a full L2 norm estimate.

For a uniformly bounded two-level Fourier companion matrix $C(\theta )$ with distinct roots $g_1,g_2$ of modulus at most one, its powers satisfyA positive frequency-uniform gap therefore gives a uniform bound on every power. The Parseval identity then supplies spatial L2 norm stability. When the gap closes at a repeated unit root, a Jordan block can produce growth proportional to the number of steps despite both roots having modulus one.

For $U_m^{n+1}=(1-2\mu )(U_m^n-U_{m+1}^n)+U_{m+1}^{n-1}$, write an amplification root as $G=e^{i\theta /2}\lambda$. Its equation is ${\lambda }^2+2i(1-2\mu )\sin (\theta /2)\lambda -1=0$. The roots have modulus one and a positive frequency-uniform separation exactly for $0<\mu <1$, giving L2 norm stability through a uniformly bounded diagonalization. At either endpoint the Nyquist symbol has a repeated unit root and a Jordan block, producing linear growth. The normalized local truncation error is generally second order; at $\mu =1/2$ and $1$ exactly sampled characteristics solve the recurrence, although the latter parameter is unstable for arbitrary perturbations.

For constant-speed advection $u_t+a{u}_x=0$ with $a>0$, the explicit upwind scheme uses $U_j^{n+1}=(1-\nu )U_j^n+\nu U_{j-1}^n$, $\nu =ak/h$. For $a<0$ the neighbour must be on the other side. When $0\le \nu \le 1$, the update is a convex combination and is contractive in the maximum norm on a periodic grid, or with appropriate controlled inflow data.

On a periodic grid, von Neumann stability analysis gives $G=1-\nu +\nu e^{-i\theta }$ and $|G{|}^2=1-2\nu (1-\nu )(1-\cos \theta )$, proving the same exact contraction range in the discrete L2 norm. Boundary estimates are still required on a finite interval; stable scalar eigenvalues of a triangular boundary update alone need not give a mesh-uniform bound.

With positive diffusion coefficient $a(x)$, this diffusion equation expresses conservation with flux $-a(x)u_x$. For a differentiable coefficient, its spatial operator is $a{u}_{xx}+a^{\prime }{u}_x$, rather than either $a{u}_x$ or $a{u}_{xx}$ alone. A symmetric half-grid diffusion consistency discretization uses the difference of the two adjacent interface fluxes, preserving that divergence structure. Suitable Dirichlet boundary conditions make the spatial operator dissipative in the L2 norm.

The vector field method commutes a wave equation with spacetime symmetry generators, estimates the resulting commuted wave energies, and converts weighted L2 norms to pointwise decay using a Klainerman-Sobolev inequality. The commutation vector fields for the wave equation include translations, rotations, Lorentz boost vector fields and the scaling vector field. It is useful for almost global existence for wave equations and for small-data global existence under the classical null condition for wave equations.

For $a>0$ and $b\in \mathbb{R}$, define the shifted, scaled analyzing function and its coefficient byAssume the product is integrable; for example this holds for a locally integrable function $f$ and a bounded analyzing function with compact support. The bar is complex conjugation. This normalization preserves the L2 norm of the analyzing function. If $p$ is a polynomial of degree of a polynomial less than $q$, then $p(b+au)$ is a sum of monomials $u^r$ with $r<q$. Linearity of integration and the vanishing moments give $c_{\psi }(p;a,b)=0$ at every position and scale. Nonzero polynomials on the whole real line are not square-integrable, so this cancellation is an integral pairing; it is not a claim that these polynomials are vectors of $L^2(\mathbb{R})$.

Here is a quantitative reason that smooth regions yield small details. Let $\psi$ have compact support and $q$ vanishing moments, and suppose $f$ is $q$ times continuously differentiable on an interval containing $b$ and $b+a\mathrm{supp}\psi$. Subtract its degree-$(q-1)$ Taylor polynomial at $b$. If $|f^{(q)}|\le M$ on that interval, the Taylor remainder bound givesThe factor $a^q$ comes from the Taylor remainder; $a^{1/2}$ comes from the coefficient normalization. Thus the analyzing function measures the departure from the local polynomial trend, rather than the height of $f$ itself. The wavelet coefficient decay for Hölder functions uses the same argument with a Hölder-Taylor remainder bound. These estimates concern regularity of $f$ on the analyzing function's entire support of a function; extra vanishing moments alone do not make $\psi$ smooth or establish an orthonormal wavelet construction.^[ref]

For an orthonormal wavelet and its scaling function, set ${\varphi }_{j,k}(x)=2^{j/2}\varphi (2^{j}x-k)$ and similarly for $\psi$. The coarse scaling functions together with nonnegative-resolution wavelets form an orthonormal basis of $L^2(\mathbb{R})$. The coefficients are the corresponding L2 inner products. The series converges in the L2 norm, and the Parseval identity for a Hilbertian basis equates squared norm with the sum of squared coefficients. Pointwise convergence requires additional hypotheses.