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For the nonlinear Schrödinger equationwith real-valued $F$, a sufficiently regular, decaying solution with finite L2 norm conservesDifferentiating gives $M^{\prime }(t)=2\mathrm{Re}\int u_t\overline{u}$. Substitute $u_t=i\Delta u+iF(|u{|}^2)u$. Integration by parts makes $\int \Delta u\overline{u}=-\int |\nabla u{|}^2$ real, and $\int F(|u{|}^2)|u{|}^2$ is real as well, so multiplication by $i$ gives zero real part.

The real coefficient is essential to this calculation; phase covariance alone does not imply mass conservation. For an $H^1$ solution obtained by a well-posed approximation theory, the invariant extends from smooth solutions by continuity in the L2 norm.^[ref]

The mass matrix is the Gram matrix of a finite element basis in the L2 inner product. Its coefficient quadratic form is the squared L2 norm of the represented trial function, so it is a positive-definite matrix for a linearly independent basis.

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 an approximation $a$ is compared with a target $f$ through an intermediate approximation $b$, with both error functions in the same Lp space. The identity $f-a=(f-b)+(b-a)$ and the Minkowski inequality combine two error budgets without any independence assumption. The L1 norm controls aggregate absolute error, the L2 norm controls the square root of aggregate quadratic loss, and the L-infinity norm controls the largest error almost everywhere. For equally weighted samples with the probability measure $1/n$ on each coordinate, the L2 norm is the root mean square error; the inequality still applies when this normalization is used consistently.

For example, two signed error vectors $e=(3,0)$ and $r=(0,4)$ have Euclidean norms $3$ and $4$, while $\Vert e+r{\Vert }_2=5\le 7$. Reinforcing errors $e=(3,0)$ and $r=(4,0)$ attain the bound $7$; opposing errors $e=(3,0)$ and $r=(-3,0)$ cancel completely. The Minkowski inequality always allows reinforcement. A smaller combined budget requires additional information about the errors, not just their separate norms.

Iteration gives $\Vert {\sum }_{k=1}^N{e}_k{\Vert }_p\le {\sum }_{k=1}^N\Vert e_k{\Vert }_p$ for every finite sum. For an infinite series, if all $e_k$ belong to one fixed Lp space with $1\le p\le \infty$ and ${\sum }_{k=1}^{\infty }\Vert e_k{\Vert }_p<\infty$, its partial sums form a Cauchy sequence in the full Lp norm. Completeness of this Banach space then gives norm convergence to an element of the same Lp space. A finite sum bound alone does not establish convergence, and completeness is an additional ingredient beyond the Minkowski inequality. ^[ref]

For $0<a(x)\le \beta$, the explicit conservative diffusion update has weights $k{a}_{m-1/2}/h^2$, $1-k(a_{m-1/2}+a_{m+1/2})/h^2$, and $k{a}_{m+1/2}/h^2$. Under the displayed restriction they are nonnegative and sum to one. With zero Dirichlet boundary conditions, each new value is a convex combination of old interior and boundary values, proving contraction in the discrete maximum norm. The spatial matrix is also symmetric with $v^T{L}_{h}v=-h^{-2}\sum a_{m+1/2}(v_{m+1}-v_m{)}^2$. This puts its eigenvalues in $[-4\beta /h^2,0]$, so the same restriction gives L2 norm contraction. A stable recurrence need not be consistent with a different proposed differential equation.

A multiresolution analysis organizes a function into successively finer approximation spaces. In the usual orthonormal construction on $L^2(\mathbb{R})$, these are closed linear subspaces $(V_j{)}_{j\in \mathbb{Z}}$ withRescaling relates the levels: $f(x)\in V_j$ if and only if $f(2x)\in V_{j+1}$. Integer translates of a scaling function form an orthonormal basis of $V_0$. The detail space is the orthogonal complement of $V_j$ inside $V_{j+1}$, giving $V_{j+1}=V_j\oplus W_j$. Coarse information and new details therefore fit together without overlap in squared L2 norm.

For the Haar wavelet, $V_j$ consists of square-integrable functions constant on cells of length $2^{-j}$. Splitting every cell into two halves creates one new contrast per old cell, the Haar refinement identity. On the bounded example $[0,1]$, the first three spaces have dimensions $1,2,4$: a whole-interval mean, two half-interval means, and four quarter-interval means. The Haar projection diagram displays those three resolutions. This bounded construction starts at level zero; the displayed intersection condition refers to the full family on the real line.

The usefulness is a controlled way to increase resolution. Earlier coarse information remains available while new wavelet coefficients identify what the finer description adds. The wavelet approximation complexity connects this extra resolution to both approximation error and the number of coefficients being estimated. ^[ref]

Nonlinear Schrödinger blowup analysis asks whether a solution remains in its local well-posedness class or concentrates so strongly that continuation fails in finite time. For the mass-critical focusing nonlinear Schrödinger equation in the standard $H^1$ theory, finite-time failure forces the gradient L2 norm to diverge, even though Mass conservation for the nonlinear Schrödinger equation keeps the total squared L2 norm constant.

Concentration and total size are different. The mass-preserving scaling $u_{\lambda }(x)={\lambda }^{d/2}u(\lambda x)$ leaves $\Vert u_{\lambda }{\Vert }_2=\Vert u{\Vert }_2$ while $\Vert \nabla u_{\lambda }{\Vert }_2=\lambda \Vert \nabla u{\Vert }_2$. Thus a fixed amount of mass can occupy smaller spatial scales with larger gradients.

The Sharp Gagliardo-Nirenberg inequality gives coercivity below the NLS ground-state mass, while a virial identity can force finite-time failure for suitable negative-energy, finite-variance data. Near the threshold, profile decomposition modulo translations separates a concentrating profile from dispersive remainder. These arguments track the structure of a solution rather than merely the maximum of a plotted wave.

For $U_x=iL(x)U$ with a self-adjoint transverse differential operator $L(x)$, differentiating the squared L2 norm gives $i⟨U,LU⟩-i⟨LU,U⟩=0$. Reality of the refractive coefficient and vanishing transverse boundary flux are required. This is exact for the scalar envelope model. Its physical interpretation as leading paraxial power uses a fixed reference impedance; a varying local impedance or a longitudinal electromagnetic polarization requires a consistent Maxwell calculation and cannot be inferred from this invariant alone.

A normed vector space is a vector space equipped with a norm, whose induced metric is $d(x,y)=\Vert x-y\Vert$. The norm must be positive on nonzero vectors, scale by the absolute value of a scalar, and satisfy the triangle inequality. These conditions make distance compatible with addition and scaling.

On $C([0,1])$, both the supremum norm $\Vert f{\Vert }_{\infty }={\max }_{0\le x\le 1}|f(x)|$ and the L2 norm $\Vert f{\Vert }_2=({\int }_0^1|f(x){|}^{2}dx{)}^{1/2}$ are genuine norms. A nonzero continuous function stays nonzero on an interval of positive length, so its squared integral cannot vanish. For $f(x)=x$, they give $1$ and $1/\sqrt{3}$, respectively. Always $\Vert f{\Vert }_2\le \Vert f{\Vert }_{\infty }$, but the shrinking peaks below show that no uniform reverse bound exists. Unlike the finite-dimensional equivalence of norms, infinite-dimensional vector spaces can support inequivalent norm topologies.

An orthonormal wavelet $\psi$ generates an orthonormal basis of $L^2(\mathbb{R})$ throughThe factor $2^{j/2}$ compensates for the shrinking support of a function, keeping the L2 norm equal to one. Orthogonality makes different coefficients separate squared-norm contributions, and completeness allows the wavelet series to recover any square-integrable function in the L2 norm.

For the Haar wavelet, wavelets at one scale have disjoint supports. At different scales their dyadic intervals are either disjoint or nested. On the support of the finer wavelet, the coarser one is constant, while the finer one has zero integral, so their L2 inner product is zero. Together with the unit L2 norm, this explains orthonormality. On the bounded interval $[0,1]$, add the constant Haar scaling function and use $j\ge 0$, $0\le k<2^j$; the constant function is not an $L^2(\mathbb{R})$ basis element on the whole line.

Take the real L2 space $\mathcal{H}=L^2((0,\infty ),dt)$ and the feature mapEach indicator function belongs to $\mathcal{H}$ because its squared L2 norm is $x<\infty$. Their inner product is the length of the overlapping intervals:This realizes $k_1$ as a positive-semidefinite kernel; it is also the Brownian covariance kernel.

Let $e=u-p$. If $p$ minimizes the squared L2 norm error, then for any $q\in A$ the real quadratic $\Vert e-tq{\Vert }_2^2=\Vert e{\Vert }_2^2-2t(e,q)+t^2\Vert q{\Vert }_2^2$ has its minimum at $t=0$. Its derivative there is zero, giving $(e,q)=0$.

Conversely, if $e$ is orthogonal to $A$, then for every $q\in A$, $p-q\in A$ and the Pythagorean theorem for inner product spaces givesThis proves the equivalence and identifies $p$ as an orthogonal projection. It characterizes a minimizer whenever one exists; a general nonclosed vector subspace need not admit a projection for every $u$.

The orthogonal projection is $p_n={\sum }_{r=0}^n(u,{\varphi }_r){\varphi }_r$. Subtract it from $u$ and use the orthonormal basis relations to obtainThis is the requested Bessel inequality.

For the infinite tail, use completeness of the orthonormal polynomial family in the usual weighted L2 space. The relevant density facts are that continuous functions are dense for the finite measure $w(x)dx$, and the Weierstrass approximation theorem makes polynomials uniformly dense in continuous functions on the compact interval. Since $\int w<\infty$, uniform approximation also gives weighted L2 norm approximation. Parseval's identity therefore givesHere finiteness of the weight integral is also necessary for the constant polynomial to belong to the stated Hilbert space. Without completeness of the polynomial family, an additional squared component orthogonal to every polynomial would remain; Bessel inequality alone does not justify equality.

For every real $r$ and integer $n\ge 1$, $|\sin r|\le \min (1,|r|)$ givesThis estimate covers both even and odd $n$, without any assumption on the sign of $u$. Since $a(t)\ge 0$, the energy estimate from part (i) becomesEquivalently, an integrating factor givesThe bracket is nonnegative and starts at zero, so it vanishes identically. This is also the zero-data case of the Gronwall inequality. The vanishing L2 norm and continuity of $u$ imply

Define the squared L2 norm $E(t)={\int }_0^{1}u(x,t{)}^{2}dx$. Multiplication of the semilinear partial differential equation by $u$, followed by integration by parts, givesAt $x=1$ the Dirichlet boundary condition makes the boundary term zero. At $x=0$ the Robin boundary condition is $u_x(0,t)=\alpha u(0,t)$, so subtraction of the lower endpoint gives $-\alpha u(0,t{)}^2$. In particular the sign is dissipative. Thus the exact energy identity iswhich implies

For a positive eigenvalue write $\mu =\sqrt{\lambda }>0$. The general solution of the limiting constant-coefficient differential equation is $y=A\cos (\mu x)+B\sin (\mu x)$. The endpoint at zero gives $A=0$. A nonzero eigenfunction then requires $B\ne 0$, and the endpoint at $\pi$ gives $\sin (\mu \pi )=0$. ConsequentlyFor real eigenfunctions the constants may be any nonzero real amplitudes. Requiring unit L2 norm would set $|C_k|=\sqrt{2/\pi }$, but that normalization is not required by the question. The lowest eigenvalue is ${\lambda }_1=1$ and its eigenspace is spanned by $\sin x$. For completeness, zero yields only an affine solution annihilated by both endpoints, and negative eigenvalues yield hyperbolic functions whose two zero endpoint values also force the zero solution.

Use a conforming finite element space $V_h\subset H_0^1(\Omega )$ with the prescribed basis, and write $u_h(x,y,t)={\sum }_j{U}_j(t){\varphi }_j(x,y)$. Testing the advection-diffusion equation by ${\varphi }_i$ and applying integration by parts to its diffusion term givesThusHere $M$ is the mass matrix; it is a symmetric positive-definite matrix because the basis is linearly independent. The initial vector represents the chosen finite element approximation to $g$, for example its orthogonal projection.

For any coefficient vector $U$, put $v_h={\sum }_j{U}_j{\varphi }_j$. The skew symmetry of Galerkin advection with constant drift $(1,1)$ givesbecause the Sobolev trace of $v_h$ vanishes. ConsequentlyAlong the solution, ${\int }_{\Omega }{u}_h^2=U^{T}MU$. The constant symmetric matrix $M$ and the semidiscrete equations therefore giveThe L2 norm of the spatial finite element solution is nonincreasing. The constant drift transports it without changing this energy; the diffusion term dissipates it. Any subsequent time-stepping method must be chosen to preserve an appropriate discrete stability estimate.

The infinitesimal strain tensor $\epsilon (v)=\frac{1}{2}(\nabla v+(\nabla v{)}^{\mathsf{T}})$ and the divergence are linear in $v$. Consequently the given $a$ is a symmetric bilinear form, andFor a matrix-valued function, $\Vert \cdot {\Vert }_0$ here uses the Frobenius inner product pointwise and then the L2 norm. All expressions are finite for $v\in H^1(\Omega ;{\mathbb{R}}^3)$.

Expand the quadratic term at $\theta v+(1-\theta )w$. The linearity of $\ell$ cancels its contribution to the difference, givingFor $0\le \theta \le 1$, this is nonnegative. Thereforeso $J$ is a convex functional on the whole Sobolev space. No boundary condition is needed for this conclusion. The argument uses nonnegativity of the quadratic term; uniqueness on the zero-boundary Sobolev space will instead use the Korn inequality in part (b).

Set $\tilde{\varphi }(x)=\overline{\varphi (-x)}$. This is a Schwartz function and its Fourier transform is $\hat{\tilde{\varphi }}(\xi )=\overline{\hat{\varphi }(\xi )}$. The convolution theorem givesThe convolution of two Schwartz functions is again a Schwartz function: differentiating under the integral and expanding $x^m=(y+(x-y){)}^m$ bounds every weighted derivative by products of integrable and bounded weighted derivatives. Evaluate Fourier inversion for this convolution at zero:The Schwartz space is dense in the L2 space. For $f\in L^2(\mathbb{R})$ choose Schwartz functions ${\varphi }_j\to f$ in the L2 norm. The identity makes ${\hat{\varphi }}_j$ a Cauchy sequence in the L2 space, and its limit defines $\mathcal{F}f$, independently of the approximation. It agrees with the distributional Fourier transform, because Cauchy-Schwarz inequality makes convergence in the L2 norm imply convergence when paired with every Schwartz function. For $f\in L^1\cap L^2$ it therefore agrees almost everywhere with the ordinary Fourier transform.

The extension is onto: its image is closed by the norm identity and contains the Schwartz space, since Fourier inversion maps that space to itself. This proves Plancherel theorem:Its inverse is the extension of the inverse Fourier transform, with limits understood in the L2 norm.

False. A typewriter sequence sweeps successively through finer dyadic partitions. For $k\ge 0$ and $j=0,\dots ,2^k-1$, define the indicator functionIts L2 norm satisfiesAs $n$ tends to infinity so does its block index $k$, proving norm convergence to zero. However, every $x\in (0,1)$ belongs to exactly one of the half-open intervals in each block, so $f_n(x)=1$ infinitely often. For $k\ge 1$ there is also an interval in each block that does not contain $x$, so $f_n(x)=0$ infinitely often. Thus there is no pointwise convergence at any $x\in (0,1)$, let alone convergence almost everywhere.