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With $x=2z-1$, the chain rule gives $T_{xx}=V_{zz}/4$. ThusSeparation of variables gives the eigenfunctions $\sin (n\pi z)$ with time factors $e^{-n^2{\pi }^{2}t/4}$. The Fourier sine coefficients of the initial constant areOnly odd modes survive, soReturning to $x$, $\sin ((2k+1)\pi (x+1)/2)=(-1{)}^k\cos ((2k+1)\pi x/2)$, henceFor every $t>0$ the exponential coefficients give uniform convergence of the differentiated series on later-time compact sets, verifying the PDE and boundary conditions. As $t\downarrow 0$, the solution tends to $T_0$ at every interior point and in L2 norm, while remaining zero at the endpoints. The expression is visibly even in $x$, consistent with the corrected symmetry argument for $b=0$.

The truncated Fourier cosine series gives the better approximation on the closed interval, in the senses of uniform convergence and the asymptotic L2 norm error. To make the comparison precise, let $C_N$ and $S_N$ retain the frequencies $1\le n\le N$, with the constant term also included in $C_N$.

The cosine Fourier coefficients are $O(n^{-2})$. In fact, absolute convergence gives the uniform tail boundFor the Fourier sine series, every Fourier partial sum satisfies $S_N(0)=S_N(\pi )=0$. Thusso there is no uniform convergence to $f$ on the closed interval. Its odd extension has jump discontinuities, producing the Gibbs phenomenon near the endpoints; its Fourier coefficients decrease only like $1/n$.

The advantage persists in the L2 norm, which ignores individual endpoint values. By Parseval's identity and Fourier orthogonality,The nonzero odd cosine coefficients and both parity classes of sine coefficients also give matching lower bounds of these orders by comparison with the corresponding p-series. Hence their L2 norm errors decrease respectively like $N^{-3/2}$ and $N^{-1/2}$. This is a comparison of approximation quality in specified norms, not a claim that one finite sum has smaller error at every individual interior point.

Take $\boxed{f(x)=\sin x}$. This function is not a polynomial: it is nonzero and satisfies $f^{\prime \prime }=-f$, whereas a nonzero polynomial cannot satisfy that identity, since taking two derivatives reduces its degree or gives zero.

By Fourier orthogonality, the Fourier sine series consists of the single term $\sin x$. Thus every truncation with $N\ge 1$ is exact, including at the endpoints. In contrast, using the product-to-sum trigonometric identity in its Fourier cosine coefficients givesFor example, ${\int }_0^{\pi }\sin x\cos (2mx)dx=-2/(4m^2-1)$ follows by integrating the two resulting sine terms. Therefore the Fourier cosine series isEvery finite cosine truncation omits infinitely many nonzero Fourier coefficients, so Parseval's identity gives it strictly positive squared L2 norm error. The one-term sine representation has zero error in both the L2 norm and the supremum norm, proving the requested advantage.

For a real $2L$-periodic square-integrable function with the convention $f\sim a_0/2+{\sum }_{n\ge 1}(a_n\cos (n\pi x/L)+b_n\sin (n\pi x/L))$, Parseval's identity isIt states that the squared L2 norm equals the sum of the squared components in the orthogonal Fourier basis, with the normalization factors shown. In the present even-function case every $b_n$ vanishes.

For a separated nonzero solution $T=f(x)g(y)$, substitution and division by $fg$ givesThe homogeneous Dirichlet boundary conditions impose $g(0)=g(1)=0$. For a nonzero spatial factor, integration by parts gives ${\lambda }^2{\int }_0^1{g}^{2}dy={\int }_0^1(g^{\prime }{)}^{2}dy$, so the separation value is nonnegative. The zero value gives only a trivial linear $g$ under these boundary conditions. The nontrivial separated factors, for $a\ne 0$, are thereforeFor the intended forward diffusion equation, assume $a>0$. The superposition principle gives a Fourier sine series, whose coefficients must expand the initial ramp in the interior:Consequently the standard heat-flow solution isIt satisfies both lateral boundary conditions for $x>0$, and tends to $y$ as $x\downarrow 0$ for $0<y<1$, also converging to the ramp in the L2 norm. The exponential coefficient decay justifies differentiating the series for positive $x$.

There is a genuine incompatibility in the printed data: at $(0,1)$ the lateral condition demands zero while the initial condition demands one. Thus no function satisfies every stated value at that corner. The displayed series solves the usual consistent interpretation with initial data on $0<y<1$ and lateral data for $x>0$; it cannot be continuous at the upper initial corner. This discrepancy is present in the original PDF.

The sign of $a$ is also unstated. If $a=0$, the spatial equation with two zero endpoints forces $T=0$, incompatible with the ramp. If $a<0$, the separated Fourier coefficients grow like $e^{n^2{\pi }^{2}x/|a|}/n$ and fail to be square summable for every $x>0$; these data therefore have no forward solution in the standard L2 space class. The positive-$a$ assumption is essential to the stated heat-flow interpretation.

Let $S_N$ be the Fourier partial sum through frequency $N$. By orthogonality and the coefficient formulas,Expanding the nonnegative squared L2 norm of the error gives the exact finite identityFor $f$ in the L2 space, completeness of the Fourier basis makes the left-hand side tend to zero. Dividing the resulting equality by $2L$ proves Parseval's identity:This argument uses convergence in L2 rather than an unqualified exchange of an infinite sum with the integral of its square. The function in part (c) is continuous and piecewise polynomial, so these hypotheses hold.

For a nonzero separated solution $v=X(x)T(t)$, separation of variables gives $X^{\prime \prime }+{\kappa }^{2}X=0$ and $T^{\prime }=-\nu {\kappa }^{2}T$. The endpoint conditions require $X=\sin \kappa x$ and $\cos \kappa L=0$. Thus the mixed Dirichlet-Neumann modes on an interval have ${\kappa }_n=(n+1/2)\pi /L$, $n\ge 0$. Their squared L2 norms are $L/2$, so the initial Fourier coefficients areThe integration uses $\cos {\kappa }_{n}L=0$ and $\sin {\kappa }_{n}L=(-1{)}^n$. ThereforeAt positive time the series can be differentiated termwise and satisfies both homogeneous endpoint conditions for $v$. At time zero it recovers $-Hx/k$, so $u$ recovers the initial zero temperature. If $H\ne 0$, the prescribed derivative at $(L,0)$ is incompatible with differentiating the initial zero profile; the temperature is still continuous there, while its spatial derivative has a corner layer. At long times the decaying modes leave $u\to Hx/k$.

The direct squared L2 norm is the length of the interval, namely $2$. The Fourier transform isFor the transformed norm, let $I={\int }_0^{\infty }{\sin }^{2}k/k^{2}dk$. Integration by parts, with vanishing endpoint term $-{\sin }^{2}k/k$, givesThe supplied oscillatory integral is understood as an improper integral. Thereforewhich agrees with the direct norm and verifies the Parseval identity with the specified normalization.

Project the initial condition onto the orthogonal spatial eigenfunctions. Their squared norms in the ordinary inner product areThe latter follows from the integrals of ${\cos }^2(n\pi x)$ and ${\sin }^2(n\pi x)$ being $1/2$, with the sine-cosine cross term zero. ThereforeFor square-integrable data the expansion converges to the data in the L2 norm at $t=0$; for $t>0$ the decaying high modes are smoothed. Initial data with zero $e^x$ projection have no growing term.

The norm induced by the specified inner product is the L2 normA polynomial $p_n\in {\Pi }_n$ is a best polynomial least-squares approximation whenMinimizing this norm is equivalent to minimizing its square, the integrated squared error. Continuity ensures that a function with zero squared norm is identically zero: a nonzero value would give a neighbourhood on which the square has a positive integral.

For a grid sequence define its semidiscrete Fourier transform in the dimensionless frequency variable $\theta \in [-\pi ,\pi ]$ byFor general square-summable data the transform is interpreted as an limit in the L2 norm, rather than an absolutely convergent pointwise sum. The Parseval identity in this normalization isThe Fourier symbols of $D_x^0$ and $D_x^{+}{D}_x^{-}$ are $i\sin \theta /\Delta x$ and $-4{\sin }^2(\theta /2)/(\Delta x{)}^2$. Transforming the implicit equation givesIts amplification factor is consequentlyThe denominator has real part at least one because $a\ge 0$. It never vanishes, so multiplication by $G_I$ defines a unique next value in the l2 sequence space, andThe Parseval identity now givesThis is unconditional practical stability of a finite difference scheme: in the course's terminology the numerical evolution does not increase this norm. Linearity gives the same estimate for perturbations of the initial data. This contraction convention is stronger than a finite-time mesh-uniform bound allowing physical exponential growth.^[ref]

The semidiscrete system $U^{\prime }=D_{xx}U+\alpha D_{0}U$, for real fixed $\alpha$, is contractive in the periodic discrete L2 norm. Discrete summation by parts makes the drift term's real energy contribution zero and the diffusion term nonpositive. Equivalently its Fourier amplification symbol generator has real part $-4h^{-2}{\sin }^2(\theta /2)$. This continuous-time estimate does not prove stability of an arbitrarily chosen time integrator.

For a function space on a set $X$ and $x\in X$, the point evaluation functional is ${\delta }_x(f)=f(x)$. On $C(K)$ for a nonempty compact space $K$, it is a bounded linear functional for the supremum norm:The constant function one attains the bound. The same argument works for bounded continuous functions on any $X$.

Boundedness depends on the chosen norm. On $C([0,1])$ with the L2 norm, fix $x_0\in (0,1)$ and take $f_n(t)=\max (1-n|t-x_0|,0)$ for sufficiently large $n$. Then $f_n(x_0)=1$ but $\Vert f_n{\Vert }_2^2=2/(3n)$, so point evaluation is unbounded. On L2 space itself, it is not even well-defined on equivalence classes: changing a representative at the single point $x_0$ does not change its Lebesgue almost everywhere class.

On continuous functions on a compact interval, evaluation at an interior point is not continuous in the integral norm. Narrow triangular peaks have fixed point value and integral tending to zero. Subtracting such a peak from the constant function one gives functions vanishing at the point and converging in the integral norm to one. Thus the evaluation kernel is not closed, although it is closed in the uniform norm.

Here is a fully computed endpoint version, now using the L2 norm on the actual continuous functions $C([0,1])$. For each integer $n\ge 1$, set $f_n(x)=\max (1-nx,0)$. Its point evaluation at zero is $E(f_n)=f_n(0)=1$, whileConsequently $f_n\to 0$ in the L2 norm but $E(f_n)\nrightarrow 0$: point evaluation is discontinuous on this normed space. Quantitatively $|E(f_n)|/\Vert f_n{\Vert }_2=\sqrt{3n}$ is unbounded. At $n=100$, the L2 norm is approximately $0.057735$, the integral is $0.005$, and the point value is still $1$. In the supremum norm, $\Vert f_n{\Vert }_{\infty }=1$ and $|E(f)|\le \Vert f{\Vert }_{\infty }$, so $E$ has operator norm one instead.

Set $g_n=1-f_n$. Then $E(g_n)=0$ and $\Vert g_n-1{\Vert }_2=1/\sqrt{3n}\to 0$, although $E(1)=1$. Thus the kernel of $E$ fails to be a closed set. On the full L2 space, a further distinction is essential: its elements are equivalence classes under equality almost everywhere, and changing a representative at zero does not change its class. Arbitrary point evaluation is therefore not even well-defined there. Restricting first to continuous functions makes the discontinuity statement precise.^[ref]

The squared L2 norm of a probability density function derivative measures roughness and appears in optimal smoothing constants. For $r=2$, its directional derivative along a smooth compactly supported perturbation $a$ is $2\int f^{\prime \prime }{a}^{\prime \prime }$. When integration by parts is valid twice, this equals $2\int f^{(4)}a$. Thus regular root-$n$ estimation requires substantially more regularity than merely estimating a smooth probability density function.

A reliable residual estimator certifies an error tolerance only when its proved upper bound is below that tolerance. Reliability alone does not prove that a particular refinement strategy will eventually pass the stopping test. Occasional uniform bisection provides a simple safeguard in a setting where the estimator vanishes as the maximum cell length tends to zero.

For example, consider one-dimensional $-(p{u}^{\prime }{)}^{\prime }+qu=f$, with $p\in W^{1,\infty }$, $p\ge p_{\ast }>0$, bounded $q\ge q_{\ast }>0$ and $f\in L^2$, using continuous piecewise-linear finite elements and homogeneous compatible boundary data. Discrete energy stability bounds $u_h$ in $H^1$, so the cell residual $R_h=f+p^{\prime }{u}_h^{\prime }-q{u}_h$ is bounded in L2 norm uniformly over refinements. The explicit one-dimensional dual residual bound in L2 givesHere $C$ depends on coefficient bounds, not on the mesh. If the refinement schedule forces $h_{\max }\to 0$, then $C{\eta }_h\to 0$ and every positive tolerance is eventually certified. This is a termination argument for the stated residual/regularity setting; it does not claim optimal complexity for an arbitrary adaptive marking rule.

For a real or complex Hilbert space $H$ with inner product linear in its first argument, every bounded linear functional $f:H\to \mathbb{F}$ has a unique representing vector $y\in H$ such thatThe Cauchy-Schwarz inequality gives $\Vert f\Vert \le \Vert y\Vert$, and evaluation at $x=y/\Vert y\Vert$ gives equality when $y\ne 0$. Uniqueness follows because $⟨x,y-z⟩=0$ for every $x$ implies $y=z$. Existence is proved by the Kernel-orthogonal proof of the Riesz representation theorem.^[ref]

For example, on $L^2(0,1)$, $f(g)={\int }_0^{1}tg(t)dt$ is represented by $y(t)=t$, with $\Vert f\Vert =1/\sqrt{3}$. By contrast, point evaluation $g\mapsto g(0)$ is not a bounded functional on this L2 space: elements are equivalence classes modulo equality almost everywhere, so point values are not even well-defined. On continuous functions, triangular peaks can also have value one at zero and arbitrarily small L2 norm.

On a complex Hilbert space, $y\mapsto ⟨\cdot ,y⟩$ is conjugate-linear. Thus the representation is an isometric conjugate-linear identification with the continuous dual space, under the declared convention; it should not silently be called a complex-linear map.

The Haar wavelet provides an explicit one-vanishing moment comparison:It cancels a constant polynomial, but not a nonconstant affine function. To obtain two vanishing moments by an equally transparent construction, take three adjacent unit intervals with heights proportional to $1,-2,1$:The L2 norm normalization follows from $1^2+(-2{)}^2+1^2=6$. Direct integration givesThus $\eta$ has exactly two vanishing moments. In fact its coefficient is a finite difference of adjacent local integrals:The weights are those of the second-order central difference. The first moment balances the centres of the three intervals; it is not enough merely to balance their signed areas.

This $\eta$ is a normalized analyzing function, with no orthonormal basis assertion. Indeed, its unit translate has nonzero L2 inner product:Hence these translates already fail orthogonality. The Daubechies wavelet construction imposes additional refinement and orthogonality requirements; two vanishing moments and unit L2 norm by themselves do not supply those requirements.

The spatial Galerkin method for the heat equation gives a mass matrix $M$ and a diffusion stiffness matrix $S$. With zero forcing, $\frac{d}{dt}(c^{T}Mc)=-2c^{T}Sc\le 0$. The quadratic form $c^{T}Mc$ is the squared L2 norm of the represented finite-element function, proving an energy contraction. For nonzero forcing, the corresponding identity has the additional term $2c^{T}F$, which can be estimated using the Cauchy-Schwarz inequality.

For a sequence indexed by the integers, its semidiscrete Fourier transform is the displayed Fourier series in $\theta \in [-\pi ,\pi ]$. For square-summable data the series is interpreted in the L2 norm. Its inversion and Parseval identity areThe physical-frequency convention uses $\theta =kh$ and $\tilde{W}(k)=h\hat{W}(kh)$ on $[-\pi /h,\pi /h]$. This is distinct from the discrete Fourier transform of a finite vector. Translation-invariant finite difference operators become Fourier multipliers, making Fourier stability analysis possible.