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A discrete wavelet transform computes multiscale approximation and detail coefficients from sampled data. For the Haar wavelet, replace each adjacent pair $(u,v)$ byThe approximation coefficient $a$ is $\sqrt{2}$ times the pair's arithmetic mean; it is not the ordinary mean. The detail coefficient $d$ records their signed contrast. Repeat on the approximation coefficients until only one remains, keeping every detail coefficient.

For a worked example, start with the vector $(2,4,6,8)$. The first pass gives approximation coefficients $(3\sqrt{2},7\sqrt{2})$ and detail coefficients $(-\sqrt{2},-\sqrt{2})$. Transforming the remaining pair gives the coarse coefficient $10$ and the half-to-half detail $-4$. In the order coarse value, coarse detail, then fine details, the transform isThis is an orthogonal matrix transformation. Each pair preserves $u^2+v^2=a^2+d^2$, so the full transform preserves the squared Euclidean norm:The inverse function for each pair is $u=(a+d)/\sqrt{2}$, $v=(a-d)/\sqrt{2}$. Reversing the passes therefore recovers every original value. This squared-norm bookkeeping is often called signal energy; it need not represent physical energy. ^[ref]

Continuous and discrete normalizations differ. If $n=2^J$ data values are interpreted as the heights of a step function on $n$ equal cells of $[0,1]$, its normalized L2 inner product coefficients are the discrete coefficients divided by $\sqrt{n}$. For our four values they are $(5,-2,-1/\sqrt{2},-1/\sqrt{2})$, as the Haar projection example shows. The same normalization is used by the wavelet regression estimator on $[0,1]$.

Under independent Gaussian noise with a common variance, orthogonality preserves that coefficient noise variance and independence. It does not make every arbitrary correlated signal independent: this statistical conclusion uses the Gaussian model and its equal-variance independent inputs.

For an integrable function on $[0,1]$, average $f$ on each dyadic interval of length $2^{-n}$. The resulting step function $f_n$ converges to $f$ almost everywhere and in the L1 norm. If $U$ has the uniform distribution on $[0,1]$, then $f_n(U)=\mathbb{E}[f(U)\mid \sigma (b_n(U))]$, where $b_n$ rounds down to a dyadic endpoint. Refinement makes these conditional expectations a uniformly integrable martingale. The uniformly integrable martingale convergence theorem gives almost-sure and convergence in L1 to a limit $Y$.

Since $b_n(U)\to U$, the generated limiting sigma-algebra contains $U$ up to null sets. For every event $A$ in any finite-stage sigma-algebra, convergence in the L1 norm gives $\mathbb{E}[Y{1}_A]=\mathbb{E}[f(U)1_A]$. The pi-lambda theorem extends this equality to the limiting sigma-algebra, identifying $Y=f(U)$. The uniform distribution of $U$ translates the two convergence conclusions into the assertions for $f_n$. Choosing a Borel representative of $f$ and assigning arbitrary values at the endpoint one makes all random variables well defined without changing either conclusion.

The empirical distribution function counts the fraction of observations at or below a threshold. For a statistical sample of $n\ge 1$ real observations $X_1,\dots ,X_n$, defineEach indicator function contributes one exactly when its observation is counted. Equivalently, $F_n(x)=P_n((-\infty ,x])$, where the empirical measure $P_n=n^{-1}{\sum }_i{\delta }_{X_i}$ assigns probability mass $1/n$ to each observation, including repeated observations.^[ref]

Choose the illustrative statistical sample $[1,1,2,4,6]$. Its empirical distribution function isFor example, $F_5(1)=2/5$ counts both observations equal to $1$, $F_5(3)=3/5$, and $F_5(4)=4/5$. There is no observation at $3$, so there is no jump there. The five observations produce only four distinct jump locations.

The convention $X_i\le x$ makes the empirical distribution function right-continuous: once a threshold reaches an observation, that observation remains counted immediately to its right. For a finite statistical sample, the count is constant from any $x$ up to the next larger observed value, proving right continuity. Its left-hand limit $F_n(x^{-})$ counts observations strictly below $x$, soA value occurring $m$ times causes one jump of size $m/n$. At $1$ the jump is $2/5$; at $2$, $4$, and $6$ each jump is $1/5$. All jumps sum to one. Thus the empirical distribution function is a nondecreasing step function with limits zero and one, and is itself a valid cumulative distribution function.

This turns many questions about data into simple cumulative distribution function operations. The observed fraction in $(a,b]$ is $F_n(b)-F_n(a)$: for $(1,4]$ it is $4/5-2/5=2/5$, counting $2$ and $4$ but excluding both $1$s. Including the left endpoint gives $F_n(4)-F_n(1^{-})=4/5$. A sample quantile reads the same staircase in reverse: the smallest threshold reaching $1/2$ is $2$. Unlike a histogram, this representation requires no choice of bins; its jumps retain every observation and its multiplicity.

The exact description of a finite statistical sample must be distinguished from a statistical model for how it was obtained. The empirical distribution function is defined for any finite list, with no independence assumption. Once the list is observed, its counts are fixed. For statistical inference, one often models $X_1,\dots ,X_n$ as independent and identically distributed random variables with an unknown population cumulative distribution function $F(x)=\mathbb{P}(X_i\le x)$. Then $F$ describes the sampling law, while $F_n$ varies across repeated statistical samples. The value $F_5(1)=0.4$ is an observed fraction and an estimate of $F(1)$; it does not establish that the population probability is $0.4$. Even when $F$ is continuous, every finite $F_n$ is discrete.

For any threshold $x$ fixed before sampling, the indicator functions are independent Bernoulli random variables with success probability $F(x)$. Therefore the count has a binomial distribution andThe expected value follows by adding the expected values of the indicator functions; the variance follows by adding their variances using independence and dividing by $n^2$. Thus $F_n(x)$ is an unbiased estimator of $F(x)$ at a fixed threshold. Controlling the error at every threshold at once requires more than a fixed-threshold calculation: the DKW inequality supplies such control, and the Glivenko-Cantelli theorem describes its uniform convergence almost surely as $n$ grows.

Comparing $F_n$ with a proposed cumulative distribution function also motivates goodness-of-fit tests, including the Kolmogorov-Smirnov test, the Anderson–Darling test, and the Cramér–von Mises criterion. A standard Kolmogorov-Smirnov test calibration assumes a fully specified continuous population law; fitting its statistical parameters from the same statistical sample changes that calibration.^[ref]

The Haar projection at level $J$ is the orthogonal projection onto the functions constant on level-$J$ dyadic intervals. On each cell it replaces $f$ by its cell average. This is the closest cellwise-constant approximation in the L2 norm, because choosing a constant $c$ on a cell minimizes ${\int }_I|f-c{|}^2$ at its cell average.

On $[0,1]$, take the step function with values $2,4,6,8$ on successive quarter-intervals. At level zero, its Haar projection is the constant $5$. At level one, it has values $3$ and $7$ on the two halves. At level two, it recovers the four original values exactly. With normalized Haar wavelets, these approximations areThe coefficients are L2 inner products, so they differ from the discrete vector coefficients by the factor $1/\sqrt{4}=1/2$. Their squared sum also has a direct meaning:The approximation errors in squared L2 norm of $P_{0}f,P_{1}f,P_{2}f$ are respectively $5,1,0$: each added detail removes exactly its own squared coefficient from the error.

The Haar projection can be expressed using level-$J$ Haar scaling functions, or through a coarse scaling family and the Haar wavelets at intermediate levels. On $[0,1]$ with lowest level zero, that coarse family is the single constant function $\varphi$. On the whole real line, use the integer translates of a coarse Haar scaling function instead; one global constant is not square-integrable. Refinement improves the representation while retaining the earlier coefficients.

The Haar wavelet is a signed contrast between two adjacent half-intervals:It has compact support, zero integral, and unit L2 norm: $\int \psi =1/2-1/2=0$ and $\int |\psi {|}^2=1$. The zero integral is its one vanishing moment. Its Haar scaling function is $\varphi =1_{[0,1)}$, which records a coarse value rather than a contrast. Endpoint conventions do not alter the L2 inner products.

The normalized copies ${\psi }_{j,k}(x)=2^{j/2}\psi (2^{j}x-k)$ compare the halves of the dyadic interval $[k{2}^{-j},(k+1)2^{-j})$. Narrower copies are taller so that their L2 norms stay one. Thus a coefficient identifies both a location, through $k$, and a scale, through $j$.

For a simple step function $f$ equal to $2$ on $[0,1/2)$ and $6$ on $[1/2,1)$, its coarse coefficient is $⟨f,\varphi ⟩=4$, while its first contrast is $⟨f,\psi ⟩=-2$. Thus $f=4\varphi -2\psi$. The negative coefficient raises the second half and lowers the first. Finer contrasts vanish because $f$ is constant on each half. A jump aligned with this dyadic interval boundary therefore needs only two nonzero coefficients; jumps at other positions can require details at many scales.

Its jump discontinuities make the Haar wavelet particularly transparent for changes between neighbouring regions, while limiting approximation of smooth variation. A Daubechies wavelet uses a more elaborate pattern to obtain additional vanishing moments and, at sufficiently high order, smoother basis functions. The elementary Haar refinement identity has coefficients $h_0=h_1=1/\sqrt{2}$ and explains how the coarse and fine descriptions fit together.

The historical source is Alfred Haar's ^[ref]. Chapter III, beginning on page 361, constructs the constant function and these localized piecewise-constant orthogonal functions. Its original endpoint assignments differ from the half-open convention used here, without changing L2 inner products.

A real-valued function $f:D\to \mathbb{R}$, with domain of a function $D\subseteq \mathbb{R}$ in the usual order of the real numbers, is a monotonic function if it is either a nondecreasing function throughout $D$ or a nonincreasing function throughout $D$. The same direction must work for every pair $x<y$ in $D$. Being a strictly increasing function on one part and a strictly decreasing function on another does not make the whole function monotone: for example, $x^2$ on the real line changes direction at zero. A constant function is both a nondecreasing function and a nonincreasing function.

The word "increasing" is ambiguous across textbooks: some allow equal outputs and some require unequal outputs. Use nondecreasing function for the weak inequality and strictly increasing function for the strict inequality; use nonincreasing function and strictly decreasing function in the other direction. Neither continuity nor differentiability is part of the definition of a monotonic function.

Strict monotonicity gives injectivity, so an equation $f(x)=b$ has at most one solution and an inverse function exists on the image of a function. Weak monotonicity permits repeated outputs and jumps. In probability theory, a cumulative distribution function is a nondecreasing function; its quantile function uses a generalized inverse to handle those flat parts and jumps. Monotonicity alone does not guarantee continuity, even for a strictly increasing function.

A real-valued function $f:D\to \mathbb{R}$ is a nondecreasing function if every pair $x<y$ in its domain of a function satisfies $f(x)\le f(y)$. Flat intervals are allowed, unlike for a strictly increasing function. For example,is a continuous nondecreasing function, but it repeats its output on each flat interval.

A nondecreasing function can also be discontinuous. The step function $F(x)=0$ for $x<0$ and $F(x)=1$ for $x\ge 0$ has a jump discontinuity at zero without ever decreasing. It is the cumulative distribution function of a random variable that equals zero with probability one; quantile functions handle such jumps through a generalized inverse. Nondecreasing allocation functions permit critical-value payments in single-parameter mechanisms.

Put $G(t)={\int }_a^{t}f(x)dx$. At an interior point and for $h\ne 0$ small enough,By continuity at $t$, the modulus is bounded by the supremum of $|f(x)-f(t)|$ between $t$ and $t+h$, which tends to zero. Thus the fundamental theorem of calculus givesAt the interval endpoints the same argument supplies the corresponding one-sided derivatives.

For an integrable discontinuous function, the single-point spike from part (i) has $G\equiv 0$, so $G^{\prime }(c)=0\ne f(c)=1$; this disproves the derivative identity even when the primitive is differentiable. To see that differentiability itself may fail, take the step function $f(x)=0$ for $x<c$ and $f(x)=1$ for $x\ge c$. It is Riemann integrable, but its primitive is $G(t)=\max (0,t-c)$ and has left derivative zero and right derivative one at $c$. Hence the full assertion is false for general Riemann-integrable functions.

Take the step functionIt is bounded between zero and one, and is discontinuous at $x=1/2$.

Choose a step function $s$ with $\Vert f-s{\Vert }_1<\epsilon$, by part (i). The Fourier transform satisfiesBy part (ii), ${\mathrm{limsup}}_{|t|\to \infty }|\hat{f}(t)|\le \epsilon$. Since $\epsilon$ is arbitrary,This proves the Riemann-Lebesgue lemma with the transform convention in the question.

Work in the real $L^1$ space, consistent with the specified real step coefficients. First approximate an integrable function by a finite-valued measurable function with bounded support. Truncate its domain to $[-M,M]$ and its values to $[-M,M]$; the $L^1$ error tends to zero by integrability. Quantize those remaining values in intervals of mesh $\delta$, producing a simple function $u={\sum }_{j=1}^m{c}_j{1}_{B_j}$ with finite-measure Borel sets $B_j$ and error at most $2M\delta$ on the truncated domain. Thus $\Vert f-u{\Vert }_1$ can be made arbitrarily small.

For each $B_j$, use the allowed approximation by a finite union of intervals $A_j$, choosing the symmetric-difference errors soThese unions can be written as finite disjoint bounded intervals, and changing endpoints to the required half-open convention changes only a null set. The function $s={\sum }_j{c}_j{1}_{A_j}$ is therefore a step function of the specified form, andCombining the two approximations proves the step functions are dense in $L^1(\mathbb{R})$. In the complex version of $L^1$, the same argument applies to real and imaginary parts when complex step coefficients are allowed.

First apply Lebesgue decomposition to the derivative measure relative to ${\mathcal{L}}^n$:Here $\nabla u$ is the almost-everywhere approximate gradient, rather than an assertion that $u$ belongs to $W^{1,1}$. Split the singular part into its jump part of a BV derivative and Cantor part of a BV derivative:An approximate jump point has a unit normal ${\nu }_u$ and distinct finite BV traces on a hypersurface $u^{+},u^{-}$, obtained as mean limits on the corresponding two half-balls. Their set $J_u$ is the jump set of a BV function, countably $(n-1)$-rectifiable. Reversing the normal swaps the BV traces on a hypersurface and leaves the displayed measure unchanged. The approximate discontinuity set $S_u$ differs from $J_u$ only by an ${\mathcal{H}}^{n-1}$-null set.

The remaining $D^{c}u=D^{s}u-D^{j}u$ is singular to Lebesgue measure and gives zero mass to every set with sigma-finite ${\mathcal{H}}^{n-1}$ measure. It is diffuse rather than a second jump contribution. In one dimension the three parts are illustrated by an affine function, a step function and the Cantor function, respectively. Countably many jumps therefore do not imply that the singular derivative has no Cantor part of a BV derivative.

The stated bound is Lipschitz continuity with constant $1$. For every $x$ and every $\epsilon >0$, taking $\delta =\epsilon$ ensures $|f(x)-f(y)|<\epsilon$ whenever $|x-y|<\delta$. The same $\delta$ works at every point, so $f$ is both continuous and uniformly continuous.

For an integer $N\ge 1$, partition $[0,1]$ into $N$ equal intervals and define a step functionAt the remaining endpoint set $g_N(1)=f((N-1)/N)$, so the final interval may be taken closed on the right. The Lipschitz bound givesChoosing $N\ge 1/\epsilon$ proves the requested uniform step approximation on a compact interval:A sufficiently fine partition gives a piecewise constant approximation uniformly over the whole interval.

The continuous function $f$ and every finite step function $g_N$ are Riemann integrable. Put $c_j=f(j/N)$. Integrating each constant piece explicitly givesSince the sine differences have absolute value at most $2$, the triangle inequality and uniform approximation giveTo prove convergence, fix $\epsilon >0$ and first choose $N$ with $1/N<\epsilon /2$. With this $N$ held fixed, the finite number ${\sum }_j|c_j|$ is independent of $n$, so the second term is also less than $\epsilon /2$ for sufficiently large $n$. ThereforeFirst make the approximation error small, then let the oscillation frequency grow. This is the step-function proof of the Riemann-Lebesgue lemma; it does not require assuming that a Lipschitz function has an everywhere-defined derivative.

For deterministic step functions, define directly from Brownian motion incrementswith $s_0$ the left endpoint of the first interval and $f=0$ outside the listed intervals. The printed endpoint list omits $s_0$; the intended convention is $0\le s_0<s_1<\cdots <s_n$, usually $s_0=0$.

Given finitely many observation times, refine the deterministic partition of an interval to include all observation times and all $s_i$. Every $Z_{t_j}$ is a linear combination of the same finite family of independent centered normal Brownian motion increments. Thus every linear combination of $(Z_{t_1},\dots ,Z_{t_m})$ is normal, including possibly a degenerate constant: $Z$ is a Gaussian process. By independence, only increments in the common interval $[0,s\wedge t]$ contribute to the covariance, givingThe same calculation for two step functions $f,g$ proves, rather than assumes, the Itô isometryFor $f\in L^2({\mathbb{R}}_{+})$, choose deterministic step functions $f_n\to f$ in the L2 space. This identity makes $Z_t(f_n)$ a Cauchy sequence in the L2 space; the Riesz-Fischer theorem defines $Z_t(f)$ by mean-square convergence, independently of the approximation. The covariance formula passes to the limit by the Cauchy-Schwarz inequality. For finitely many times and coefficients $a_j$, the characteristic functions converge toso the extension remains a Gaussian process.

For completeness, the construction has a continuous version. Each approximating $Z(f_n)$ is a martingale, directly by the centered independent increments of Brownian motion; the Doob L2 maximal inequality givesTo prove the inequality used here without assuming Itô integral properties, a continuous martingale $N$ satisfies $\lambda \mathbb{P}(N_T^{\ast }\ge \lambda )\le \mathbb{E}(|N_T|1_{\{N_T^{\ast }\ge \lambda \}})$, by conditioning at the first crossing on finite deterministic grids and then using continuity; integrating in $\lambda$ and using the Cauchy-Schwarz inequality yields $\Vert N_T^{\ast }{\Vert }_2\le 2\Vert N_T{\Vert }_2$. A sufficiently fast subsequence therefore converges uniformly almost surely on every compact time interval, giving the continuous version of the stochastic integral.

The space is not compact because it is not complete. For $n\ge 2$, define the continuous functionThese functions converge in the L1 norm to the step function $h=1_{(1/2,1]}$, because $f_n-h$ is supported on an interval of length $2/n$ and bounded in absolute value by one. They are consequently Cauchy for the metric $d$, since for sufficiently close pairs the outer minimum with $1$ does nothing.

If $(f_n)$ converged in this metric to some $f\in C[0,1]$, it would also converge to $f$ in $L^1$. Uniqueness of an $L^1$ limit would give $f=h$ almost everywhere. Continuity would then force $f=0$ on $[0,1/2)$ and $f=1$ on $(1/2,1]$, which is impossible at $1/2$. Thus this Cauchy sequence has no limit in $C[0,1]$, while every compact metric space is complete.

The chemical potential is the change in internal energy when one particle is added at fixed entropy and volume:Equivalently, $\mu =(\partial F/\partial N{)}_{T,V}$ for the Helmholtz free energy.

Let a microstate $s$ have energy $E_s$ and particle number $N_s$. Maximizing the Gibbs entropy subject to normalization and fixed mean values of $E$ and $N$ is the maximum-entropy derivation of equilibrium ensembles. With Lagrange multipliers $\alpha$, $\beta$, and $-\beta \mu$, varyThe stationarity equation is $-\log p_s-1-\alpha -\beta (E_s-\mu N_s)=0$. Normalization therefore gives the grand canonical ensembleHere $\mathcal{Z}$ is the grand canonical partition function and $k_B$ is the Boltzmann constant.

For one fermionic quantum state of energy $\epsilon$, the Pauli exclusion principle permits occupation numbers only $n=0,1$. Its two grand-canonical weights are $1$ and $e^{-\beta (\epsilon -\mu )}$, so its mean occupation is the Fermi-Dirac distribution

For a free nonrelativistic particle, $\epsilon ={\hslash }^2{k}^2/(2m)$. In a region of area $A$, the number of wave-vector states in the annulus from $k$ to $k+dk$, including the two spin angular momentum states, isSince $kdk=(m/{\hslash }^2)d\epsilon$, the two-dimensional free-electron density of states is constant:

We now use units in which $k_B=1$, as in the question. At zero temperature, the Fermi-Dirac distribution is a step function, and henceAt positive temperature the same fixed particle number satisfies the exact relationThusThis is the low-temperature particle-number cancellation for constant density of states. Therefore, for $T\ll {\epsilon }_F$, the mean number of particles in the energy interval $[\epsilon ,\epsilon +d\epsilon ]$ is

Finally, compare the internal energy with its zero-temperature value. The thermally excited particles above ${\epsilon }_F$ and holes below ${\epsilon }_F$ have equal leading particle numbers, so their terms proportional to ${\epsilon }_F$ cancel. With $x=|\epsilon -{\epsilon }_F|/T$,where the integral follows from the fermion-to-boson thermal integral ratio. Differentiation gives the linear low-temperature heat capacity of a two-dimensional Fermi gas:The requested power law is therefore linear in $T$, with exponent one.