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Estimate the signal by one-dimensional total variation denoising, which penalizes adjacent differences rather than the constant level:This is a fused Lasso objective. Its L1 norm penalty makes many adjacent differences exactly zero, producing a piecewise-constant fit. For change-point detection, estimate the locations by $\hat{S}=\{i:{\hat{\mu }}_i\ne {\hat{\mu }}_{i+1}\}$. The tuning parameter controls the strength of the penalty; it can be chosen, for example, using a validation criterion appropriate to the noise model.

For a general family of elementary nulls $H_1,\dots ,H_m$, the closed testing procedure forms every nonempty intersection $H_B={\bigcap }_{j\in B}{H}_j$ and gives it a local test of level $\alpha$. Reject $H_j$ precisely when all intersections with $j\in B$ are locally rejected. This gives strong control of the familywise error rate: let $T_0$ be the indices of the true elementary nulls. If $T_0$ is empty there is no false rejection. Otherwise, rejecting any $H_j$ with $j\in T_0$ requires rejection of the true intersection $H_{T_0}$. Its local test has rejection probability at most $\alpha$, henceThis proof of closed-testing control of the familywise error rate needs no independence of the tests.

For the interval procedure, a true tested interval is contained in one maximal constant block $B\in \mathcal{T}$. If $I\subseteq B$, the intervals over which $p_B$ takes its maximum are a subset of those used for $p_I$, so $p_I\ge p_B$. Therefore rejection of a true $I$ forces rejection of its containing block. Conversely, a rejected block is itself a rejected true null.

There is a small indexing issue in the PDF: its testing family excludes singleton intervals, while $\mathcal{T}$ may contain singleton blocks. Put ${\mathcal{T}}_{\ast }=\{B\in \mathcal{T}:|B|\ge 2\}$. Every true tested interval lies in one of these blocks, so the exact statement, using only hypotheses defined in the paper, isSingleton blocks contain no eligible tested interval and can be ignored. Equivalently, one may define their nulls as never rejected.

For each $B\in {\mathcal{T}}_{\ast }$, the term $J=B$ occurs in the maximum defining $p_B$, giving $p_B\ge n{q}_B/|B|$. A valid p-value is a super-uniform random variable under its null, so for $0\le \alpha \le 1$,The union bound and disjointness of the constant blocks now giveThis establishes weighted interval testing for a piecewise-constant mean even when the interval p-values are dependent. Values $p_I>1$ do not affect rejection at levels at most one.

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.

Let $P_{+}$ be the maximizing projection in the variational characterization of trace distance. Use the binary POVM $\{P_{+},I-P_{+}\}$ and its measurement channel. The difference between its two output probability vectors iswhere $D=D(\rho ,\sigma )$ and the second equality uses $\mathrm{Tr}(\rho -\sigma )=0$. Since trace distance between diagonal density operators is half the L1 norm of their probability-vector difference,Thus a binary quantum measurement can preserve the distinguishability of this particular pair exactly. This is the trace-distance-preserving binary measurement, which may depend on the two states.

A simple choice is an independent Laplace distribution at each pixel, giving a Laplace prior for sparse images:Its negative log density penalizes the L1 norm. Relative to a normal distribution, it combines a sharp peak at zero with heavier tails, allowing most pixels to lie near the background level while a few have appreciable intensity. Quantitatively, $\mathbb{P}(|u_j|>\tau )=e^{-\lambda \tau }$, so the expected number of appreciable pixels is $d{e}^{-\lambda \tau }$. Choosing $\lambda \tau$ large makes this small. This is a simple sparsity model: it does not enforce that nearby active pixels form connected objects, nor does its continuous density give exact zero pixels positive probability. Additional spatial modelling would be needed to insist on contiguous shapes.

To sample, draw independent $U_j\sim \mathcal{U}(0,1)$ and use inverse transform sampling:The endpoint events have probability zero and can be assigned arbitrary finite values. To justify the rule, the cumulative distribution function of a centered Laplace distribution isThe displayed rule is $F^{-1}(U_j)$, so $\mathbb{P}(u_j\le x)=\mathbb{P}(U_j\le F(x))=F(x)$. Independence of random variables then gives the stated product prior distribution.

HereThe L1 norm and uniform norm satisfyso the identity from the $d_{\infty }$ metric to the $d_1$ metric is one-Lipschitz and therefore continuous. The two metrics do not induce the same topology on all of $C[0,1]$. Forone has $d_1(f_n,0)=1/(2n)\to 0$ but $d_{\infty }(f_n,0)=1$. Thus $d_1$ convergence need not imply uniform convergence.

A map $F:(X,d_X)\to (Y,d_Y)$ is Lipschitz continuous if there is $L<\infty$ such thatfor all $x,x^{\prime }\in X$. Evaluation at $a\in [0,1]$ is one-Lipschitz because

Choose distinct $a_0,\dots ,a_n\in [0,1]$ and defineThe Vandermonde determinant is nonzero, so $T$ is a bijection. It is one-Lipschitz for the two uniform metrics. If ${\ell }_i$ are the associated Lagrange cardinal polynomials, thenand henceThus $T^{-1}$ is also Lipschitz.

Let ${\hat{\mathcal{P}}}_n$ be the polynomials whose values lie in $[-1,1]$. Its image under $T$ lies in the bounded cube $[-1,1{]}^{n+1}$. It is closed: if $T(p_j)\to v$, then the Lipschitz inverse gives uniform convergence $p_j\to p=T^{-1}(v)$, and passing to the limit pointwise preserves $|p(x)|\le 1$. By the Heine-Borel theorem, $T({\hat{\mathcal{P}}}_n)$ is compact, and the continuous inverse carries compactness back to ${\hat{\mathcal{P}}}_n$. Therefore

The softplus functionis convex becausePart (b) therefore shows that every $\beta \mapsto h(a_i^T\beta )$ is convex. The absolute value function is convex, so $\beta \mapsto |{\beta }_j|$ is convex for every coordinate $j$. Finally, part (a) says that a finite sum of convex functions is convex. Thereforeis convex. The second sum is the L1 norm regularizer.

The scalar logistic loss $u\mapsto -u+\log (1+e^u)$ is strictly convex because its second derivative is $e^u/(1+e^u{)}^2>0$. If $\hat{\beta }$ and $\tilde{\beta }$ are minimizers but $X\hat{\beta }\ne X\tilde{\beta }$, strict convexity of the loss as a function of the fitted vector and convexity of the L1 norm make the objective at their midpoint strictly smaller than the common minimum. This contradiction proves thatfor every pair of solutions.

The subdifferential of the L1 norm consists of vectors $z$ with $z_j=\mathrm{sgn}({\beta }_j)$ when ${\beta }_j\ne 0$ and $z_j\in [-1,1]$ when ${\beta }_j=0$. These coordinate conditions give the Karush-Kuhn-Tucker conditions for Lasso and Square-root Lasso.

Taxicab geometry equips Euclidean coordinate space with the distance induced by the L1 norm, so the distance between two points is the sum of their coordinatewise absolute differences.

For $f\in L^1(\mathbb{T},m)$ with $m$ normalized Lebesgue measure, translation on the circle group is continuous in the L1 norm: $\Vert f(\cdot +t)-f{\Vert }_1\to 0$ as $t\to 0$ modulo one. Approximate $f$ in the L1 norm by a continuous function, use its uniform continuity, and use translation invariance to preserve the approximation error. In particular, $m(B△(B-t))\to 0$ for every measurable $B$.