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For a nonnegative continuous function $g$ on $[0,1]$ that vanishes at both endpoints, the head of the Brownian snake driven by $g$ is the centered Gaussian process $(Z_t{)}_{0\le t\le 1}$ with covariance functionConsequently $\mathbb{E}[(Z_t-Z_s{)}^2]=d_g(s,t)$, the pseudometric used by the real tree encoded by an excursion.

The canonical distance is the L2 norm of an increment of a Gaussian process. It is a pseudometric, since distinct parameters may represent equal random variables almost surely. Its metric covering number appears in the Dudley entropy integral. For standard fractional Brownian motion, $d(s,t)=|s-t{|}^H$; a normalization with twice the covariance function multiplies this distance by $\sqrt{2}$.

For two Cauchy sequences, $|d(x_n,y_n)-d(x_m,y_m)|\le d(x_n,x_m)+d(y_n,y_m)$, so the displayed limit exists. It defines a pseudometric on the Cauchy sequences, and its metric quotient identifies sequences whose mutual distance tends to zero. Constant sequences embed the original metric space isometrically. Their image is dense, because the class of $(x_n)$ is approached by the constant classes of $x_n$. The quotient is complete: from a Cauchy sequence of its points select a subsequence whose successive distances are below $2^{-j}$ and choose original-space points within $2^{-j}$ of these subsequence points. Those original-space points form a Cauchy sequence and its class is the limit of the subsequence, hence of the whole sequence. The original embedding is onto exactly when the original space is complete.

A Fréchet space is a Hausdorff locally convex space whose topology is induced by a complete translation-invariant metric. Translation invariance means $d(x+z,y+z)=d(x,y)$; completeness means every Cauchy sequence converges in the space. Use a translation-invariant metric for this condition: an arbitrary compatible metric can change which sequences are Cauchy sequences. ^[ref].

Equivalently, the topology comes from a countable separating family of seminorms $(p_k{)}_{k\ge 0}$ and is complete in their common sense: if $p_k(x_n-x_m)\to 0$ as $n,m\to \infty$ for every fixed $k$, then some $x\in E$ satisfies $p_k(x_n-x)\to 0$ for every $k$. A useful compatible translation-invariant metric isIt is finite because the weights sum to one. The separating condition makes it a metric; without that condition the formula gives only a pseudometric. Its convergence in a metric space and Cauchy sequence conditions are exactly the conditions for all the $p_k$ separately: control finitely many terms, then bound the remaining tail by $2^{-(K+1)}$. Replacing $p_k$ by ${\max }_{0\le j\le k}{p}_j$ gives an increasing family with the same topology.

Every Banach space is a Fréchet space, using its norm as the single measurement. The space of smooth functions on a compact interval is a Fréchet space with derivative seminorms but with no single norm inducing its smooth topology. This is why Fréchet spaces are useful in differential equations: one can require accuracy in every order of derivative, while each particular estimate uses only finitely many orders.

The name honors Maurice Fréchet. His doctoral thesis was published as ^[ref]; an ^[ref] is available. This identifies a historical source without attributing the modern locally convex definition to that paper.

The kernel of a seminorm is its zero set, a linear subspace by absolute homogeneity and the triangle inequality. The estimate $|p(x)-p(y)|\le p(x-y)$ shows that adding an element of $\ker p$ does not change $p$. Hence $\Vert x+\ker p\Vert =p(x)$ defines a norm on the quotient vector space $E/\ker p$.

On $E$ itself, $d_p(x,y)=p(x-y)$ is a pseudometric; it is a metric exactly when $p$ is a norm. In particular, zero distance in one pseudometric need not mean equality of vectors. A separating family of seminorms removes the common zero-distance ambiguity.

A metric on a set $X$ is a function $d:X\times X\to [0,\infty )$ satisfying, for every $x,y,z\in X$:

A metric equips its underlying set with a metric space structure; the distance function and the space carrying it are distinct concepts. The second requirement distinguishes a metric from a pseudometric, which may identify distinct elements at zero distance. The symmetry requirement also explains why directed travel times, such as uphill versus downhill times, need not define a metric.

A tempting alternative on the real numbers is $d(x,y)=(x-y{)}^2$. It is nonnegative, symmetric, and zero exactly on equal endpoints, but fails the triangle inequality: $d(0,2)=4>1+1=d(0,1)+d(1,2)$. Checking only the first three requirements would miss the failure.

For a metric space or pseudometric space $(T,d)$, this is the minimum number of balls of radius $\epsilon >0$, with centers in $T$, needed to cover $T$. The displayed grid bound uses closed balls. For open balls, replacing the radius by a fixed constant factor gives the same entropy estimates. For $T=[0,1]$ and $d(s,t)=c|s-t{|}^H$, a uniform grid gives $N(\epsilon ,T,d)\le 1+\lceil (c/\epsilon {)}^{1/H}\rceil$. These polynomial growth bounds make the Dudley entropy integral finite.

The displayed distance is the normalized Hamming distance. Each coordinate mismatch between $x,z$ implies a mismatch between $x,y$ or $y,z$, so summing the indicator inequalities proves the triangle inequality. Nonnegativity, symmetry and separation follow immediately. It is a metric on profiles; on separately labelled individuals with identical profiles it is a pseudometric.

Zero distance in a pseudometric defines an equivalence relation: the triangle inequality proves transitivity. If $x$ and $x^{\prime }$ have zero distance, as do $y$ and $y^{\prime }$, the triangle inequality in both directions gives $d(x,y)=d(x^{\prime },y^{\prime })$. Thus distance descends to the quotient and becomes a metric, since zero quotient distance means the classes coincide.

A metric space is a set $X$ with a nonnegative symmetric distance $d$, zero exactly for equal points, satisfying $d(x,z)\le d(x,y)+d(y,z)$. A Cauchy sequence is a sequence for which, given any $ϵ>0$, there is $N$ such that $d(x_n,x_m)<ϵ$ whenever $n,m\ge N$.

For a pseudometric, zero distance defines an equivalence relation. Reflexivity and symmetry are immediate; transitivity follows from $d(x,z)\le d(x,y)+d(y,z)=0$ when both terms vanish. Define $d_Q([x],[y])=d(x,y)$. If $x\sim x^{\prime }$ and $y\sim y^{\prime }$, the triangle inequality gives $d(x,y)\le d(x^{\prime },y^{\prime })$ and its reversed version gives equality. Thus the definition is independent of representatives. The nonnegative, symmetric and triangle properties descend, while zero distance means precisely that the two classes are equal. It is therefore a metric quotient of a pseudometric.

Now take two Cauchy sequences in the original metric space. Two applications of the triangle inequality giveThe right-hand side is arbitrarily small when both indices are large, so $(d(x_n,y_n))$ is a real Cauchy sequence and has a finite real limit. Consequentlyis well defined. Nonnegativity, symmetry and zero self-distance follow by passing to the limit in the corresponding properties of $d$. Passing to the limit in $d(x_n,z_n)\le d(x_n,y_n)+d(y_n,z_n)$ gives its triangle inequality. Thus $\overline{d}$ is a pseudometric on $C$, and the preceding quotient construction makes $C/R$ a metric space.

Map $x\in X$ to the class of its constant sequence. The distance between two such classes is $d(x,y)$, so this is an injective isometric embedding. It is surjective exactly when every Cauchy sequence $(x_n)$ is equivalent to a constant sequence $(x)$, that is, when $d(x_n,x)\to 0$ for some $x\in X$. This is precisely completeness. HenceThe quotient is the Cauchy-sequence construction of a metric completion; the conclusion requires no assumption that $X$ is compact.

A metric dissimilarity $d$ must satisfy nonnegativity, separation, symmetry and the triangle inequality:The underlying objects must be specified: here they are binary measurement profiles, rather than labels attached to individuals.

Let $n_{11},n_{00},n_{10},n_{01}$ count coordinate pairs of the indicated types. The simple matching coefficient is $S=(n_{11}+n_{00})/4$, so its complementary dissimilarity is $(n_{10}+n_{01})/4$. Counting mismatches in every pair gives the dissimilarity matrixFor the Jaccard coefficient, joint absences are excluded: $J=n_{11}/(n_{11}+n_{10}+n_{01})$. Thus the Jaccard distance is $1-J$, givingOnly the pair of individuals two and four has a joint absence. Their profiles have one shared presence, two mismatched presences and one joint absence, giving dissimilarities $2/4$ and $2/3$, respectively. All other pairs have a presence in at least one profile at every coordinate, so the two denominators coincide.

For the requested metric property of simple matching dissimilarity, write a general pair of $p$-coordinate binary profiles as $x,y$. Thenwhich is the normalized Hamming distance. It is nonnegative and symmetric, and equals zero exactly when all coordinates match. For each coordinate, a mismatch between $x_j$ and $z_j$ forces a mismatch between $x_j,y_j$ or between $y_j,z_j$. ThereforeSumming and dividing by $p$ proves the triangle inequality, establishing all four metric properties. Simple matching dissimilarity is a metric on binary profiles. If two separately labelled individuals have identical profiles, it gives zero distance between them, so on those labels it is only a pseudometric. The four profiles here are distinct, so this distinction does not prevent $D_S$ from defining a metric on the given sample.

For the unit-modulus prime values in this question, the Granville-Soundararajan distance isThe second identity uses $|f(p)|=|g(p)|=1$. Thus it is the Euclidean distance between the finite weighted vectors $(f(p)/\sqrt{2p}{)}_{p\le X}$ and $(g(p)/\sqrt{2p}{)}_{p\le X}$. Nonnegativity, symmetry and the triangle inequality follow from that norm representation, and it vanishes exactly when the two prime-value vectors agree. This proves the prime-restriction metric for pretentious distance.

On whole arithmetic functions at fixed $X$ it is a pseudometric. For example $f=1$ and $g={\mu }^2$ have identical values at every prime and zero distance for every $X$, but different values at four. Thus the literal identity-of-indiscernibles assertion needs the prime-restriction quotient.

Since $\mu (p)=-1$,For completeness, partial summation and the prime-counting form of the Prime number theorem giveTo verify the constant and the error, replace $\pi$ by $\mathrm{Li}+E$ in the first expression. Differentiating $\mathrm{Li}(u)/u$ shows its two contributions combine to $\log \log X$ plus a constant. The error integral converges absolutely because $|E(u)|/u^2\ll e^{-c\sqrt{\log u}}/u$, and $E(X)/X\to 0$. This proves the Mertens second theorem in the form needed here. In particular the pretentious distance between one and the Möbius function satisfies

For the last request, retain the earlier prime normalization $|f(p)|=1$; the proof also works with the weaker assumption $|f(p)|\le 1$. Without a bound on prime values, the original definition and the absolute convergence of the displayed measure need not apply to an unrestricted multiplicative arithmetic function. For example $f(n)=\mu (n{)}^{2}n$ is multiplicative and squarefree-supported, but for $X>e$ the displayed measure has infinite positive mass: its prime terms alone are ${\sum }_p{p}^{-1/\log X}$, which diverges by the Prime number theorem.

Put $L=\log X$, $\sigma =1+1/L$, and use the Fourier transform convention $\hat{\nu }(t)=\int e^{-itu}d\nu (u)$. Since $f$ is supported on squarefree integers, the defining property of a multiplicative arithmetic function gives the absolutely convergent Euler productThe bound on prime values gives $|f(n)|\le 1$ for squarefree $n$, so the total variation norm of a measure is bounded by $\zeta (\sigma )$; thus $\nu$ is a finite measure. Expanding the logarithm of the modulus, with an absolute $O(1)$ remainder since ${\sum }_p{p}^{-2\sigma }\ll 1$, givesAll constants here are uniform in $t$. Moreoverusing the pole of the Riemann zeta function. The nonnegative deficits $1-\mathrm{Re}(f(p)p^{-it})$ giveIndeed, the replacement of $p^{-\sigma }$ by $1/p$ for $p\le X$ costs at mostby the Mertens first theorem, while discarding the $p>X$ terms costs nothing in this lower bound. Thus the large Fourier coefficient criterion for pretentiousness isIn particular,The comparison function is the Archimedean character $n^{it}$ with this sign because our Fourier transform uses $e^{-itu}$.

For $s,t\in [0,1]$, defineThe function $d_g$ is a pseudometric. Declare $s\sim t$ when $d_g(s,t)=0$, and give the quotient set $T_g=[0,1]/\sim$ the induced metric, again denoted $d_g$. This is the real tree encoded by an excursion $g$.

PutThe numerator is the linear kernel, while the reciprocal of the denominator is the kernel from part c applied to $\Vert x{\Vert }^2$ and $\Vert y{\Vert }^2$. The product of positive-semidefinite kernels therefore shows that $a$ is a positive-semidefinite kernel. The Cauchy-Schwarz inequality and the arithmetic-geometric mean inequality give $|a(x,y)|\le 1/2$, soEach power is positive semidefinite by the Schur product theorem, and the convergent sum is positive semidefinite.

Moreover $k_2(x,x)=1$. If $\Phi$ is its canonical feature map, thenThe feature-space norm gives symmetry and the triangle inequality. Finally, $d(x,y)=0$ implies $k_2(x,y)=1$, hence $a(x,y)=1/2$ andwhich is equivalent to $\Vert x-y{\Vert }^2=0$. Thus $d$ is a metric rather than merely a pseudometric.

For unit-modulus prime values, pretentious distance is the Euclidean metric on the finite vectors $(f(p)/\sqrt{2p}{)}_{p\le X}$. On whole arithmetic functions it is only a pseudometric: it sees neither values on prime powers nor primes above the cutoff. Even for every cutoff, the functions one and ${\mu }^2$ have distance zero but differ on nonsquarefree integers. The norm representation proves the triangle inequality and identifies exactly the quotient on which separation holds.

Let $g:[0,1]\to {\mathbb{R}}_{+}$ be continuous with $g(0)=g(1)=0$, and setThe function $d_g$ is a pseudometric. Its metric quotientis the real tree encoded by $g$.