Integrate the joint density in $u$. Symmetry gives $f(-u)g(-uv)=f(u)g(uv)$, so the integrand $|u|f(u)g(uv)$ is even. HenceTonelli's theorem justifies marginalization for these nonnegative densities.

The inverse change of variables is $x=u$, $y=uv$, whose Jacobian determinant is $u$. Since $X,Y$ are independent, the original joint density is $f(x)g(y)$. ThusThe transformation is nonsingular away from $u=0$; this line has probability zero because $X$ has a density, so no extra mass is omitted.

Set $A={\sum }_i{p}_i{a}_i>0$ and $x_i=a_i/A$. Then ${\sum }_i{p}_i{x}_i=1$, so part 1 gives ${\prod }_i(a_i/A{)}^{p_i}\le 1$. Since ${\sum }_i{p}_i=1$, this becomesEquality holds exactly when $x_i=1$ for all $i$, equivalently all the $a_i$ are equal. This is the weighted arithmetic–geometric mean inequality.

Work on the positive orthant, where the real powers defining $g$ are smooth. Its closure under the constraint is the compact weighted simplex $\overline{M}=\{x_i\ge 0,\sum p_i{x}_i=1\}$, since $0\le x_i\le 1/p_i$. The product extends continuously there, is zero on its boundary when $n>1$, and is one at $(1,\dots ,1)$. Thus a maximum is attained in the positive interior.

At an interior maximizer the constraint gradient $(p_1,\dots ,p_n)$ is nonzero. The multiplier equations giveAll coordinates are equal, and the weighted constraint forces each to equal one. Therefore the maximum is one and $\boxed{g(x)\le 1}$ on $M$. For $n=1$ the constraint itself gives $x_1=1$, so the same conclusion holds directly. Equality occurs exactly at the all-ones point.

Continuity and nonvanishing on the compact interval implyHence $m\Vert f-g{\Vert }_{\infty }\le d_{\alpha }(f,g)\le M\Vert f-g{\Vert }_{\infty }$. A $d_{\alpha }$-Cauchy sequence is therefore uniformly Cauchy, so (i) gives a continuous uniform limit; the upper bound then gives convergence in $d_{\alpha }$. Thus $\boxed{(C([0,1]),d_{\alpha })\text{ is complete}}$. The positive lower bound is what makes the argument work.

Yes. A Cauchy sequence of continuous functions in the supremum norm is Cauchy in the given complete space of bounded functions, so it has a uniform limit there. A uniform limit of continuous functions is continuous: at a point, use two uniform errors smaller than $\epsilon /3$ and continuity of one sufficiently late term for the remaining error. Thus the limit is in $C([0,1])$, proving completeness. Equivalently this is a closed subspace of the given complete space.

A metric space is complete if every Cauchy sequence of its points converges to a point of that same space. The requirement on the location of the limit distinguishes, for example, $\mathbb{R}$ from its incomplete subspace $\mathbb{Q}$.

A sequence is Cauchy ifThe comparison is between sufficiently late terms, without presupposing a limiting point in the space.

The image $f(K)$ is compact: pull any open cover back by the continuous map, use a finite subcover of $K$, then map forward. For a general topological target, closedness is not guaranteed; a continuous image into the indiscrete two-point space can be a nonclosed singleton. No connectedness assertion follows merely from compactness of $K$.

If $Y$ is metric, $f(K)$ is compact, therefore closed in $Y$, bounded and totally bounded. It is also complete in its induced metric and sequentially compact. For the latter properties, compact metric spaces have a convergent subsequence of every sequence; a Cauchy sequence with one convergent subsequence converges to that same limit. Total boundedness follows by selecting finitely many balls of each prescribed radius. These are properties of the image itself, irrespective of whether $Y$ is complete.

By the Heine–Borel theorem a subset of $\mathbb{R}$ is compact exactly when it is closed and bounded. Consequently the nonempty compact intervals are exactly $\boxed{[a,b](-\infty <a\le b<\infty )}$. This includes the degenerate interval $[a,a]$. The empty interval is compact as well; nonclosed bounded intervals and all unbounded intervals fail the criterion.

In any topological space a closed subset of a compact set is compact: add its open complement to an open cover and take a finite subcover of the compact set. However, compact subsets need not be closed in a general space. In the indiscrete two-point space a singleton is compact but not closed. Conversely a closed subset need not be compact: $\mathbb{R}$ itself is closed in $\mathbb{R}$ but the cover $\{(-n,n):n\ge 1\}$ has no finite subcover.

In a Hausdorff space every compact subset $K$ is closed. Given $x\notin K$, separate $x$ and each $k\in K$ by disjoint neighborhoods; compactness selects finitely many neighborhoods covering $K$, and the intersection of their companion neighborhoods of $x$ misses $K$. Thus the complement is open. Closedness still need not imply compactness, as the same real-line example shows. It does imply compactness if the ambient space is compact.

In a metric space compactness implies boundedness: finitely many radius-one balls cover the set, so their centers have a finite maximal distance from any fixed center and the triangle inequality gives a global bound. The converse is false. An infinite set with discrete metric $d(x,y)=1$ for $x\ne y$ is bounded, but its singleton cover has no finite subcover. It is even closed and complete, so closedness plus boundedness alone does not suffice in general metric spaces.

A subset $B$ of a metric space is bounded if it lies in a ball of finite radius: there exist a point $x_0$ and $R<\infty$ such that $d(x,x_0)\le R$ for every $x\in B$. Equivalently its diameter ${\sup }_{x,y\in B}d(x,y)$ is finite, with the empty set bounded by convention.

A subset $K$ is compact if every cover of $K$ by sets open in the ambient space has a finite subcover. Equivalently, every relatively open cover of the subspace $K$ has a finite subcover. These agree because relatively open sets are intersections with ambient open sets.

The components cover $X$ and are pairwise disjoint by (b)(ii); by (iii) they are open intervals. Each nonempty interval contains a rational number. Enumerate the rationals and assign to each interval the first rational it contains. Disjointness makes these assigned rationals distinct, giving an injection from the set of components into $\mathbb{N}$. Thus $\boxed{X\text{ is a countable union of pairwise disjoint open intervals}}$, with the empty union for $X=\varnothing$.

The inclusion of $X$ into $\mathbb{R}$ is continuous, so a connected component $K$ of $X$ is an interval in $\mathbb{R}$. For $x\in K$, openness of $X$ gives $\epsilon >0$ with $(x-\epsilon ,x+\epsilon )\subseteq X$. This interval is connected and meets $K$, so their union is connected and maximality forces the interval into $K$. Thus $K$ is open in $\mathbb{R}$. A nonempty open connected subset of the line is an open interval, allowing an infinite endpoint or the whole line. Hence every component is an open interval.

If $K_a\cap K_b\ne \varnothing$, their union is connected by (a)(iii). It contains $a$, so maximality gives $K_a\cup K_b\subseteq K_a$ and hence $K_b\subseteq K_a$. Interchanging the roles gives the reverse inclusion. Thus the two components coincide; otherwise they are disjoint.

Let $K_a$ be the union of all connected subsets containing $a$. This collection is nonempty because it includes $\{a\}$. Any two of its members meet at $a$, so (a)(iii) makes their full union connected. It contains $a$ and by its definition contains every connected subset through $a$. Thushas exactly the required largest-set property.

Use the separation definition in (i). If $A={\bigcup }_i{A}_i$ had a separation $A=U\cup V$, the intersections with each connected $A_i$ show that $A_i$ lies wholly in one side. Choose one nonempty $A_{i_0}$; every other $A_i$ intersects it and must lie on that same side. Thus the other side is empty, a contradiction. If the union is empty it is connected by definition. Therefore $\boxed{\bigcup _i{A}_i\text{ is connected}}$.

The connected subsets of the real line are exactly the intervals, including singletons, unbounded intervals and the empty set. Equivalently, whenever $x<y$ belong to the set, every $z$ with $x<z<y$ also belongs to it. Endpoints may or may not be included.

A subset $A$ is connected if, with its subspace topology, it cannot be written as $A=U\cup V$ with $U,V$ disjoint nonempty relatively open sets. Equivalently its only subsets both open and closed relative to $A$ are $\varnothing ,A$. This definition includes the empty set and singletons as connected sets.