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We use the Bolzano-Weierstrass theorem: every bounded real sequence has a convergent subsequence. If $f$ were unbounded on $[a,b]$, there would be $x_n\in [a,b]$ with $|f(x_n)|\ge n$. The Bolzano-Weierstrass theorem gives a subsequence $x_{n_k}\to c$, and the closed interval contains $c$. Since $f$ is a continuous function, $f(x_{n_k})\to f(c)$, contradicting $|f(x_{n_k})|\ge n_k\to \infty$. Thus $f$ is bounded.

Let $s={\sup }_{x\in [a,b]}f(x)$. By the definition of the supremum, choose $y_n$ with $s-1/n<f(y_n)\le s$. Again the Bolzano-Weierstrass theorem gives $y_{n_k}\to d\in [a,b]$. Continuity yields $f(d)=s$. Applying the same argument to $-f$ gives a point where $f$ attains its infimum. Therefore both the supremum and the infimum are attained. This proves the extreme value theorem for a closed bounded interval.

Away from integer multiples of $\pi$, the function is a composition of continuous functions. At zero, the bound $|g(x)|\le \sqrt{|x|}$ gives $g(x)\to 0=g(0)$ by the squeeze theorem. This is oscillation multiplied by a vanishing amplitude.

Fix $n\ne 0$ and set $t_k=\pi /2+2\pi k$ and $x_k=n\pi +(-1{)}^n\arcsin (1/t_k)$. Then $x_k\to n\pi$, $\sin x_k=1/t_k$, andSo $g$ is discontinuous at every nonzero integer multiple of $\pi$. Its set of continuity points is

On $[0,\pi ]$, $g(x)\le \sqrt{x}<\sqrt{\pi }$ for $x<\pi$, while $g(\pi )=0$. With $x_k=\pi -\arcsin (1/t_k)$ as above, $g(x_k)=\sqrt{x_k}\to \sqrt{\pi }$. Consequently the supremum is $\sqrt{\pi }$ and is not attained. This illustrates that supremum can fail to be attained at an oscillatory endpoint.

On $[\pi ,3\pi /2]$, set $x_0=\pi +\arcsin (2/(3\pi ))$. It lies strictly inside the interval, and $1/\sin x_0=-3\pi /2$, so $g(x_0)=\sqrt{x_0}>\sqrt{\pi }$. Choose $0<\delta <x_0-\pi$. On $[\pi ,\pi +\delta ]$,On $[\pi +\delta ,3\pi /2]$, $g$ is a continuous function, so the extreme value theorem gives a maximum there, at least $g(x_0)$. It exceeds every value on the discarded interval. Therefore $g$ does attain its supremum on $[\pi ,3\pi /2]$. This is the interior-value criterion for an attained maximum.

For a partition $P$, write $U(f,P)$ and $L(f,P)$ for the upper Darboux sum and lower Darboux sum. The Riemann integrability criterion for bounded $f$ is that for every $\epsilon >0$ some partition satisfies $U(f,P)-L(f,P)<\epsilon$. Thus if the gaps for $D_n$ tend to zero, the Riemann integrability criterion immediately proves that $f$ is Riemann integrable.

Conversely, suppose $f$ is Riemann integrable, and choose a fixed partition $P$ with $U(f,P)-L(f,P)<\epsilon /2$. Let $r$ be the number of its interior division points and choose $M\ge 1$ with $|f|\le M$. Call a cell of $D_n$ bad when its interior contains a division point of $P$. There are at most $r$ bad cells, and their total length is at most $r/n$.

Every other cell lies in a single cell of $P$, so its oscillation is no larger than that of the containing cell. Summing the contributions of these good cells gives at most $U(f,P)-L(f,P)$, while each bad cell has oscillation at most $2M$. Therefore the finite bad-cell estimate for Darboux sums givesFor large enough $n$, the right-hand side is less than $\epsilon$. This proves the uniform-mesh Darboux criterionThis argument does not assume that the partitions $D_n$ are nested; in general they are not.

For the composition, $f$ takes values in $[-M,M]$. Since $g$ is continuously differentiable, the extreme value theorem bounds both $|g|$ and $|g^{\prime }|$ on that interval. In particular, let $K={\sup }_{[-M,M]}|g^{\prime }|<\infty$. The mean value theorem states that a function continuous on the interval between $s,t$ and differentiable in its interior satisfies $g(s)-g(t)=g^{\prime }(\xi )(s-t)$ for some intermediate $\xi$. HenceOn each partition cell, this bounds the oscillation of $g\circ f$ by $K$ times the oscillation of $f$, whether or not the local extrema are attained. It follows thatThe composition is bounded, so the proved uniform-mesh Darboux criterion applies. Thus $g\circ f$ is Riemann integrable. The reusable fact is Lipschitz composition preserves Riemann integrability; continuous differentiability supplies the required bound on the range of $f$.

First prove nowhere monotonicity of Brownian motion. On a fixed interval $[a,b]$ with rational endpoints and $a<b$, the $2^m$ increments over its equal subdivision are independent centered normal random variables. If the Brownian motion path were nondecreasing, all these increments would be nonnegative, an event with probability $2^{-2^m}$. Letting $m\to \infty$ gives probability zero. The same argument excludes nonincreasing paths. A countable union over rational intervals shows that, almost surely, no nontrivial interval supports a monotone function restriction of the path, since every such interval contains one with rational endpoints.

Work on this event and on the event of continuous paths. Inside any open interval $I\subset (0,\infty )$, choose two separated smaller intervals, the first to the left of the second. The first contains $r<s$ with $B_r<B_s$, because its restriction is not nonincreasing. The second contains $u<v$ with $B_u>B_v$, because its restriction is not nondecreasing. Thus $r<s<u<v$, all in $I$.

By the extreme value theorem, the path attains its maximum on $[r,v]$. This value exceeds $B_r$, since it is at least $B_s$, and exceeds $B_v$, since it is at least $B_u$. A maximizing time therefore lies in $(r,v)$ and is a local maximum of Brownian motion. Every open interval in the half-line contains a positive-time interval of this kind. Consequently the set of local maxima of Brownian motion is dense in $[0,\infty )$ almost surely.

The closure is closed and bounded in ${\mathbb{R}}^2$, hence compact by the Heine-Borel theorem. It is nonempty. The extreme value theorem therefore gives a point of the closure at which the continuous function attains its maximum. The closure, rather than the open domain, is essential.

The Bolzano-Weierstrass theorem states that every bounded sequence of real numbers has a convergent subsequence. We first use it to show that a continuous function on $[a,b]$ is bounded above. Otherwise, choose $x_n\in [a,b]$ with $f(x_n)>n$. A convergent subsequence $x_{n_k}\to c$ exists, with $c\in [a,b]$ because the interval is closed. Continuity would give $f(x_{n_k})\to f(c)$, contradicting $f(x_{n_k})>n_k\to \infty$.

Let $L=\sup \{f(x):x\in [a,b]\}$, which is finite. By the definition of supremum, there are $y_n\in [a,b]$ with $L-1/n<f(y_n)\le L$. Another application of the Bolzano-Weierstrass theorem gives a subsequence $y_{n_k}\to c\in [a,b]$. By continuity,Thus the supremum is actually attained:This proves the maximum part of the extreme value theorem from sequential compactness rather than assuming it.

Now suppose $f^{\prime \prime }<0$ on $\mathbb{R}$. Applying the mean value theorem to $f^{\prime }$ shows that $f^{\prime }$ is strictly decreasing. At a local maximum of a differentiable function, the left difference quotients are nonnegative and the right difference quotients are nonpositive; existence of the derivative forces $f^{\prime }=0$. A strictly decreasing function has at most one zero. ThereforeIf that zero exists, $f^{\prime }$ is positive to its left and negative to its right, so the local maximum is also a global maximum.

For the uniform bound by $K<0$, defineIts derivative is $h^{\prime }(x)=f^{\prime \prime }(x)-K<0$, so $h$ is strictly decreasing by the mean value theorem. Comparing with $h(0)=f^{\prime }(0)$ givesBecause $K<0$, this makes $f^{\prime }$ negative at sufficiently large positive $x$ and positive at sufficiently large negative $x$. The function $f^{\prime }$ is continuous, since it is differentiable; the intermediate value theorem supplies a zero $c$ between such points. Its strict decrease makes this zero unique. Again using the mean value theorem for $f$, the derivative signs imply that $f$ increases up to $c$ and decreases after $c$:This illustrates uniform negative curvature and global maximization.

Merely requiring $f^{\prime \prime }<0$ does not force existence. TakeThen $f^{\prime }(x)=e^{-x}>0$ and $f^{\prime \prime }(x)=-e^{-x}<0$, but $f$ is strictly increasing, approaches $0$ as $x\to +\infty$, and never equals $0$. The answer to the final question is no, even for a function bounded above:The stronger uniform curvature assumption prevents this behaviour. Here $f^{\prime \prime }(x)\to 0$ at the end where the supremum is approached, so no uniform negative bound of that kind is available.

Set $h(x)=f(x)-yx$. ThenThe extreme value theorem gives a minimum of $h$ on $[a,b]$. If it occurs in $(a,b)$, the two-sided difference quotient gives $h^{\prime }(c)=0$. If it occurs at $a$, either $h^{\prime }(a)=0$ or a negative right derivative contradicts minimality; the endpoint $b$ is analogous. Hence some $c\in [a,b]$ satisfiesThis proves the Darboux theorem for derivatives in the stated case without assuming that $f^{\prime }$ is continuous.

Because a differentiable function is continuous, defineThe fundamental theorem of calculus gives $F^{\prime }(x)=f^{\prime }(x)+f(x)$. At the endpoints,whileApplying the result just proved to $F$ yields $d\in [a,b]$ with

Let the nonconstant polynomial have degree $d$ and leading coefficient $c$. If $d$ is odd, its limits at $+\infty$ and $-\infty$ have opposite signs, so the intermediate value theorem shows that its image is $\mathbb{R}$. If $d$ is even and $c>0$, it tends to $+\infty$ at both ends. Outside a sufficiently large compact interval it exceeds $f(0)$, while on that interval the extreme value theorem gives a global minimum $a$. Continuity and the intermediate value theorem then give image $[a,\infty )$. For $c<0$, applying this argument to $-f$ gives $(-\infty ,a]$. These are exactly the three possible images.

Two norms $N_1,N_2$ are equivalent norms when constants $c,C>0$ satisfyfor every $v$.

For $x,y\in {\mathbb{R}}^n$, the reverse triangle inequality givesby the Cauchy-Schwarz inequality. Hence $g$ is Lipschitz continuous, and therefore continuous.

On the compact Euclidean unit sphere, $g$ is positive and continuous, so the extreme value theorem gives $0<m\le g\le M<\infty$. Homogeneity then yieldsThus every norm on a finite-dimensional vector space is equivalent to the Euclidean norm; comparing two such bounds proves any two norms on $V$ are Lipschitz equivalent.

Let $P_n$ be the finite-dimensional vector space of real polynomials of degree at most $n$. For fixed $a>0$,is a norm: if it vanishes, the polynomial vanishes on an interval and hence is the zero polynomial. Alsois a norm on $P_n$. By finite-dimensional equivalence of norms, there is $C>0$, depending only on $n$ and $a$, such that $M(p)\le CN(p)$. ThereforeThe extreme value theorem supplies $y\in [0,a]$ at which the last supremum is attained, and hence

Let $R_s$ be the finite, nonempty set of available routes for source-sink pair $s$, and write $x_r$ for a route flow. The link-route incidence matrix has $A_{jr}=1$ when route $r$ uses link $j$, and zero otherwise. The source-sink route incidence matrix has $H_{sr}=1$ when $r\in R_s$. Thus $y=Ax$ gives link throughputs and $f=Hx$ gives aggregate source-sink flows. A route's delay isA Wardrop equilibrium uses only routes of minimum delay for their source-sink pair:An infinitesimal user cannot improve its delay by changing route.

For fixed $f_s\ge 0$, the feasible route flows form a nonempty compact convex set: for each $s$, $x_r\ge 0$ and ${\sum }_{r\in R_s}{x}_r=f_s$. Define the Beckmann potentialIt is continuous, so the extreme value theorem gives a minimizer. Since every $D_j$ is strictly increasing, $V$ is a strictly convex function of $y$. Its route-flow gradient is $\partial V(Ax)/\partial x_r=d_r(x)$.

If an occupied route $r\in R_s$ has greater delay than another route $q\in R_s$, moving a small amount of flow from $r$ to $q$ gives a negative directional derivative, contradicting minimality. Conversely, at a Wardrop equilibrium, for any feasible $\tilde{x}$,The first-order condition for convex optimization therefore proves optimality. ConsequentlyIf two minimizers had different $y$, their midpoint would give a strictly smaller Beckmann potential. Thus the equilibrium link throughputs are unique. The route flows need not be unique, since $V(Ax)$ need not be a strictly convex function of $x$.

For a concrete example, take two successive stages of parallel links: $a,b$ in the first stage and $c,d$ in the second, with one source-sink pair and four routes $ac,ad,bc,bd$. Put $D_j(y)=y$ on every link and total flow $f=2$. All allocationshave the same link throughputs $(1,1,1,1)$ and every route has delay two. These distinct allocations are all Wardrop equilibria.

For elastic demand, use the standard physical convention that $D_j\ge 0$ and $B_s:[0,\infty )\to [0,\infty )$ is finite, continuous, and strictly decreasing. Set $M_s=B_s(0)$ and ${\ell }_s={\lim }_{\lambda \to \infty }{B}_s(\lambda )$, and define the inverse demand function $P_s=B_s^{-1}$ on $({\ell }_s,M_s]$. It is continuous and strictly decreasing, with $P_s(M_s)=0$ and $P_s(u)\to \infty$ as $u\downarrow {\ell }_s$. Choose any reference value $f_s^0\in ({\ell }_s,M_s)$ and setEach summand of $-G$ is a strictly convex function. The reference values only add a constant to the objective. In the usual case ${\ell }_s=0$, one may use a lower limit zero whenever the improper integral is finite; that integrability is not guaranteed merely by continuity of $B_s$.

Minimize $\Psi$ over the compact feasible set $x\ge 0$, ${\ell }_s\le (Hx{)}_s\le M_s$, interpreting the objective at $f_s={\ell }_s$ by its one-sided limit, which can be $+\infty$. This extension is sequentially lower semicontinuous, and an interior-demand feasible point has finite value, so a minimizer exists. It cannot have $f_s={\ell }_s$: adding a small amount on any route for $s$ has bounded marginal link cost, whereas its marginal demand benefit $P_s(f_s)$ tends to infinity. Thus $f_s>{\ell }_s$.

For ${\ell }_s<f_s<M_s$, the route-flow partial derivative is $d_r-P_s(f_s)$. The first-order conditions for convex optimization giveIf $f_s=M_s$, reducing any occupied route cannot improve the objective only if that route has zero delay; nonnegative link delays then give ${\lambda }_s=0=P_s(M_s)$. Therefore the conditions in every case areConversely these conditions give the nonnegative directional derivative for every feasible competitor and hence minimize the convex function $\Psi$. Allowing $f$ explicitly yields the stated form with $Hx=f$ and $Ax=y$, with inverse-demand endpoint conventions understood. In particular, $G$ is defined on its demand range, rather than evaluating $B_s^{-1}$ outside that range.

Strict convexity in the pair $(y,f)$ proves that the equilibrium aggregate source-sink flows and link throughputs are unique. Individual route flows may still be nonunique. Nonnegative, finite demand, nonnegative delays, finite route sets, and a usable route for each source-sink pair are the implicit modeling assumptions needed for the existence assertions.

Let $E=\mathbb{R}[x{]}_{\le d}$ with the supremum norm. By equivalence of norms in finite dimensions, there is $C_d$ such that ${\sum }_{j=0}^d|a_j|\le C_d\Vert p{\Vert }_{\infty }$ whenever $p={\sum }_j{a}_j{x}^j$. Henceso ${\epsilon }_n:=\Vert B_n-I{\Vert }_{\mathrm{op}}\le C_d{\max }_{0\le j\le d}\Vert B_n(x^j)-x^j{\Vert }_{\infty }\to 0$. This upgrades convergence on fixed polynomials to operator norm convergence on the fixed space.

For sufficiently large $n$, ${\epsilon }_n<1$, and a Neumann series givesThus the inverse Bernstein approximation on a fixed-degree polynomial space satisfies $\Vert g_n-f{\Vert }_{\infty }\to 0$. By the extreme value theorem, strict positivity on the compact interval gives $\mu ={\min }_{[0,1]}f>0$. Eventually $\Vert g_n-f{\Vert }_{\infty }<\mu /2$, and consequentlyApplying pointwise convergence to the varying polynomials $g_n$ without this finite-dimensional operator argument would not justify the conclusion.

For part (a), fix $\mathbf{x}\in L$ and consider the line segmentThe convex set $L$ contains this entire segment. Forwe have $g(0)=1$ andIf ${\sum }_j{x}_j>n$, then $g(t)>1$ for all sufficiently small positive $t$, contradicting the hypothesis on $L$. Thus

For part (b), $K$ is compact, and the product of the coordinates is a continuous function. The extreme value theorem therefore supplies $\mathbf{u}\in K$ maximizing that product. Assume every $u_j>0$ and apply part (a) to the convex setIt contains $1$, and maximality of $\mathbf{u}$ makes every coordinate product at most one. HenceIf $\mathbf{v}$ is another maximizer, its coordinates are positive and ${\prod }_j(v_j/u_j)=1$. The arithmetic-geometric mean inequality giveswhile the displayed inequality gives the reverse bound. Equality in the arithmetic-geometric mean inequality forces every $v_j/u_j=1$. Therefore the maximizer $\mathbf{u}$ is unique.

The set $K\cap P$ is compact because $K$ is closed and bounded, and the productis continuous. By the extreme value theorem, $G$ has a maximizer. Since $K$ meets the interior of the nonnegative orthant $P$, some feasible product is positive, so every maximizer lies in $\mathrm{Int}P$.

On $\mathrm{Int}P$, maximizing $G$ is equivalent to maximizingThe Hessian of $F$ iswhich is negative definite. Thus $F$ is a strictly concave function, and it has at most one maximizer on the convex set $K\cap P$. Denote this unique point by $x^{\ast }$.

For any $x\in K\cap P$, the line segmentlies in $K\cap \mathrm{Int}P$. Since $F(x(t))$ is maximal at $t=0$, its right derivative there is nonpositive:ThereforeThis is the supporting inequality for the product maximizer on a compact convex subset of the positive orthant.

Suppose now that $K$ is invariant under cyclic coordinate permutation. That permutation preserves the product, so uniqueness forces it to fix $x^{\ast }$. Hencefor some $r>0$. More explicitly,Indeed, cyclic symmetry averaging puts the diagonal point with coordinate equal to the mean of any $x\in K$ back in $K$, while the preceding inequality with $x_j^{\ast }=r$ gives ${\sum }_j{x}_j\le nr$.

Finally, given $a=(a_1,\dots ,a_n)\in \mathrm{Int}P$, defineThis set is nonempty, closed, convex and bounded, and it contains $a$. By the arithmetic-geometric mean inequality,Equality holds only when all $x_j/a_j$ are equal and their sum is $n$, namely only at $x=a$. Thus this $K$ has

Let $f:[a,b]\to \mathbb{R}$ be continuous. If it were unbounded, one could choose $x_n\in [a,b]$ with $|f(x_n)|>n$. The Bolzano-Weierstrass theorem gives a subsequence $x_{n_j}\to x\in [a,b]$, but continuity would then give $f(x_{n_j})\to f(x)$, contradicting unboundedness.

Now let $M=\sup f([a,b])$. Choose $x_n$ with $f(x_n)>M-1/n$. A convergent subsequence and continuity give $f(x)=M$ at its limit. Applying the same argument to $-f$ shows that the infimum is attained. This proves the extreme value theorem on a closed bounded interval.

The functionis continuous and bounded, but attains neither its infimum $0$ nor its supremum $1$. For the second example, enumerate the rationals in $[0,1]$ as $(q_n)$ and setEvery nondegenerate interval contains infinitely many rational numbers, hence some $q_n$ with arbitrarily large $n$; therefore $\psi$ is unbounded on every such interval.

For the running extrema, compactness lets us writeThe function $f$ is uniformly continuous on $[a,b]$. Given $\epsilon >0$, choose $\delta >0$ such that $|f(u)-f(v)|<\epsilon$ whenever $|u-v|<\delta$. If $x<y$ and $y-x<\delta$, every new value $f(\xi )$ with $x\le \xi \le y$ is at most $f(x)+\epsilon \le M(x)+\epsilon$. Since $M(y)\ge M(x)$,Interchanging $x,y$ handles the other order. Applying this argument to $-f$ proves continuity of $m$ as well.

Finally fix $T>0$ and put $h(x)=g(x+T)-g(x)$. For each positive integer $n$, there must be some $x_n>n$ with $|h(x_n)|<1/n$. Otherwise, for some $n$ the continuous function $h$ would satisfy $|h(x)|\ge 1/n$ on the interval $(n,\infty )$. By the intermediate value theorem, $h$ would have a constant sign there. The valueswould then increase or decrease by at least $1/n$ at every step, contradicting boundedness of $g$. Thus $x_n\to \infty$ and

Rolle theorem: if $f$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $f(a)=f(b)$, then $f^{\prime }(c)=0$ for some $c\in (a,b)$. Indeed, the extreme value theorem gives a maximum and minimum. If both occur only at the endpoints then $f$ is constant; otherwise an interior extremum $c$ satisfies $f^{\prime }(c)=0$ by comparing the two-sided difference quotients.

Mean value theorem: if $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then some $c\in (a,b)$ satisfiesApply Rolle's theorem towhich has $h(a)=h(b)=0$.

SetThe triangle inequality makes $f$ Lipschitz continuous and hence a continuous function. Since $\gamma$ is an isometric embedding,so $f(t)\to \infty$ as $|t|\to \infty$. The function is therefore coercive, and the extreme value theorem on a sufficiently large compact interval gives a minimizing parameter.

Suppose $t_1,t_2$ both minimize $f$, put $p=\gamma (t_1)$ and $q=\gamma (t_2)$, and letThe restriction of $\gamma$ between the two parameters is a geodesic from $p$ to $q$. Let $m$ be its midpoint. In the geodesic triangle with vertices $x,p,q$, the thin geodesic triangle condition gives a point $y$ on one of the other two sides with $d(m,y)\le \delta$. By symmetry suppose $y\in [x,p]$. ThensoBut $m$ lies on $\gamma (\mathbb{R})$ and $p$ is a closest point, so $R\le d(x,m)$. Therefore $L\le 4\delta$, which is stronger than the requiredThis is the closest point on a geodesic line in a hyperbolic metric space estimate.

For $\mathbf{a},\mathbf{b}\in {\mathbb{R}}^{n+1}$, the triangle inequality and the Cauchy-Schwarz inequality giveThus $S$ is a continuous function. The unit sphere in the finite-dimensional Euclidean normed vector space ${\mathbb{R}}^{n+1}$ is compact. Moreover, $S(\mathbf{a})=0$ implies that the polynomial $T\mathbf{a}$ vanishes identically, so every coefficient $a_r$ is zero. Hence $S$ is strictly positive on the unit sphere. By the extreme value theorem, it has a positive minimum

The supremum norm is homogeneous. For $\mathbf{a}\ne 0$,It follows directly that $\Vert T\mathbf{a}{\Vert }_{\infty }\to \infty$ whenever $\Vert \mathbf{a}{\Vert }_2\to \infty$.

Write each polynomial of degree at most $n$ as $T\mathbf{a}$ and defineThis is a continuous function, and the reverse triangle inequality together with part (a) givesConsequently $F$ is a coercive function. Choose $R$ so large that $F(\mathbf{a})>F(0)$ outside the closed Euclidean ball of radius $R$. That ball is compact, so the extreme value theorem supplies a minimizer ${\mathbf{a}}_{\ast }$ there. Then $P=T{\mathbf{a}}_{\ast }$ is a best uniform approximation to $f$ among polynomials of degree at most $n$.

The extreme value theorem requires continuity on the entire closed interval. A bounded function can have values tending to an endpoint envelope without ever reaching that envelope, while its defined endpoint value is smaller. To prove nonattainment, establish a strict pointwise upper bound and construct a sequence approaching that bound. An oscillatory factor can supply the approaching sequence by selecting phases where it equals $1$.