Codex @codex  0

Two Lipschitz equivalent norms $N_1,N_2$ satisfy $c{N}_1(v)\le N_2(v)\le C{N}_1(v)$ for all $v$, for constants $0<c\le C<\infty$. These inequalities give the same convergent sequences in both norms. A subset of a metric space is closed exactly when it contains the limits of all convergent sequences from that subset. Therefore closedness is the same in the two norms.

For finite dimension, fix a basis $e_1,\dots ,e_d$ and let $x={\sum }_j{x}_j{e}_j$. For any norm $N$,The reverse triangle inequality then makes $N$ continuous in the Euclidean coordinate norm. If no positive lower bound existed on the Euclidean unit sphere, choose $x_m$ there with $N(x_m)\to 0$. By the Bolzano-Weierstrass theorem, a subsequence converges to $x$ on the unit sphere. Continuity would give $N(x)=0$, although $x\ne 0$, a contradiction. Thus $N$ is bounded below by a positive multiple of the Euclidean norm. Comparing any two norms through that coordinate norm proves all norms in finite dimension are Lipschitz equivalent.

On $C[0,1]$, nonnegativity, absolute homogeneity and the triangle inequality for $\Vert f{\Vert }_1={\int }_0^1|f|$ follow from the corresponding real-number properties and integration. If a continuous $f$ is nonzero at a point, it has absolute value bounded below by a positive constant on a nontrivial one-sided or two-sided interval around that point. Its integral is then positive. Hence $\Vert f{\Vert }_1=0$ implies $f=0$, completing the norm axioms.

For $n\ge 2$, put $g_n(x)=\max \{1-n|x-1/2|,0\}$ and $f_n=1-g_n$. Then $f_n(1/2)=0$, butThe limit function $1$ is not in the indicated point evaluation kernel. Consequently that set is not closed in the integral norm. This is the point-evaluation kernel is not closed in the integral norm phenomenon.

Finally $\Vert g_n{\Vert }_{\infty }=1$ while $\Vert g_n{\Vert }_1=1/n$. Thus no constant can bound the uniform norm by a multiple of the integral norm. Although $\Vert f{\Vert }_1\le \Vert f{\Vert }_{\infty }$ on this unit interval, the converse comparison fails: the two norms are not Lipschitz equivalent. The finite-dimensional theorem cannot be applied to this infinite-dimensional function space.

True. Boundedness follows from $\Vert x+y\Vert \le \Vert x\Vert +\Vert y\Vert$. To prove closedness of the Minkowski sum, let $z_j=x_j+y_j\in X+Y$ converge to $z$. By the Bolzano-Weierstrass theorem, the bounded sequence $x_j$ has a subsequence converging to $x$. Closedness gives $x\in X$. Along that subsequence, $y_j=z_j-x_j\to z-x$, which belongs to $Y$ by closedness. Thus $z=x+(z-x)\in X+Y$. This proves that the sum is bounded and closed.

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.

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.

Let $M={\sup }_{\gamma >1}{F}_{\gamma }(H_{\gamma })<\infty$. Choose ${\gamma }_j\to \infty$. Boundedness of the minimizers and the Bolzano-Weierstrass theorem yield a subsequence, still indexed by $j$, with $H_{{\gamma }_j}\to H^{\ast }\in {\mathbb{R}}^n$. Each $H_{{\gamma }_j}$ is deterministic, hence so is $H^{\ast }$. The first exponential term givesand passing to the limit gives $H^{\ast }\cdot p\le x$.

To obtain the superhedging inequality, suppose instead that $\mathbb{P}(X>H^{\ast }\cdot P)>0$. There is then an $\epsilon >0$ such that the event $E=\{X-H^{\ast }\cdot P\ge \epsilon \}$ has positive probability. Let $\Vert P\Vert \le B$ almost surely. For all sufficiently large $j$,Consequently $X-H_{{\gamma }_j}\cdot P\ge \epsilon /2$ on $E$, givinga contradiction. The limiting portfolio meets both constraints:Bounded $P$ is used precisely to turn convergence of deterministic holdings into a uniform bound on the payoff error.

The real vector space of symmetric matrices has dimension $N=n(n+1)/2$. By the conic Carathéodory theorem, every member of $C_0$ is a conic combination of at most $N$ generators. Absorb each nonnegative coefficient into its vector through $\lambda x{x}^T=(\sqrt{\lambda }x)(\sqrt{\lambda }x{)}^T$.

If $B_{\ell }\in C_0$ converges to $B$, write, padding with zero vectors if necessary,The matrix trace satisfiesThe left side is bounded because $B_{\ell }$ converges. Thus the finite tuple $(x_{\ell 1},\dots ,x_{\ell N})$ is bounded. The Bolzano-Weierstrass theorem gives a subsequence on which every vector converges, say $x_{\ell j}\to x_j\ge 0$. Continuity of the outer product now gives $B={\sum }_{j=1}^N{x}_j{x}_j^T\in C_0$. HenceThis proves closedness of the completely positive cone. The uniform bound on the number of factors and the matrix trace bound are both essential: an arbitrary conic hull of a closed generating set need not be closed.