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The Green-function representation givesThe bracket simplifies to $x(x-1)/2$, soIt vanishes at both endpoints, and applying $(D-2{)}^2$ returns $e^{2x}$.

Write $G(x;\xi )=e^{2x}H(x;\xi )$. Since $(D-2{)}^2(e^{2x}H)=e^{2x}{H}^{\prime \prime }$, the Green function must have $H(0;\xi )=H(1;\xi )=0$, be continuous at $x=\xi$, and satisfy the jump condition for an impulse $H_x({\xi }^{+};\xi )-H_x({\xi }^{-};\xi )=e^{-2\xi }$. Solving the two linear pieces gives

On $\mathbb{C}\setminus [-1,1]$, the chosen branch of a multivalued function is characterized by $f(z{)}^2=z^2-1$ and $f(x)>0$ for $x>1$; equivalently $f(z)/z\to 1$ as $z\to \infty$. Define $h(z)=-f(-z)$. Then $h(z{)}^2=z^2-1$ and $h(z)/z\to 1$ at infinity, so uniqueness of the normalized holomorphic square root gives $h=f$. Therefore $f(-z)=-f(z)$.

Let $U={\sum }_i{u}_i$. The candidate strategies areAt this pair, A's partial derivatives $2u_i{y}_i/(x_i+y_i{)}^2$ are independent of $i$, as are the negatives of B's partial derivatives. Since $F$ is concave in $x$ for fixed $y$ and convex in $y$ for fixed $x$, the Lagrange sufficiency theorem provesfor all feasible $x,y$. Hence $(x^{\ast },y^{\ast })$ is a saddle point and gives the optimal strategies in this zero-sum game. Its game value is

The Lagrange sufficiency theorem says that if a feasible point $x^{\ast }$ and nonnegative multipliers satisfy complementary slackness and $x^{\ast }$ globally maximizes the Lagrangian function in constrained optimization, then $x^{\ast }$ globally solves the constrained maximization problem. Indeed, for every feasible $x$, the signs of the constraints give $f(x)\le L(x)\le L(x^{\ast })=f(x^{\ast })$. Concavity and the Karush-Kuhn-Tucker conditions are a common way to verify the global Lagrangian maximum.

For fixed $y$, A maximizessubject to $x_i\ge 0$ and ${\sum }_i{x}_i=X$. This is a concave function of $x$. On every active coordinate, the multiplier equation isThus A's optimal allocation is given by the water-filling algorithm:where $\lambda >0$ is chosen so that ${\sum }_i{x}_i=X$. Coordinates with zero expression receive no resource. This is a water-filling algorithm. The formula gives the optimum when every $y_i>0$. If $y_i=0$, any $x_i>0$ already gives A the whole market $u_i$ in that region, so A assigns such coordinates arbitrarily small positive amounts and applies water filling to the remaining budget; without a convention for $0/0$, the resulting value can be a supremum rather than an attained maximum.

The algebraic norm on $\mathbb{Z}[\sqrt{-5}]$ is $N(a+b\sqrt{-5})=a^2+5b^2$ and satisfies multiplicativity of the algebraic norm. The calculation in part (a) therefore gives all elements of norm $9$:If one of these factored into two nonunits, multiplicativity would force both factors to have norm $3$. But the Diophantine equation $a^2+5b^2=3$ has no solution. Hence all six displayed elements are irreducible. This is a standard example in which an irreducible element need not have prime norm.

Since $5y^2\le 9$, we have $y\in \{-1,0,1\}$. For $y=0$, $x^2=9$; for $y=\pm 1$, $x^2=4$. Thus the solutions of the Diophantine equation arewhere the two signs in the second family are independent.

Since the odd sine part of the complex exponential integrates to zero,Part (e)(ii) therefore yieldsin the sense of distributions. This is the Fourier representation of the Dirac delta function.

Interpreting the displayed function aswe have ${\gamma }_n\ge 0$ and, after $u=nx$, ${\int }_{\mathbb{R}}{\gamma }_n=1$. Moreover its mass outside any fixed neighborhood of zero tends to zero. Therefore $({\gamma }_n)$ is an approximate identity, and for every test function $f$,Thus ${\gamma }_n\to \delta$ in the sense of distributions.

Repeated integration by parts, or the definition of the $m$th distributional derivative, gives

Using the definition of a distributional derivative,Hence $x{\delta }^{\prime }(x)=-\delta (x)$ as distributions.

For every test function $f$, the change of variables $u=\alpha x$ givesEquality of distributions therefore gives

No. The nonvanishing holomorphic functions $f_n(z)=e^{nz}$ satisfy $f_n(0)=1$, butAvoiding zero alone permits an unbounded holomorphic logarithm's real part; the omitted ray in part (c) was what constrained the image after taking a square root.

Because $f(D)$ avoids the closed negative real axis, its principal square root $g=\sqrt{f}$ is holomorphic, satisfies $g(0)=1$, and has $\mathrm{Re}g>0$. The Cayley transform between the half-plane and diskmaps the right half-plane into the unit disk and has $h(0)=0$. By Schwarz lemma, $|h(z)|\le |z|$. When $|z|\le 1/2$,Thusso one may take the universal constant $C=9$.

This is Schwarz lemma. Define $g(z)=f(z)/z$ for $z\ne 0$ and $g(0)=f^{\prime }(0)$. The removable singularity makes $g$ holomorphic. Applying the maximum modulus principle to $g$ on disks of radius $r<1$, then letting $r\uparrow 1$, gives $|g(z)|\le 1$. Hence

The hypothesis says that $|f|$ attains a global, hence local, maximum at the interior point $a$. By the maximum modulus principle, a nonconstant holomorphic function cannot do this. Therefore $f$ is constant.