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Non-characteristic. The first-order principal symbol is ${\nu }_t-i{\nu }_x$, which equals one on $(0,1)$.

Set $z=x+it$. The equation is equivalent to ${\partial }_{\bar{z}}\varphi =({\varphi }_x+i{\varphi }_t)/2=0$. Because $\varphi$ is continuously differentiable, the Cauchy-Riemann equations imply that it is a holomorphic function of $z$. Its convergent complex Taylor series restricts to a real power series along $t=0$, so $g$ is a real analytic function.

For smooth-data instability of the Cauchy-Riemann Cauchy problem, chooseThese are entire functions and satisfy the equation. Their initial derivatives obey ${\sup }_x|{\partial }_x^j{g}_n|\le n^j{e}^{-\sqrt{n}}$, so every finite sum tends to zero. ButThus analytic solutions can exist while continuous dependence on initial data fails for this smooth-data topology.

The Cauchy estimate follows directly from the differentiated Cauchy integral formula:The circle has length $2\pi r$, giving the bound for every integer $j\ge 0$.

Put $w=\varphi -\psi$ and integrate on the straight segment $z=\tau t/|t|$. For fixed $|x|<sR$, a disc centred at $x$ with radius approaching $R(\sigma (\tau )-s)$ stays inside $|\xi |\le \sigma (\tau )R$. Applying the first-derivative Cauchy estimate there givesIntegrating its modulus and taking the supremum over $|x|<sR$ is exactly the asserted integral estimate. The source cancels from $T\varphi -T\psi$.

To obtain a contraction mapping, use the weighted holomorphic norm on a shrinking time domain. Its finite-norm space consists of holomorphic functions on ${\mathcal{C}}_{\alpha }$ with zero initial value; the apparent quotient at $t=0$ is interpreted by a limit. Completeness follows because convergence in this norm implies locally uniform convergence of holomorphic functions, including near $t=0$, and the limiting pointwise bounds give convergence in the norm.

Here is the explicit contraction estimate on a shrinking holomorphic domain. Write $N=\Vert w{\Vert }_{\alpha }$, $r=|t|$, $A=\alpha (1-s)$, and chooseThen $\sigma -s=(A-\tau )/(2\alpha )$ and $\alpha (1-\sigma )-\tau =(A-\tau )/2$. The whole auxiliary polydisc lies inside ${\mathcal{C}}_{\alpha }$, and its supremum of $|w|$ is at most $2N\tau /(A-\tau )$. ConsequentlyThus $\Vert T\varphi -T\psi {\Vert }_{\alpha }\le (4\alpha /R)\Vert \varphi -\psi {\Vert }_{\alpha }$. Choose $0<\alpha <\min (\eta ,R/4)$. The source is bounded on the closed polydisc, say by $M_f$, and $\Vert T0{\Vert }_{\alpha }\le \alpha M_f$. Therefore $T$ maps the Banach space into itself and is a strict contraction mapping. The Banach fixed-point theorem gives a holomorphic fixed point with $\varphi (x,0)=0$. Differentiating the integral identity yields ${\varphi }_t-i{\varphi }_x=f$, establishing this case of the Cauchy-Kovalevskaya theorem.

Finally, near each real initial point, a real analytic function $g$ has a holomorphic extension $G$. Apply the same construction to $\varphi =G(x)+v$ with $v(x,0)=0$ and source $i{G}^{\prime }(x)$, restricting the discs if necessary. This gives a local real analytic function of $(x,t)$ with the prescribed initial value. Equivalently the solution is $\boxed{\varphi (x,t)=G(x+it)}$ wherever the extension is defined.

Fix $R>|z|$. The Cauchy integral formula and the uniformly convergent geometric series on $|\zeta |=R$ giveDefiningthe remainder after degree $N$ has modulus at most $M_R(|z|/R{)}^{N+1}/(1-|z|/R)$, which tends to zero uniformly on each smaller disk. Differentiating the integral formula at zero identifies $c_j=f^{(j)}(0)/j!$, independently of $R$. Since $f$ is an entire function, $R$ can be chosen larger than any given $|z|$, proving the global Taylor series.

For the requested radius $r$, the same coefficient integral gives the Cauchy estimateTo characterize equality in the Cauchy coefficient estimate, parametrize its circle:If $M_r=0$, the identity theorem gives $f=0$. Otherwise equality in the modulus bound requires the continuous integrand to have both constant modulus $M_r$ and constant complex argument. To see this directly, rotate its average to the positive real axis: its real part is bounded above by $M_r$, and equality of its average with $M_r$ forces that real part to equal $M_r$ everywhere. Thus $f(r{e}^{i\theta })=c_n{r}^n{e}^{in\theta }$ on the circle. The identity theorem extends that equality to the plane. Conversely every such monomial attains the bound. Thereforeincluding the zero function and, when $n=0$, arbitrary constant functions.

The coordinate change has inverse $x=\tilde{x}-{\tilde{y}}^2+\epsilon$, $y=\tilde{y}$. Thus the coefficient function needed after changing variables isThe PDF's displayed formula has an undefined subscript $f_{\alpha }$ and writes the forward change in the argument. It is not the coefficient pullback by the stated coordinates; the inverse expression above is the one used in transforming the equation. The forward polynomial composition, if intended instead, has the same type of uniform bounds, but is not that pullback.

For $|(\tilde{x},\tilde{y})|\le 1/2$ and $0\le \epsilon \le 1/32$, the inverse image lies in the fixed compact subset specified by $|x|\le 25/32$, $|y|\le 1/2$, whose Euclidean radius is less than one. A real analytic function extends holomorphically to a complex neighborhood of this compact set: local extensions agree on overlaps by uniqueness, and a finite covering gives a common smaller neighborhood. The polynomial inverse maps a fixed small complex neighborhood of the closed half-ball into that neighborhood, uniformly in $\epsilon$. The extensions are bounded there by a common constant $A$. Applying the multivariable Cauchy estimate on equal-radius polydiscs givesThis is uniform analyticity under polynomial coordinate changes. The argument gives a compact margin and a common complex radius, not merely separate analyticity for each parameter.

Fix $z_0\in \mathcal{U}$. We prove the Riemann mapping theorem in this case by maximizing a normalized derivative. Choose $a\notin \mathcal{U}$. Since $\mathcal{U}$ is a simply connected domain and $z-a$ never vanishes, it has a holomorphic square root $g$. This $g$ is injective, and $g(\mathcal{U})$ is disjoint from $-g(\mathcal{U})$: equality up to sign would first force the original points to coincide. Choose $b\in g(\mathcal{U})$ and $r>0$ with $B(b,r)\subset g(\mathcal{U})$. The ball $B(-b,r)$ is omitted, so $1/(g+b)$ is a bounded univalent function. Scaling and composing with an automorphism of the unit disk produces an injective holomorphic function $\varphi :\mathcal{U}\to \mathbb{D}$ with $\varphi (z_0)=0$ and ${\varphi }^{\prime }(z_0)>0$.

Let $\mathcal{F}$ be the family of all such normalized univalent functions. A Cauchy estimate in a small disk about $z_0$ bounds their derivatives, so $M={\sup }_{\varphi \in \mathcal{F}}{\varphi }^{\prime }(z_0)$ is finite and positive. Choose a maximizing sequence in this normal family. Montel theorem gives a subsequence converging uniformly on compact subsets to $\varphi$. Its derivative at $z_0$ is $M>0$. The maximum modulus principle puts its image in $\mathbb{D}$, and Hurwitz's theorem implies that a nonconstant limit of injective holomorphic functions is injective. Thus the maximum is attained.

If $a_1\in \mathbb{D}$ is omitted, then $a_1\ne 0$. Put $T_a(w)=(w-a)/(1-\overline{a}w)$. A holomorphic square root $v$ of $T_{a_1}\circ \varphi$ exists, is injective, and takes values in $\mathbb{D}$. Compose $v$ with $T_{v(z_0)}$ and a rotation to normalize it. Writing $s=|a_1|\in (0,1)$, the new derivative has magnitudea contradiction. Therefore $\varphi$ maps onto $\mathbb{D}$. The Cayley transform between the half-plane and disk now givesIts inverse is holomorphic by the holomorphic inverse function theorem, so this is a biholomorphism. The proper-subset hypothesis was used to choose $a\notin \mathcal{U}$; the whole complex plane cannot be mapped this way, by Liouville theorem.

The Cauchy integral formula and its differentiated form areIf $f$ is entire and $|f|\le M$, the Cauchy estimate gives $|f^{\prime }(a)|\le M/\rho$. Letting $\rho \to \infty$ yields $f^{\prime }(a)=0$ for every $a$. Thus $f$ is constant, which is Liouville theorem.

Write the Taylor series of the entire function asThe Cauchy estimate on the circle $|z|=R$ givesIf $k>n/2$, letting $R\to \infty$ gives $c_k=0$. Since $n$ is odd,

For the second question, suppose such an $f$ existed. It never vanishes, so $g=1/f$ is analytic on $\mathbb{C}\setminus \{0\}$ andThe Riemann removable singularity theorem extends $g$ analytically across zero with $g(0)=0$. Applying the result just proved with $n=1$, $a=1$, and $b=0$ makes $g$ a polynomial of degree at most zero. It must then be identically zero, contradicting $g=1/f$ away from zero. Therefore no such function exists.

Fix $a\notin [\gamma ]$ and letFor $|z-a|<d$ and $\lambda \in [\gamma ]$, the geometric series givesThe series converges uniformly on every smaller closed disc, so it may be integrated term by term along the curve. Hencewhich is a power series about $a$.

If $f$ is holomorphic on a neighbourhood of the closed disc $\overline{D}(a,r)$, the Cauchy integral formula saysDifferentiation under the integral is valid uniformly on smaller discs and gives the Cauchy derivative formulaThis proves inductively that every holomorphic function has complex derivatives of every order. In particular,

Now suppose $f_n\to f$ locally uniformly on $U$. Given a compact set $K\subset U$, choose finitely many closed discs whose slightly larger concentric discs remain in $U$ and whose interiors cover $K$. Applying the derivative formula to $f_n-f$ on the larger boundary circles gives a uniform Cauchy estimatewhere $L\subset U$ is the compact union of those circles. The right-hand side tends to zero, proving that

Finally, choose open neighbourhoods $U_j$ of the closed discs $D_j$ so small thatlies in the given neighbourhood on which $f$ is holomorphic. Inside that neighbourhood choose a positively oriented piecewise smooth contour $\Gamma$ surrounding $D_1\cap D_2$. It may be chosen as the boundary of a slightly enlarged lens and split into arcsso that ${\Gamma }_1$ stays away from $D_1$ and ${\Gamma }_2$ stays away from $D_2$. DefineandBecause each arc avoids the corresponding disc, these formulas define holomorphic functions on possibly smaller neighbourhoods $U_1,U_2$. On their overlap, the Cauchy integral formula for the full contour gives

By the spectral radius formula, $r(T)<1$. ChooseThe formula gives a constant $C$ such that $\Vert T^n\Vert \le C{r}^n$ for every $n$. Since $f$ is holomorphic on the unit disc, the Cauchy estimate on the circle of radius $s$ gives $|a_n|\le M_s{s}^{-n}$. ThereforeThus ${\sum }_{n\ge 0}{a}_n{T}^n$ converges absolutely in operator norm.

Suppose a family of polynomial coordinate maps has a fixed bound on its degrees and coefficients depending continuously on a compact parameter set, and sends a fixed real compact set into a compact subset of the domain of a real analytic function. A common complex neighborhood can then be chosen so the compositions are holomorphic and uniformly bounded there. The multivariable Cauchy estimate gives uniform factorial derivative bounds. For the change $\tilde{x}=x+y^2-\epsilon$, the coefficient pullback uses its inverse: $f_{\epsilon }(\tilde{x},\tilde{y})=f(\tilde{x}-{\tilde{y}}^2+\epsilon ,\tilde{y})$.