The only possible finite singularities are $z_{\pm }=\pm i\pi /2$, the zeros of $4z^2+{\pi }^2$. They are simple zeros of the denominator, because its derivative $8z_{\pm }$ is nonzero. They are also simple zeros of the numerator: $\cosh z_{\pm }=0$ and $\sinh z_{\pm }=\pm i\ne 0$.

The common simple factors cancel, so both points are removable singularities, with extension valuesThere are no other finite singularities; after inserting these two values the function is entire.

The expression is $F(z)=e^{z^2+1/z}=e^{z^2}{e}^{1/z}$, which is holomorphic for every $z\ne 0$. At zero,has infinitely many nonzero negative-power terms, so it has an essential singularity. Multiplication by the nonvanishing entire function $e^{z^2}$ preserves that classification: if $F$ had a removable singularity or a pole, multiplying by $e^{-z^2}$ would give the same type for $e^{1/z}$, a contradiction. Hence

Let $F(z)=\overline{\sinh z}$. More generally, for a holomorphic function $h$, the increment expansion at $z_0$ givesAlong real increments $w$ the first factor $\overline{w}/w$ is $1$, while along purely imaginary increments it is $-1$. Thus complex differentiability of $\overline{h}$ requires $h^{\prime }(z_0)=0$. Conversely, if $h^{\prime }(z_0)=0$, the displayed expression tends to zero for arbitrary complex increments, so the complex derivative exists and equals zero.

For $h=\sinh$, this condition is $\cosh z_0=0$, equivalent to $e^{2z_0}=-1$. ConsequentlyThese points form a discrete subset of the complex plane and contain no nonempty open set. Every neighbourhood of each one contains points where $F$ is not complex differentiable. Therefore $F$ is holomorphic on no nonempty open set, even though it has complex derivatives at the specified points.

The denominator vanishes exactly when $e^z=-1$, namely at $z=(2k+1)\pi i$, $k\in \mathbb{Z}$. Its derivative there is $e^z=-1\ne 0$, so each such zero is simple and the numerator does not cancel it. Henceand every omitted point is a simple pole.

The zeros of the hyperbolic sine are exactly $z=k\pi i$, $k\in \mathbb{Z}$, since $\sinh z=0$ is equivalent to $e^{2z}=1$. Each zero is simple because $\cosh (k\pi i)=(-1{)}^k\ne 0$. The quotient is therefore holomorphic precisely onAt zero, the Taylor expansion $\sinh z=z+O(z^3)$ gives $z/{\sinh }^{2}z=1/z+O(z)$, so zero is a simple pole. At $k\pi i\ne 0$, the denominator has a double zero and the numerator is nonzero, giving a pole of order two. Thus none of the omitted points admits a holomorphic extension.

The hyperbolic sine satisfies $\sinh z=(e^z-e^{-z})/2$. The complex exponential function is entire, and sums and compositions of entire functions are entire. Therefore

Apply the local bound from condition (1) separately to $f$ and $-f$. For any $a\in X$ and $\epsilon >0$, there are $r_{+},r_{-}>0$ such thatWith $\delta =\min \{r_{+},r_{-}\}$, both inequalities hold when $d(x,a)<\delta$, soThis is exactly continuity at $a$. Since $a$ was arbitrary, $f$ is continuous on $X$: upper semicontinuity of $f$ and of $-f$ is equivalent to simultaneous upper semicontinuity and lower semicontinuity of $f$.

If $f$ is continuous at $a$, then for every $\epsilon >0$ there is $r_0>0$ such that $d(x,a)<r_0$ implies $|f(x)-f(a)|<\epsilon$. In particular, for $0<r\le r_0$,The squeeze theorem giveswhich implies condition (1). Thus continuity implies upper semicontinuity.

We prove uniform continuity by contradiction. If it fails, there is ${\epsilon }_0>0$ such that, for every $n\ge 1$, we can choose $x_n,y_n\in X$ withBy sequential compactness, pass to a subsequence with $x_{n_k}\to x\in X$. The triangle inequality givesBy continuity, both $f(x_{n_k})$ and $f(y_{n_k})$ converge to $f(x)$. A second triangle inequality therefore givescontradicting the lower bound. Thus the requested $\delta$, independent of $x,y$, exists for every $\epsilon >0$.

Both properties are preserved by a continuous map.

For sequential compactness, take any sequence $(y_n)$ in $f(X)$ and choose $x_n\in X$ with $f(x_n)=y_n$. If $X$ is sequentially compact, some subsequence satisfies $x_{n_k}\to x\in X$. By continuity, $y_{n_k}=f(x_{n_k})\to f(x)\in f(X)$. Hence $f(X)$ is a sequentially compact space.

For connectedness, suppose instead that $f(X)=U\cup V$ is a separation into disjoint nonempty relatively open sets. Then $f^{-1}(U)$ and $f^{-1}(V)$ are disjoint nonempty open sets of $X$ whose union is $X$, contradicting connectedness. Therefore $f(X)$ is a connected space.

First verify that $V_0$ is a vector space. Sums and scalar multiples preserve continuity and the condition at zero. For every partition of an interval $D$, the triangle inequality and absolute homogeneity of a norm of the absolute value implyTaking suprema yieldsFor $\lambda =0$ the second equality follows directly from $V(0)=0$. In particular, finite variation is preserved, so $V_0$ is closed under these operations.

Nonnegativity is immediate. For any $t\in [0,1]$, use the partition of an interval $(0,t,1)$, omitting repeated endpoints if necessary, to getThus $V(f)=0$ forces $f=0$, and conversely $f=0$ has zero total variation of a function. These observations prove that $\Vert f\Vert =V(f)$ is a norm on $V_0$.

Applying the same bound to $f_n-f$ givesTherefore convergence in the anchored variation norm implies uniform convergence on $[0,1]$. The anchoring condition $f(0)=0$ removes the nonzero constant functions which otherwise have zero total variation of a function.

A norm on a real vector space $X$ is a map $\Vert \cdot \Vert :X\to [0,\infty )$ such that, for $x,y\in X$ and $\lambda \in \mathbb{R}$,The last property is the triangle inequality; the middle property is absolute homogeneity of a norm. The induced metric isPositivity and separation follow from the norm axioms, symmetry from absolute homogeneity of a norm with $\lambda =-1$, and the triangle inequality from $x-z=(x-y)+(y-z)$.

For $0\le t\le 1$, the triangle inequality givesTaking the supremum proves $\boxed{d(p_n,p)\le \rho (p_n,p)}$. Hence convergence for $\rho$ implies convergence for $d$.

The converse is false. For $n\ge 1$, take $q_n(t)=(1-t{)}^n/n$ and $p=0$. Its supremum norm is $d(q_n,0)=1/n\to 0$, whereas the binomial theorem givesThe alternating coefficients cancel on $[0,1]$; the polynomial coefficient sum norm measures their absolute values before any cancellation.

The space of polynomials is not complete for the supremum norm metric $d$. ConsiderThe geometric series has sum $F(t)=1/(1-t/2)$, and for $0\le t\le 1$ its tail satisfiesThus $(p_n)$ converges uniformly to $F$ and is a Cauchy sequence for $d$.

There is no polynomial $p$ agreeing with $F$ on $[0,1]$: otherwise the polynomial $(1-t/2)p(t)-1$ would vanish at infinitely many points and therefore be the zero polynomial. Evaluating this identity at $t=2$ gives $-1=0$. By uniqueness of the uniform limit, $(p_n)$ has no limit in $X$. This proves the claimed failure of completeness.

Extend the coefficient lists of each polynomial by zeros. Then $\rho (p,q)={\sum }_{k\ge 0}|a_k-b_k|$ is finite because only finitely many terms are nonzero. It is nonnegative and symmetric. Moreover, $\rho (p,q)=0$ holds exactly when every coefficient agrees, which by uniqueness of polynomial coefficients is equivalent to $p=q$.

If $r(t)=\sum c_k{t}^k$, the triangle inequality for the absolute value givesThus $\rho$ satisfies all the metric axioms. It is the metric induced by the polynomial coefficient sum norm.

The arctangent pullback metric is not complete. Take $x_n=n$. Since $\arctan n\to \pi /2$, its transformed sequence is Cauchy, and hence $(x_n)$ is a Cauchy sequence for $\rho$. If it converged in $\rho$ to $x\in \mathbb{R}$, thenwould force $\arctan x=\pi /2$, which is impossible. The missing endpoint of $I$ explains the failure of completeness, even though part (i) shows that the metric topologies agree.

The inverse tangent $T(x)=\arctan x$ is a homeomorphism from the real line with its Euclidean metric onto $I=(-\pi /2,\pi /2)$ with its subspace topology: $T$ and its inverse $\tan$ are continuous. It is also an isometry from $(\mathbb{R},\rho )$ onto $I$, because $\rho (x,y)=|T(x)-T(y)|$. Consequently, for $U\subseteq \mathbb{R}$,For a direct open ball interpretation, $B_{\rho }(x,r)=T^{-1}((T(x)-r,T(x)+r)\cap I)$ is $d$-open by continuity of $T$. Conversely, continuity of $\tan$ at $T(x)$ supplies a sufficiently small $\rho$-ball inside any prescribed $d$-ball about $x$. Thus both metrics generate exactly the same open sets.