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Use the source's notation for the vanishing-at-zero ideal $I=\{f\in C[0,1]:f(0)=0\}$. It is the kernel of bounded evaluation at zero, hence closed. It is stable under pointwise multiplication and complex conjugation, so inherits the C-star identity and completeness. If it had an identity $e$, multiplication against $f(t)=t$ would give $e(t)=1$ for $t>0$, contradicting continuity and $e(0)=0$. Thus $I$ is a non-unital C-star subalgebra. This explicit meaning of the notation matters because the usual notation $C_0$ for a compact space has a different convention.

Define the unitization map by $\Phi (f,\lambda )=f+\lambda 1$. Its inverse is $h\mapsto (h-h(0),h(0))$. The specified multiplication and involution are preserved by direct substitution. The pullback norm $\Vert (f,\lambda )\Vert =\Vert f+\lambda 1{\Vert }_{\infty }$ is a C-star norm, so equals the unique unitization norm supplied in the question. Therefore $\Phi$ is a unital isometric C-star isomorphism onto $C[0,1]$.

For the approximation put $h(s)=a(\sqrt{s})$ on $[0,1]$, so $h$ is continuous and $h(0)=0$. The Stone-Weierstrass theorem supplies a polynomial $p(s)$ with $\Vert h-p{\Vert }_{\infty }<\epsilon /2$. Then $b(t)=p(t^2)-p(0)$ is a polynomial in the positive even powers of $t$, with zero constant term, andThis gives exactly the required finite form.

A C-star algebra is a complex Banach algebra with a conjugate-linear involution satisfying $(ab{)}^{\ast }=b^{\ast }{a}^{\ast }$, $(a^{\ast }{)}^{\ast }=a$ and $\Vert a^{\ast }a\Vert =\Vert a{\Vert }^2$. It is unital when it has an identity for multiplication. For $B(H)$ use composition and the Hilbert adjoint. Completeness in the operator norm is given; composition is submultiplicative and the adjoint has the required algebraic properties. For a unit vector $\xi$,Taking the supremum and using the opposite bound $\Vert T^{\ast }T\Vert \le \Vert T^{\ast }\Vert \Vert T\Vert =\Vert T{\Vert }^2$ proves the C-star identity. Its identity is $I_H$, proving $B(H)$ is a unital C-star algebra.

Compact operators form a norm-closed linear space and are stable under composition with bounded operators. They are also adjoint stable. To see this directly, choose a finite $\epsilon$-net of the compact image of the unit ball under $T$, and let $P$ project orthogonally onto the span of its centers. Then $\Vert (I-P)T\Vert \le \epsilon$. The finite-rank operator $T^{\ast }P$ approximates $T^{\ast }$ with the same error, so $T^{\ast }$ is compact by norm closure. Thus $\mathcal{K}(H)$ is a C-star algebra, with the inherited norm and operations.

If it has a multiplicative identity $E$, apply $ER=R$ to every rank-one operator with arbitrary prescribed range vector $\xi$. Evaluating that equation on a vector sent to $\xi$ gives $E\xi =\xi$. Hence $E=I_H$. This identity is compact exactly when $H$ is finite dimensional: finite dimension gives compactness of the unit ball, while in infinite dimension an infinite orthonormal sequence has pairwise distance $\sqrt{2}$ and no convergent subsequence. Therefore $\mathcal{K}(H)$ is unital exactly in finite dimension, with the zero-algebra convention understood if $H=0$.

A linear operator has closed graph when $\{(x,Tx):x\in X\}$ is closed in $X\times Y$. The closed graph theorem states that an everywhere-defined linear operator between Banach spaces is bounded if and only if it has closed graph. Boundedness gives the closed-graph property by continuity. Conversely a closed graph is Banach in the product norm $\Vert (x,y)\Vert =\Vert x\Vert +\Vert y\Vert$. Projection of the graph onto $X$ is a bounded linear bijection, so the open mapping theorem makes its inverse bounded. Composing that inverse with the second-coordinate projection proves $T$ is bounded.

For the composition assertion, suppose $x_n\to x$ and $T{x}_n\to y$. Boundedness of $S$ gives $ST{x}_n\to Sy$, and boundedness of $ST$ gives $ST{x}_n\to STx$. Injectivity of $S$ forces $y=Tx$. Hence $T$ has closed graph and is bounded by the theorem.

For the derivative bound take $Y=C^1[0,1]$ with norm $\Vert v{\Vert }_{C^1}=\Vert v{\Vert }_{\infty }+\Vert v^{\prime }{\Vert }_{\infty }$, and $Z=C[0,1]$ with its supremum norm; both are Banach. Regard $T(x)=Ux$ as a map into $Y$, and let $S:Y\to Z$ be inclusion. This inclusion is bounded and injective, and $ST=U$ is bounded by hypothesis. The preceding result gives a constant $c$ with $\Vert Ux{\Vert }_{C^1}\le c\Vert x\Vert$. In particular $|(Ux{)}^{\prime }(t)|\le c\Vert x\Vert$ uniformly in $t$.

Regard the spaces as real and set $u=\mathrm{Re}\varphi$. Its real norm equals $\Vert \varphi \Vert$, since rotating any test vector makes its functional value real and nonnegative without changing its norm. By the allowed real Hahn-Banach theorem extend $u$ to a real-linear $v:X\to \mathbb{R}$ with that norm, and define $\psi (x)=v(x)-iv(ix)$. Real linearity gives $\psi (ix)=i\psi (x)$, so $\psi$ is complex linear. On $Y$ this formula recovers $\varphi$. Also $\mathrm{Re}\psi =v$; rotating a test vector again gives $|\psi (x)|\le \Vert v\Vert \Vert x\Vert$. The reverse norm inequality follows by restriction to $Y$, proving a norm-preserving complex extension.

We use two standard consequences: a point outside a closed subspace can be separated by a bounded functional annihilating that subspace, and the norm of a vector is the supremum of its pairings with dual unit-ball functionals. If $\psi \in X^{\ast }$ annihilates $Y$, the scalar holomorphic function $\psi \circ f$ vanishes at the prescribed sequence tending to zero. Its identity theorem makes it identically zero. Separation then gives $f(z)\in Y$ for every $z\in D$.

For $\varphi \in Y^{\ast }$ extend to $X^{\ast }$ with the same norm and put $F=\varphi \circ f$. Then $F$ is holomorphic with $|F|\le \Vert \varphi \Vert$. For the understood nonzero increments $h,k$, the Cauchy integral formula on $|\zeta |=r$ givesTaking absolute values, bounding the contour length by $2\pi r$, and each denominator below by $r-|z|$, $r-|z|-|h|$, $r-|z|-|k|$ proves exactly the stated estimate.

Fix $z,r$ and restrict $|h|,|k|<(r-|z|)/2$. Taking the supremum over $\Vert \varphi \Vert \le 1$ bounds the norm of the difference of the two $Y$-valued difference quotients by a fixed constant times $|h-k|$. They are therefore Cauchy as $h\to 0$. Since $Y$ is a closed subspace of a Banach space, it is complete. Consequently the derivative exists as a norm limit in $Y$. The labelled (a)-(c) are the simultaneous hypotheses, not separate claims to prove.

Consider the nonresonant traveling forced solution, with $k>0$ and $c\ne 0$. Put $x=\bar{x}+\xi$, $y=\bar{y}+\upsilon$, where $(\bar{x},\bar{y})$ is the mean position of a fluid element. At linear order, evaluate the velocity field at this mean position: substituting $\xi ,\upsilon$ into its spatial arguments would change the already first-order velocity only at second order. With $\Theta =k(\bar{x}-ct)$, differentiation of the velocity potential givesSince $\dot{\Theta }=-kc$, integration, with zero mean displacements, yieldsIf both coefficients are nonzero, eliminating the wave phase proves that the pathline is an ellipse:Its principal axes of an ellipse are horizontal and vertical, with respective semi-axesThus the full axis lengths are $2|A|$ and $2|B|$. The signed coefficients also determine the direction of traversal. The useful checks $B(0)=b$ and $B(-d)=a$ reproduce the vertical displacements of the free surface and bottom.

To identify the major axis without comparing the two expressions separately, use ${\cosh }^{2}s-{\sinh }^{2}s=1$:Consequently the major axis is horizontal if $c^2<g/k$ and vertical if $c^2>g/k$; equality gives circles. If $A$ or $B$ vanishes at a particular depth, the first-order orbit degenerates to a line segment. These are linearized particle orbits of a bottom-forced gravity wave, not exact closed trajectories at all orders: a higher-order calculation may introduce a mean drift.