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Use the convention that the inner product is linear in its first argument. If $v\ne 0$, nonnegativity of $\Vert u-⟨u,v⟩v/\Vert v{\Vert }^2{\Vert }^2$ givesThe case $v=0$ is immediate. This Cauchy-Schwarz inequality shows that $f_v$ is bounded linear with norm at most $\Vert v\Vert$. For $v\ne 0$, evaluation at $v/\Vert v\Vert$ gives equality; for $v=0$ both norms vanish. The Banach dual is the space of all bounded linear functionals with the operator norm.

The Riesz representation theorem asserts that every $f\in V^{\prime }$ is uniquely $f(u)=⟨u,v⟩$ for some $v\in V$, with $\Vert f\Vert =\Vert v\Vert$. If $f=0$, take $v=0$. Otherwise put $W=\ker f$, a closed subspace, and use the projection theorem to obtain $w\in W^{\perp }$ with $f(w)=1$. Since $u-f(u)w\in W$, $⟨u,w⟩=f(u)\Vert w{\Vert }^2$. Thus $v=w/\Vert w{\Vert }^2$ represents $f$. If two vectors represent it, their difference has inner product zero with every vector, in particular itself, giving uniqueness. The norm statement was already proved.

The space $P_n$ is finite-dimensional and hence complete; evaluation at zero is bounded there by equivalence of finite-dimensional norms. Riesz therefore supplies exactly one $q_n\in P_n$. In any orthonormal polynomial basis,For the real Legendre functions the bars may be omitted.

The printed even-degree normalization has an error. Orthogonality forces a degree-two polynomial to be a scalar multiple of $3t^2-1$, whose squared integral on $[-1,1]$ is $8/5$. Hence any orthonormal choice has $|p_2(0){|}^2=5/8$, whereas the printed formula gives $3/8$. The factor under its square root must be $4m+1$ instead of $2m+1$. Signs of the individual polynomials can be chosen so that the values at zero are positive; these signs do not affect $q_n$ or its norm.

Put $M=\lfloor n/2\rfloor$ and $c_m=\left(\genfrac{}{}{0ex}{}{2m}{m}\right)/4^m$. With the consistent orthonormal normalization the norm isThe finite sum telescopes because $c_m=(2m-1)c_{m-1}/(2m)$, so $(2m+1{)}^2{c}_m^2-(2m-1{)}^2{c}_{m-1}^2=(4m+1)c_m^2$. If one substitutes the erroneous printed values formally instead, one obtains $\frac{1}{2}{\sum }_{m=0}^M(2m+1)c_m^2$; that is not the norm of the actual orthonormal Legendre evaluation representer.

For a polynomial $r$ of degree $d$, the defining identity gives $⟨r,q_n⟩=r(0)$ for every $n\ge d$, so the requested limit is eventually constant. It does not extend to all of $L^2$. First, evaluation at a point is not well-defined on equivalence classes. More strongly, the functional norms $\Vert q_n\Vert$ are unbounded. Indeed the largest binomial coefficient in row $2m$ is at least the average, so $c_m\ge 1/(2m+1)$, and the displayed sum dominates a divergent constant multiple of the harmonic series. The Uniform boundedness principle then provides an $L^2$ function for which ${\sup }_n|⟨r,q_n⟩|=\infty$. Thus no choice of point representative can restore the assertion for all $L^2$ functions.

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.

A set is totally bounded if for every $\epsilon >0$ it can be covered by finitely many open balls of radius $\epsilon$.

Here is the needed separated-sequence argument, including the relevant Riesz lemma. For a proper closed subspace $Y$, choose $z\notin Y$, put $d=\mathrm{dist}(z,Y)>0$, and choose $y\in Y$ with $\Vert z-y\Vert <2d$. Then $w=(z-y)/\Vert z-y\Vert$ is a unit vector at distance greater than $1/2$ from $Y$. Starting with one unit vector, apply this to the span of the vectors already chosen. These spans are closed and proper because they are finite-dimensional while $X$ is infinite-dimensional. The resulting unit vectors satisfy $\Vert w_m-w_n\Vert >1/2$ for $m\ne n$. A ball of radius $1/4$ contains at most one of them. Hence the closed unit ball is not totally bounded.

Yes. A Cauchy sequence of continuous functions in the supremum norm is Cauchy in the given complete space of bounded functions, so it has a uniform limit there. A uniform limit of continuous functions is continuous: at a point, use two uniform errors smaller than $\epsilon /3$ and continuity of one sufficiently late term for the remaining error. Thus the limit is in $C([0,1])$, proving completeness. Equivalently this is a closed subspace of the given complete space.

If a topological space is the union of finitely many closed subspaces and continuous maps on those subspaces agree on every overlap, they combine into a continuous map on the whole space. For a closed set in the target, its preimage is a finite union of closed sets, proving continuity.