= Solution
Use the convention that the <inner product> is linear in its first argument. If $v\ne0$, nonnegativity of $\|u-\langle u,v\rangle v/\|v\|^2\|^2$ gives
$$
\boxed{|\langle u,v\rangle|\leq\|u\|\|v\|.}
$$
The case $v=0$ is immediate. This <Cauchy-Schwarz inequality> shows that $f_v$ is bounded linear with <norm> at most $\|v\|$. For $v\ne0$, evaluation at $v/\|v\|$ 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 $f\in V'$ is uniquely $f(u)=\langle u,v\rangle$ for some $v\in V$, with $\|f\|=\|v\|$. 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 $u-f(u)w\in W$, $\langle u,w\rangle=f(u)\|w\|^2$. Thus $v=w/\|w\|^2$ 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,
$$
\boxed{q_n(t)=\sum_{j=0}^n\overline{p_j(0)}p_j(t),\qquad\|q_n\|^2=\sum_{j=0}^n|p_j(0)|^2.}
$$
For the real Legendre functions the bars may be omitted.
\b[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 $c_m=\binom{2m}{m}/4^m$. With the consistent <orthonormal> normalization the <norm> is
$$
\boxed{\|q_n\|^2=\frac12\sum_{m=0}^M(4m+1)c_m^2=\frac{(2M+1)^2c_M^2}{2},\qquad\|q_n\|=\frac{(2M+1)c_M}{\sqrt2}.}
$$
The finite sum telescopes because $c_m=(2m-1)c_{m-1}/(2m)$, so $(2m+1)^2c_m^2-(2m-1)^2c_{m-1}^2=(4m+1)c_m^2$. If one substitutes the erroneous printed values formally instead, one obtains $\frac12\sum_{m=0}^M(2m+1)c_m^2$; that is not the <norm> of the actual <orthonormal> Legendre evaluation representer.
For a <polynomial> $r$ of degree $d$, the defining identity gives $\langle r,q_n\rangle=r(0)$ for every $n\geq d$, so the requested limit is eventually constant. It does \b[not] extend to all of $L^2$. First, evaluation at a point is not well-defined on equivalence classes. More strongly, the functional norms $\|q_n\|$ are unbounded. Indeed the largest binomial coefficient in row $2m$ is at least the average, so $c_m\geq1/(2m+1)$, 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 $\sup_n|\langle r,q_n\rangle|=\infty$. Thus no choice of point representative can restore the assertion for all $L^2$ functions.
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