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Multiplication by $x$ transforms to $id/d\lambda$. Write $p_n(x)={\sum }_{j=0}^n{a}_j{x}^j$ and use $\hat{g}=cg$ for the Gaussian $g=e^{-x^2/2}$. ThenSuccessive differentiation of the Gaussian gives a polynomial of degree $j$ with leading term $(-1{)}^j{\lambda }^j$ multiplying that same Gaussian. Thus $q_n$ has leading coefficient $c{a}_n(-i{)}^n\ne 0$. This proves $q_n$ is a polynomial of degree exactly $n$, rather than merely at most $n$.

Set $f=p(x)e^{-x^2/2}$. Differentiating givesThus the polynomial must satisfy $p^{\prime \prime }-2x{p}^{\prime }-(\mu +1)p=0$. If it has degree $n$ and nonzero leading coefficient, comparison of highest powers gives $\mu =-(2n+1)$. For this value the equation becomes the Hermite differential equationWrite $p={\sum }_{j=0}^n{a}_j{x}^j$, with coefficients beyond $n$ zero. Comparing powers yieldsThe equation at $j=n-1$ forces $a_{n-1}=0$, and downward recurrence removes all terms of the opposite parity and determines all same-parity terms from $a_n$. Taking $a_n=1$ produces the explicit polynomialIt satisfies every coefficient equation, so existence as well as uniqueness up to a nonzero scalar is proved. HenceThe monic $p_n$ is $2^{-n}$ times the usual physicists' Hermite polynomial.

Use $\hat{f}(\lambda )=(2\pi {)}^{-1/2}{\int }_{\mathbb{R}}{e}^{-i\lambda x}f(x)dx$. The transform statement requires sufficient decay and regularity for the transform and its derivatives; functions in the Schwartz space suffice, and all the functions constructed below belong to that space. Integration by parts twice and differentiation under the integral giveTransforming the differential equation therefore givesThe differential equation by itself does not ensure that an ordinary Fourier transform exists: $f=e^{x^2/2}$ solves it with $\mu =1$ but grows too rapidly. Thus the usual transformability assumptions are implicit in this part.

Choose a fixed reference basis and let $V,W$ have columns $v_i,w_i$. Both matrices are invertible. The vectors $v_i+\lambda w_i$ are dependent exactly when $\det (V+\lambda W)=0$. This is a polynomial in $\lambda$ of degree $m$, whose leading coefficient is $\det W\ne 0$. A nonzero degree-$m$ polynomial has at most $m$ distinct real roots. Thus there are at most $m$ exceptional parameters. This is the determinant argument for the matrix pencil $V+\lambda W$.

Assume for contradiction that $n>m$. Apply part b to the matrix of the given relation coefficients, whose columns are nonzero. For its first dependent prefix choose $b_1,\dots ,b_k$ withMultiply the $j$th given vector relation by $b_j$ and sum. Interchanging the finite sums givesThe $w_i$ are a basis, so all $z_i$ must be zero, contradicting part b. Therefore $n\le m$.

Choose any nonzero vector $b$ in the one-dimensional kernel of $A_k$. It gives ${\sum }_{j=1}^k{a}_{ij}{b}_j=0$ for every row. Because no column is zero, $b$ must have at least two nonzero entries: a relation supported at a single index would make that column zero.

Suppose, contrary to the claim, that $A_k\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_k)b=0$. Its kernel is spanned by $b$, so there is a scalar $c$ with ${\lambda }_j{b}_j=c{b}_j$ for every $j$. At the two nonzero entries this gives two equal weights ${\lambda }_j=c$, contradicting their distinctness. Thus the weighted sum is nonzero in at least one row. This proves that distinct diagonal weights break a first column dependence. Some entries of $b$ may be zero; only two nonzero entries are needed.

By the rank-nullity theorem, $\mathrm{nullity}A=n-\mathrm{rank}A\ge n-m>0$. Thus a first dependent column prefix exists. Minimality of $k$ makes the first $k-1$ columns independent, so $\mathrm{rank}{A}_{k-1}=k-1$. Adding one column cannot reduce this rank, while nonzero nullity of $A_k$ implies rank at most $k-1$. HenceFor $k=1$ the earlier prefix is the zero-dimensional empty matrix, and the same argument applies.

The error $e(f)=f^{\prime \prime }(0)-f(-1)+2f(0)-f(1)$ is linear and annihilates all degree-at-most-two polynomials. The Peano kernel theorem with derivative order three givesFor $t\ne 0$, the second derivative of the truncated quadratic at $x=0$ is $1_{\{t<0\}}$. Its values at $x=-1,0,1$ then giveThe value at the single point $t=0$ is irrelevant to the integral. This is the centered second-derivative Peano kernel. Its absolute integral isConsequentlyIts changing sign explains why cubic polynomials also have zero error: their third derivative is constant and ${\int }_{-1}^{1}K=0$. No fourth derivative is needed for the claimed bound.

Exactness on the polynomial basis $1,x,x^2$ requiresSolving givesThus the approximation is the unit-spacing centered second difference, $\mu (f)=f(-1)-2f(0)+f(1)$.

The set $Y$ is path-connected as the continuous image of $[1,\infty )$, hence connected. Its closure is $Y$ together with the unit circle. Indeed $g(k+\theta /(2\pi ))\to e^{i\theta }$ as $k\to \infty$, and any convergent sequence of spiral points either has bounded parameters, giving a point of $Y$, or has parameters tending to infinity along a subsequence, giving radius one. The closure of a connected set is connected: a separation of its closure would induce a separation of the dense original set, with neither open part missing that set. The closed disk is connected and meets this closure in the unit circle; their union $Y\cup \overline{D}$ is therefore connected.

To show it is not path-connected, suppose a path $\alpha :[0,1]\to Y\cup \overline{D}$ starts in $Y$ and ends in the disk. Let $u_{\ast }$ be its first contact with the disk. The disk is closed, so this first contact exists, and $u_{\ast }>0$. Before it, the path belongs to $Y$ and has a continuous uniquely determined parameterContinuity at contact forces $|\alpha (u_{\ast })|=1$, and hence $t(u)\to \infty$ as $u\uparrow u_{\ast }$. The intermediate value theorem supplies times approaching $u_{\ast }$ at which $t(u)$ is an integer, and other times at which it is an integer plus $1/2$. These times must approach contact because $t$ is bounded on every compact interval strictly before it. Along the first sequence $\alpha (u)\to 1$; along the second it tends to $-1$. This contradicts continuity at $u_{\ast }$. Thus no path joins the spiral to the disk, proving failure of path-connectedness for the spiral accumulating on a circle.

Define $F:[0,1]\to \mathbb{C}$ by $F(s)=s{e}^{2\pi i/s}$ for $s>0$ and $F(0)=0$. It is continuous away from zero, and $|F(s)-F(0)|=s\to 0$, so it is continuous at zero too. Its image is exactly $X\cup \{0\}$, using $s=1/t$. The interval is compact, so its continuous image is compact.

For any two image points $F(s_0),F(s_1)$, the explicit pathis continuous and joins them. Thus the reciprocal spiral with its endpoint is also path-connected. Infinite winding causes no discontinuity at the endpoint because its radius tends to zero.

Assume every continuous function $E\to \mathbb{R}$ is bounded. The Euclidean norm function $x\mapsto \Vert x\Vert$ is continuous, so its boundedness proves that $E$ is a bounded set.

If $E$ were not a closed set, there would be $a\in \overline{E}\setminus E$. Choose $x_j\in E$ with $\Vert x_j-a\Vert <1/j$. The function $f(x)=1/\Vert x-a\Vert$ is continuous everywhere on $E$, because its denominator is nonzero there. Yet $f(x_j)>j$, contradicting the hypothesis. Thus $E$ is closed. Together with the two previous implications this proves all three conditions are equivalent. The empty set satisfies all three conditions trivially. This is the Euclidean case of the bounded-continuous-function characterization of compact metric spaces.

Assume the subsequence property, and let $f:E\to \mathbb{R}$ be a continuous function. If $f$ were unbounded, for each positive integer $j$ we could choose $x_j\in E$ with $|f(x_j)|>j$. There is a subsequence $x_{j_k}\to x$ with $x\in E$. By continuity, $f(x_{j_k})\to f(x)$, so that real sequence is bounded. But $|f(x_{j_k})|>j_k\to \infty$, a contradiction. Hence every continuous real-valued function on $E$ is bounded. This is the bounded-function consequence of sequential compactness.

We prove the equivalence by a cycle of implications. First assume closedness and boundedness. Given a sequence $(x_j)$ in $E$, boundedness puts each coordinate in a fixed closed bounded real interval. Apply the one-dimensional Bolzano-Weierstrass theorem to its first coordinate to extract a convergent subsequence, then apply it to the second coordinate of that subsequence, and continue through the finitely many coordinates. The final subsequence converges coordinatewise to some $x\in {\mathbb{R}}^n$. It also converges in the Euclidean norm, since the sum of the squares of its finitely many coordinate differences tends to zero. Since $E$ is a closed set, this limit belongs to $E$. Thus closedness and boundedness imply the required subsequence property. No higher-dimensional compactness theorem has been assumed.

Inversion permutes the nonzero residues excluding $-1$. Hence the result of part (c) changes the requested sum intoThere are exactly $(p-1)/2$ nonzero squares: the map $x\mapsto x^2$ on nonzero residues identifies exactly the pairs $x,-x$, because $x^2=y^2$ implies $x=\pm y$ in the finite field. Thus there are equally many nonzero squares and nonsquares, and the complete Legendre-symbol sum is zero. Removing the $j=1$ term leaves

Since $n$ is nonzero modulo $p$, it has a unique multiplicative inverse $m$ in the nonzero residues. The inverse cannot be $-1$ because that would imply $n=-1$, excluded by the range. Thus $1\le m\le p-2$ as claimed. The congruence $mn=1$ givesThe Legendre symbol of the nonzero square $n^2$ is one. Using multiplicativity of the Legendre symbol therefore gives

Apply Euler's criterion to $-1$. It givesFor $p\nmid ab$, the same criterion gives $(ab/p)\equiv (ab{)}^{(p-1)/2}=a^{(p-1)/2}{b}^{(p-1)/2}\equiv (a/p)(b/p)(\mathrm{mod}p)$. Both sides are integers in $\{1,-1\}$, which are distinct modulo an odd prime, so congruence implies equality. This proves multiplicativity of the Legendre symbol:

For a nonzero residue $a$ modulo an odd prime $p$, the Legendre symbol $(a/p)$ equals $+1$ if $a$ is a nonzero square modulo $p$ and $-1$ otherwise. The extension to multiples of $p$ has value zero. Euler's criterion states

Use the standard basis $e_1,\dots ,e_n$. For the smallest possible rank, set $\alpha e_j=e_j$ for $j\le r$ and zero otherwise, and set $\beta e_j=-e_j$ for $j\le s$ and zero otherwise. Then the maps have ranks $r,s$, while their sum has nonzero diagonal entries only for $r<j\le s$. HenceFor full rank when $r+s\ge n$, instead take $\alpha e_j=e_j$ for $j\le r$ and zero otherwise, and $\beta e_j=e_j$ for $j\ge n-s+1$ and zero otherwise. Their supports cover every basis index because $n-s+1\le r+1$. The sum has diagonal entries one where just one map is nonzero and two where both are nonzero. None vanish over $\mathbb{R}$, so these linear maps have ranks $r,s$ and

Apply the preceding rank bounds for a sum of linear maps to $\alpha =(\alpha +\beta )+(-\beta )$. Since multiplying a linear map by $-1$ does not change its image, $\mathrm{rank}\alpha \le \mathrm{rank}(\alpha +\beta )+\mathrm{rank}\beta$. Thus $\mathrm{rank}(\alpha +\beta )\ge \mathrm{rank}\alpha -\mathrm{rank}\beta$. Interchanging the maps gives the other difference, proving