Choose a minimizing sequence $(u_n)$ withSincethe sequence is bounded in $H_0^1(\Omega )$. Passing to a subsequence, weak compactness givesin $H_0^1(\Omega )$. Rellich compactness gives $u_n\to w$ strongly in $L^2$, and hencePart (i) now yieldsThe definition of $\lambda$ gives the reverse inequality because $w$ is admissible. ThereforeThis is the constrained ground-state minimizer on a bounded domain obtained by the direct method.

Weak convergence $u_n\rightharpoonup u$ in $H_0^1(\Omega )$ impliesin $L^2$. The $L^2$ norm is weakly lower semicontinuous, so

The Rellich-Kondrashov compactness theorem for H01 upgrades the weak convergence tostrongly in $L^2(\Omega )$. Indeed, compactness gives this along every subsequence after passage to a further subsequence, and the weak limit uniquely identifies every such strong limit as $u$. Consequently,Combining the two terms proves the weak lower semicontinuity of a bounded-domain Schrodinger energy:

No. In $H={\ell }^2$, let $x_n=e_n$ be the standard orthonormal sequence. For any $y=(y_j)\in {\ell }^2$,so $e_n\rightharpoonup 0$. However,for $n\ne m$. No subsequence is Cauchy, hence none converges strongly. This is a weakly null orthonormal sequence.

Strong convergence immediately implies convergence of norms by the reverse triangle inequality.

Conversely, suppose $x_n\rightharpoonup x$ and $\Vert x_n\Vert \to \Vert x\Vert$. ThenThus $x_n\to x$ strongly. This is the Radon-Riesz theorem for Hilbert spaces.

If $x=0$, the claim is immediate. Otherwise weak convergence givesBy Cauchy-Schwarz,Taking the lower limit and dividing by $\Vert x\Vert$ proves the weak lower semicontinuity of the Hilbert norm:

The cube roots of minus one modulo an odd prime are controlled byFor $p\ne 3$, the quadratic factor has roots exactly when its discriminant $-3$ is a square, namely when $p\equiv 1(\mathrm{mod}3)$. In that case there are three distinct roots; when $p\equiv 2(\mathrm{mod}3)$, only $X=-1$ remains. For $p=3$,so $X=-1$ is again the unique root. Therefore

The classsatisfies $X^3=-1$ for every prime. Thus a solution exists for every odd prime $p$, regardless of its residue class modulo $3$.

PutSince $B=U⋊T$ and $U\cap T=1$, restriction to $U$ gives its regular representation:As $U\cong ({\mathbb{F}}_p,+)$ is abelian, this is the direct sum of all $p$ one-dimensional characters of $U$, each with multiplicity one.

The diagonal subgroup $T$ permutes these one-dimensional weight spaces. It fixes the trivial character, while part (ii) shows that the nontrivial characters split into two orbits, indexed by the nonzero squares and nonsquares. Let $W_0,W_{+},W_{-}$ be the sums of the weight spaces in these three orbits. Thenas $B$-representations, and

Each summand is irreducible. Any nonzero $B$-subrepresentation decomposes into $U$-weight spaces. If it contains one of the one-dimensional weight spaces in an orbit, transitivity of the $T$-action forces it to contain every weight space in that orbit, hence the whole corresponding $W_0$, $W_{+}$, or $W_{-}$. The three summands are pairwise nonisomorphic because their restrictions to $U$ have disjoint character supports. Therefore the induction from the diagonal subgroup of upper-triangular SL2 decomposition has the required three irreducible constituents of dimensions

For $a\in {\mathbb{F}}_p^{\ast }$ putConjugation givesSuppose the $\chi$-weight spaceis nonzero. If $0\ne z\in V_{\chi }$, then $t(a)z$ lies in the $\chi (a^{-2}\cdot )$-weight space, becauseHence every character indexed by a nonzero square occurs in ${\mathrm{Res}}_U^{B}V$.

These characters are distinct. Indeed, if $\chi (v\cdot )=\chi (v^{\prime }\cdot )$ with $v\ne v^{\prime }$, multiplication by $v-v^{\prime }$ would show that $\chi$ is trivial on all of $U$, contrary to hypothesis. There are $(p-1)/2$ nonzero squares, sofor at least $(p-1)/2$ values of $v\in {\mathbb{F}}_p^{\ast }$. This is the square orbits of additive characters of a finite field argument.

WriteSince $u(x)u(y)=u(x+y)$,Thus $\chi (v\cdot )$ is a one-dimensional representation of $U$; for $v=0$ it is the trivial representation.

Apply Frobenius reciprocity and then the decomposition from part (i):Since all the representations in the last line are one-dimensional,It is therefore $2$ when both equalities hold, $1$ when exactly one holds, and $0$ when neither holds. This is the mackey inner product for the finite SL2 principal series.

The Mackey restriction formula says that for subgroups $H,K\le G$ and a $K$-representation $W$,where ${}^{x}W(xk{x}^{-1})=W(k)$.

For $H=K=B$, the Bruhat decomposition of SL2 over a finite field givesThus $B\backslash G/B$ has representatives $1,w$. The identity double coset contributes $\theta$, whilecontributes an induced representation from the diagonal subgroup. ThereforewhereFor $t(a)=\mathrm{diag}(a,a^{-1})$, one has $w^{-1}t(a)w=t(a^{-1})$.

Here $H$ is the paw graph, a triangle with a pendant edge.

For the lower bound, partition the vertices of $K_6$ into two triples. Colour the edges inside each triple red and all edges between the triples blue. Each red component is only a triangle, while the blue graph is the triangle-free graph $K_{3,3}$, so there is no monochromatic copy of $H$. Hence $R(H)>6$.

For the upper bound, every colouring of $K_7$ contains a monochromatic triangle because $R(3)=6$. Suppose that a triangle $T$ is red, and let $S$ be the other four vertices. If any edge from $T$ to $S$ were red, it would be a pendant edge extending $T$ to a red $H$. Thus all edges between $T$ and $S$ are blue.

If an edge $xy$ inside $S$ were blue, then for any $u\in T$ the vertices $u,x,y$ would form a blue triangle, and an edge from a second vertex of $T$ to $x$ would extend it to a blue $H$. Therefore every edge inside $S$ is red. The resulting red $K_4$ contains a red triangle and an additional incident edge, hence a red $H$. The blue-triangle case is symmetric, so the Ramsey number of the paw graph is

Part (i) gives $R(K_{1,t})\le 2t$ for every $t$.

Suppose first that $t$ is odd. On $2t-1$ vertices, take a red $(t-1)$-regular graph; for example, label the vertices cyclically and join each vertex to the $(t-1)/2$ nearest vertices in each direction. Its blue complement is also $(t-1)$-regular. Neither colour contains a vertex of degree $t$, so there is no monochromatic $K_{1,t}$. Therefore

Now let $t$ be even. If a colouring of $K_{2t-1}$ had no monochromatic $K_{1,t}$, every vertex would have red and blue degree at most $t-1$. Since the two degrees sum to $2t-2$, each would equal $t-1$. The red graph would then be $(t-1)$-regular on $2t-1$ vertices, impossible because both numbers are odd and the sum of degrees must be even. Hence $R(K_{1,t})\le 2t-1$.

For the matching lower bound, colour a copy of $K_{t-1,t-1}$ red on $2t-2$ vertices and colour all remaining edges blue. Red degree is $t-1$ and the blue graph is two disjoint copies of $K_{t-1}$, of degree $t-2$. Again neither colour contains $K_{1,t}$. Thus the Ramsey number of a star is

Choose any vertex $v$ of $K_{2t}$. Its $2t-1$ incident edges have two colours, so at leasthave the same colour. Those edges, together with $v$, form a monochromatic $K_{1,t}$.

If the real initial displacement is even, then $a(k)$ is real and even. Assume $a(k_0)\ne 0$ and setWhen $a(k_0)>0$, the leading result from part (i) is a positive multiple ofIts upward zero crossings of an oscillatory stationary-phase tail occur when the phase is $3\pi /2$ modulo $2\pi$. ThereforeIf $a(k_0)<0$, the sign is reversed and the upward crossings instead satisfyThe slow derivative of the $t^{-1/2}$ envelope changes these times only beyond leading order.

Each Fourier mode has zero initial velocity, soAlong $x=Vt$ this isPutThe two phases have stationary points at $-k_0$ and $k_0$, respectively. At those points,

Because the initial field is real, $a(-k)=a(k{)}^{\ast }$. Applying the stationary phase method to the two conjugate contributions gives the stationary-phase asymptotic of a Klein-Gordon wave along a subluminal ray:Equivalently, if $a(k_0)=|a(k_0)|e^{i\delta }$,

Sincethe phase speed is always larger than the group speed for nonzero finite $k$. Wave crests therefore move faster than the packet envelope.

For $k>0$, the phase velocity and group velocity areAs $k$ increases from zero, $c_p$ decreases from $+\infty$ toward $c$, while $c_g$ increases from zero toward $c$.

For a plane wave $\varphi =e^{i(kx-\omega t)}$, the Klein-Gordon equation givesThus the positive-frequency dispersion relation is