For $u\in U$, define $\varphi (u)\in U^{\ast }$ by $\varphi (u)(u^{\prime })=⟨u,u^{\prime }⟩$. If $\varphi (u)=0$, then in particularso $u=0$. Thus $\varphi$ is injective. Since a finite-dimensional vector space and its dual space have the same dimension, $\varphi$ is also surjective and hence an isomorphism. This is the finite-dimensional real case of the Riesz representation theorem.

The adjoint operator of $\alpha :V\to W$ is the unique linear map ${\alpha }^{\ast }:W\to V$ satisfyingfor every $v\in V$ and $w\in W$. In the stated orthonormal bases, if $v$ and $w$ denote coordinate columns, thenTherefore the matrix of ${\alpha }^{\ast }$ is the matrix transpose $A^T$.

For $w\in W$,Hence $\ker {\alpha }^{\ast }=(\mathrm{im}\alpha {)}^{\perp }$. Taking orthogonal complements in the finite-dimensional space $W$ proves the image-kernel orthogonality for an adjoint

Put $r=\alpha (v_0)-w_0$. For any $h\in V$,This is minimized at $h=0$ exactly when $r\perp \mathrm{im}\alpha$, equivalently when ${\alpha }^{\ast }r=0$. Thus the normal equation for a linear inverse problem is

For the given linear least-squares problem,The normal equations $2x+y=3$ and $x+2y=5$ have the unique solution

For square-integrable wavefunctions with the stated decay, multiplication by the real coordinate $x$ is Hermitian becauseFor $p_x=-i\hslash {\partial }_x$, integration by parts givesand the boundary term vanishes. The same arguments apply to $y$ and $p_y$.

If $F$ and $G$ are Hermitian on a common suitable domain, then $(FG{)}^{†}=GF$, soBecause $x$ commutes with $p_y$ and $y$ with $p_x$,is Hermitian. Using $p_{x}x=x{p}_x-i\hslash$ and its $y$ analogue,so $D$ is Hermitian as well.

Finally set $R=x{\partial }_y-y{\partial }_x$ and $S=x{\partial }_x+y{\partial }_y$. Direct evaluation on a smooth test function gives $[R,S]=0$: rotations commute with radial dilations. The scalar term in $D$ also commutes with everything, and therefore

Let $\omega =e^{2\pi i/N}$. Using the conventionthe discrete Fourier transform of the supplied sequence follows from the binomial theorem:The orthogonality of roots of unity givesSince $m,k$ both lie between zero and $N-1$, it follows that

Sincefor $x>0$, the second-derivative criterion shows that $f(x)=x^a/a$ is strictly convex. Put $b=a/(a-1)$, so $1/a+1/b=1$. For dual variable $p>0$, the stationary equation in the Legendre transform isIt is the unique global maximizer of $px-f(x)$, and henceApplying the same calculation with the conjugate exponent $b$ givesUnder the Legendre-Fenchel transform on all real dual variables, the same supremum additionally gives $f^{\ast }(p)=0$ for $p\le 0$.

Now take $a=r$, $b=s$, and apply the Fenchel–Young inequality $f(x)+f^{\ast }(y)\ge xy$. This yields Young's inequality for productsfor positive $x,y$ and conjugate exponents $r,s$.

On the unit sphere, the spherical excess formula gives the area of a spherical triangle asFor a sphere of radius $R$, the right side is multiplied by $R^2$.

Join an interior point of a convex spherical $n$-gon to its vertices by geodesics. The resulting $n$ spherical triangles contain all polygon angles and angles summing to $2\pi$ at the interior point. Adding their spherical excesses givesFor a regular polygon with interior angle $\alpha$, positivity of area requires $n\alpha >(n-2)\pi$, while strict convexity requires $\alpha <\pi$. Both limiting values can be approached by regular polygons, so

The Möbius transformationmaps the unit disc conformally onto the upper half-plane. On the upper semicircle of its boundary, $w$ approaches the positive real axis; on the lower semicircle it approaches the negative real axis. If $0<\arg w<\pi$, the harmonic functionhas the required boundary limits.

For $z=x+iy$,Since the imaginary part is positive inside the disc,Therefore the solution of the Dirichlet problem isThe exceptional boundary endpoints are precisely the jump points of the prescribed data.