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The Goldstone theorem states that every spontaneously broken generator of a continuous internal global symmetry in a Lorentz-invariant quantum field theory produces a massless scalar particle.

For the classical proof, let $V(\varphi )$ be the invariant scalar potential and let $v_i$ be a vacuum. Infinitesimal invariance givesDifferentiate with respect to ${\varphi }_k$ and evaluate at the stationary point $v$, where ${\partial }_{i}V(v)=0$. The scalar mass matrix then obeysFor every broken generator, $(T^{a}v{)}_i\ne 0$, so $T^{a}v$ is a zero eigenvector of the Hessian matrix. It is a massless fluctuation tangent to the vacuum manifold.

For the quantum proof, spontaneous breaking means that some local field hasWrite $Q_a=\int d^{3}x{j}_a^0(x)$ and insert a complete set of momentum eigenstates into the current-field correlation function. Current conservation and Lorentz covariance imply that a scalar intermediate state couples asThe nonzero equal-time commutator requires a pole at $p^2=0$; otherwise the conserved-current spectral integral vanishes at zero momentum. Thus a massless Goldstone boson exists for every independent broken direction.

True. Continuity and compactness give a maximum of $\varphi$ on $\overline{\Omega }$. At an interior maximum the Hessian matrix is negative semidefinite, so its trace satisfies $\Delta \varphi \le 0$, contradicting $\Delta \varphi >0$. The maximum must therefore occur on $\partial \Omega$. This is the strict form of the maximum principle for subharmonic functions.

The multivariable chain rule gives, at $(x,y)=(a+t\cos \gamma ,b+t\sin \gamma )$,Sufficient conditions for a strict local minimum at $t=0$ are $g^{\prime }(0)=0$ and $g^{\prime \prime }(0)>0$.

At a stationary point, $g^{\prime }(0)=0$ in every direction. Its second derivative is the quadratic form of the Hessian matrix. Ifthen completing the square givesfor every nonzero direction $(u,v)$. The Hessian is therefore positive definite, and the stationary point is a strict local minimum.

The stationary equations areThus $y=-x$ and $4x(x^2-2)=0$, givingThe Hessian matrix isAt either nonzero stationary point,so both are strict local minima.

Along a line through the origin,whereWhen $\cos \gamma \ne 0$ and $r=\tan \gamma$,Consequently the origin is a local minimum along the line whenincluding the equality cases because the positive quartic term then leads. It is also a minimum on the vertical line. It is a local maximum along the line whenIn the first case the graph is locally bowl-shaped. In the second it initially bends downward from the origin, but the positive quartic term eventually turns it upward, producing the usual double-well profile.

The only positive equilibrium solves $1-y=0$ and $x-1=0$, hence isAway from a nullcline,This is separable:Integration gives the first integralThus one convenient conserved quantity is

Its gradient isso $(1,1)$ is its only stationary point in the positive quadrant. Its Hessian matrix iswhich is positive definite everywhere. Hence $(1,1)$ is a strict, indeed global, minimum.

The nearby level curves of $E$ are closed curves surrounding this minimum. Since $E$ is constant along every trajectory, a solution starting on a sufficiently small nearby level set cannot leave the region bounded by a slightly larger level set. This proves Lyapunov stability of the equilibrium: solutions initially close to $(1,1)$ remain close for all time.

Starting from $x_0\in {\mathbb{R}}^n$, gradient descent repeatedly computes the gradient and updatesstopping when the gradient norm, step, or objective decrease is sufficiently small.

The Hessian bounds say that $f$ is $\alpha$-strongly convex and has $\beta$-smooth gradient. With $\eta =1/\beta$,Thus the iteration count isconvergence becomes slower linearly with the condition number $\kappa =\beta /\alpha$.

Forthe Hessian matrix is $\mathrm{diag}(2,200,20000)$, soTakeThenwhose Hessian is $2I$ and whose condition number is $1$.

Because $\overline{\Omega }$ is compact, $\varphi$ attains a maximum. If a maximizer lay in $\Omega$, its Hessian matrix would be negative semidefinite, so its trace would satisfy $\Delta \varphi \le 0$, contradicting $\Delta \varphi >0$. Hence every strictly subharmonic function under the stated hypotheses attains a maximum on $\partial \Omega$.

Statement (i) is true. For $ϵ>0$, the functionhas $\Delta {\varphi }_{ϵ}=4ϵ>0$, soLetting $ϵ\downarrow 0$ givesThis is the maximum principle for harmonic functions.

A function $f:D\to \mathbb{R}$ on a convex set $D$ is convex whenfor all $x,y\in D$ and $0\le t\le 1$.

If $f$ is once differentiable, this is equivalent to monotonicity of its gradient:Equivalently, every tangent hyperplane supports the graph:If $f$ is twice differentiable, convexity is equivalent to the Hessian matrix being positive semidefinite throughout $D$.

Forthe Hessian isA real symmetric two-by-two matrix is positive semidefinite exactly when its two diagonal entries and its determinant are nonnegative. ThusThe largest convexity domain is thereforeFor $A\ne 0$, its boundary is the hyperbola $y=A^2/(36x)$ in the first quadrant and the domain lies above it. For $A=0$, it is the closed first quadrant. This is the convexity domain of x cubed plus y cubed plus Axy.

Write $z_i(x)=⟨a_i,x⟩+b_i$. The log-sum-exp function is convex and composition with the affine functions $z_i$ preserves convexity, so $f_{\beta }$ is convex. Directly, its Hessian matrix will also be shown positive semidefinite in part d.

Let $M=f(x)={\max }_i{z}_i(x)$. Factoring $e^{\beta M}$ out of the sum givesAt least one term in the sum is $1$, while every term is at most $1$. Thereforeand henceThus $f_{\beta }$ is a uniform smooth maximum of the affine pieces of $f$.

Define the softmax weightsThey obey $p_i\ge 0$ and ${\sum }_i{p}_i=1$. The gradient is their weighted mean,Differentiating once more gives the covariance-form Hessian matrixFor every unit vector $u$,where $G={\max }_i\Vert a_i{\Vert }_2$. The Hessian is a covariance matrix, so it is positive semidefinite; the displayed upper bound also gives ${\nabla }^2{f}_{\beta }⪯\beta G^{2}I$ in the Loewner order. Consequently $f_{\beta }$ has a Lipschitz gradient with

At $p$, the surface $\tilde{S}$ cannot cross the boundary $S=\partial \mathcal{R}$, because it lies inside $\mathcal{R}$. Consequently the two surfaces are tangent at $p$. Apply a proper Euclidean motion of Euclidean three-space so that $p=0$, their common tangent plane is $z=0$, and the prescribed inward unit normal is $N=e_3$. Locally writewith the region $\mathcal{R}$ on the side $z\ge f(x,y)$. ThenThus $g-f$ has a local minimum at zero and its Hessian matrix is positive semidefinite. The fundamental forms of a graph surface at a horizontal tangent plane giveTaking the trace of the positive-semidefinite difference proves the mean-curvature comparison at tangential contact:

There is no analogous conclusion for $K$. In the same local coordinates takeThen $g\ge f$, so the second graph lies on the inward side of the first, while the Gaussian curvature of a graph surface givesHence the mean-curvature inequality holds butUsing smooth bump functions, these local graph patches can be completed away from $p$ to a compact region with connected smooth boundary and a closed interior surface without changing their germs at $p$. This realizes the counterexample under the global hypotheses.

Forthe weak maximum principle for elliptic operators states that $Lu\ge 0$ impliesIn particular, a solution of $Lu=0$ cannot have a positive interior maximum exceeding its boundary maximum.

Because the coefficient matrix is positive definite and the closure of the smooth bounded domain is compact, strict ellipticity supplies a uniform lower bound after restricting to $\overline{\Omega }$. Rotate and translate coordinates so that $\Omega$ is bounded in the $x_1$ direction, and set $h=e^{\gamma x_1}$. For sufficiently large $\gamma$,If $u+\epsilon h$ had a positive interior maximum, its gradient would vanish and its Hessian matrix would be negative semidefinite there, giving $L(u+\epsilon h)\le 0$. This contradicts $L(u+\epsilon h)=\epsilon Lh>0$. Comparing on the boundary and sending $\epsilon \downarrow 0$ proves the assertion.

$\nabla U=(3x^2-3+y^2,2y(x+1))$, so the critical points are $(\pm 1,0)$. The Hessian matrix at $(1,0)$ is $\mathrm{diag}(6,4)$, making it the local minimum; $(-1,0)$ is degenerate and not a minimum.

For a zero of order $N$ at the origin and no other zeros,Away from the zero, the flat Taubes equation isIf $u$ had a positive interior maximum, then the second-derivative test would give $\Delta u\le 0$ there, while $e^u-1>0$, a contradiction. The boundary values are $-\infty$ at the zero and $0$ at infinity, so the maximum principle gives $u\le 0$. Hence

For $V=m^2{\varphi }^2/2+\lambda ({\varphi }^2{)}^2/4$, $m^2<0$, the vacuum manifold is a sphere of radius $v=\sqrt{-m^2/\lambda }$. At a chosen vacuum the Hessian matrix is $2\lambda v^{2}n{n}^T$, where $n$ is its unit direction. The radial scalar has mass squared $2\lambda v^2$; the $n-1$ angular modes are physical Goldstone bosons when the symmetry is global.

A Lagrangian is regular when its velocity Hessian matrix is invertible. The inverse function theorem then makes the fiber derivative $p_i=L_{v^i}$ locally invertible, allowing a local Legendre transform in mechanics to a Hamiltonian. If that fiber derivative is globally invertible, the Lagrangian is called a hyperregular Lagrangian. Regularity alone does not assert global invertibility, and singular Lagrangians may give constrained rather than ordinary unconstrained Hamilton's equations.

The scalar mass matrix is the Hessian matrix of the scalar potential at a vacuum. Its eigenvalues are the tree-level squared masses; broken symmetry directions are zero eigenvectors before a Higgs mechanism or explicit symmetry breaking removes the corresponding massless modes.

At an interior local maximum of a twice differentiable function, its Hessian matrix is negative semidefinite; at a local minimum it is positive semidefinite. For a scalar function on Euclidean space, the Laplacian is therefore nonpositive at a local maximum and nonnegative at a local minimum.

At an equilibrium, a positive-definite Hessian matrix of the potential gives stable small oscillations. If a continuous symmetry supplies zero modes, positive definiteness on the directions transverse to the symmetry gives stability of the relative configuration.

Subtracting the quadratic Taylor polynomial at the rescaling centre makes the function, gradient and Hessian matrix vanish there. Dividing by $A{r}^{2+\mu }$ preserves the normalized Hölder oscillation of the Hessian matrix while turning a small forcing-to-oscillation ratio into a harmonic limiting equation. The three vanished Taylor coefficients provide local bounds needed by the Arzela-Ascoli theorem.