Use the Minkowski metric with signature $(+---)$ and write $U=(U^0,\mathbf{u})$, $V=(V^0,\mathbf{v})$. Future-pointing timelike means $U^0>\Vert \mathbf{u}\Vert$ and $V^0>\Vert \mathbf{v}\Vert$. Thusso $U+V$ is again future-pointing timelike.

Let $a=\sqrt{g(U,U)}>0$ and $b=\sqrt{g(V,V)}>0$. In a Lorentz frame where $U=(a,0,0,0)$, the future-pointing vector $V$ has $V^0\ge b$. Consequently $g(U,V)=a{V}^0\ge ab$, with equality exactly when $V$ has no spatial component in that frame. HenceTaking positive square roots provesEquality holds exactly when $U,V$ are positive scalar multiples of each other. This is the reverse triangle inequality for future timelike vectors, not the Euclidean triangle inequality.

The sets of admissible functions are nested:All impose the same left conditions; the right height adds a restriction, and the right slope adds another. Minimizing the same functional over a smaller set cannot lower its minimum. ThereforeThe doubly clamped beam has the greatest minimum energy, and the free-ended beam the smallest. No energy integrals are needed. In fact the inequalities are strict: the positive quadratic bending term makes the minimizer unique on the left-clamped space, and the three solutions in part (b) are different. The relevant principle is monotonicity of a constrained minimum under added constraints.

Here $F_y=\rho g$, $F_{y^{\prime }}=0$, and $F_{y^{\prime \prime }}=K{y}^{\prime \prime }$. The Euler-Lagrange equation for a second-derivative functional is thereforeWrite $q=\rho g/K>0$. After imposing the clamped conditions at zero, the general solution isFree right endpoint. Both endpoint variations are free, so the natural boundary conditions for second-derivative functionals give $y^{\prime \prime }(L)=y^{\prime \prime \prime }(L)=0$. Since $y^{\prime \prime \prime }=-qx+A$ and $y^{\prime \prime }=-q{x}^2/2+Ax+B$, they give $A=qL$, $B=-q{L}^2/2$. HenceRight endpoint with fixed height and free slope. Now $y(L)=0$ is imposed, while the free slope gives $y^{\prime \prime }(L)=0$. These equations yield $A=5qL/8$ and $B=-q{L}^2/8$, soBoth endpoints clamped. The right conditions are $y(L)=y^{\prime }(L)=0$, giving $A=qL/2$, $B=-q{L}^2/12$. ThusEach formula has fourth derivative $-\rho g/K$ and satisfies all four boundary conditions of its case. The three endpoint choices therefore change the integration constants, not the interior differential equation.

Take a smooth variation $y_{\epsilon }=y+\epsilon \eta$. Differentiating the functional at zero gives the first variationIntegrate the second term once and the third term twice by integration by parts:Choose variations supported away from the endpoints. Stationarity and the fundamental lemma of the calculus of variations forceThis is the Euler-Lagrange equation for a second-derivative functional. With the interior equation imposed, the remaining boundary expression must vanish for every admissible variation.

At each endpoint, a freely varying value $y$ requires $\boxed{F_{y^{\prime }}-(F_{y^{\prime \prime }}{)}^{\prime }=0}$, and a freely varying slope $y^{\prime }$ requires $\boxed{F_{y^{\prime \prime }}=0}$. A fixed value or slope instead makes its corresponding variation zero and supplies no natural condition for that variable. These are the natural boundary conditions for second-derivative functionals, with the value and slope variations treated independently.

Apply the regular level set theorem to $F(A)=\det A-1$ on the $n^2$-dimensional vector space of real matrices. At a point with determinant one, part (b) gives $D{F}_A(H)=\mathrm{tr}(A^{-1}H)$. Choosing $H=A/n$ gives $D{F}_A(H)=1$, so this derivative is surjective.

Hence the special linear group $\mathrm{SL}(n,\mathbb{R})$ is a smooth submanifold of dimension $n^2-1$. At the identity,This is the special linear Lie algebra. When $n=1$, the group is a single point and the tangent space is zero, consistent with the same formula.

The regular level set theorem says that a $C^1$ map $f:{\mathbb{R}}^n\to {\mathbb{R}}^m$ with surjective derivative at every point of $f^{-1}(0)$ has a zero set that is an embedded $C^1$ submanifold of dimension $n-m$. At each point $a$ of that zero set,For a smooth map the submanifold is smooth. The full-rank hypothesis is essential for this kernel description when using a particular defining map.

Factor $A+H=A(I+A^{-1}H)$. The determinant expansion near the identity isthe linear terms in the permutation formula come from choosing one diagonal increment and all remaining identity entries. ThereforeThe required Fréchet derivative is the linear functionalThe quadratic remainder proves full differentiability, rather than only the existence of each directional derivative. This is the derivative of the determinant.

The Fréchet derivative at a point $a$ is the unique linear map $D{f}_a:{\mathbb{R}}^n\to {\mathbb{R}}^m$ such thatThe error divided by $\Vert h\Vert$ tends to zero as $h\to 0$. In standard coordinates this linear map is represented by the Jacobian matrix, but the derivative itself is the map acting on increments, not merely a list of directional rates.