Past exam of the short options course of the University of Oxford / 2015 / 4

A map $g$ is differentiable at $a$ when a linear map $D{g}_a:{\mathbb{R}}^n\to {\mathbb{R}}^k$ approximates its increment to first order:This Fréchet derivative is unique, and its matrix in the standard bases is the Jacobian matrix. Differentiability on a domain means differentiability at each point there.

A standard sufficient condition for a zero locus to be a submanifold is the regular level set theorem: if $g$ is $C^1$ on an open domain and $D{g}_a$ is surjective at every $a\in g^{-1}(0)$, thenFor smooth $g$, the submanifold is smooth. The full-rank condition is the relevant constraint qualification.

Under this regularity hypothesis, its tangent space isIndeed, if a differentiable curve $\gamma$ lies in $M$ and passes through $a$, the chain rule gives $D{g}_a({\gamma }^{\prime }(0))=0$. Thus all tangent velocities lie in the kernel of a linear map. The rank-nullity theorem makes this kernel $(n-k)$-dimensional, the same dimension as the tangent space, proving equality. In coordinates, solve the homogeneous linear equations given by the Jacobian matrix.

The regularity qualification matters: $g(x)=x^2$ has zero set $\{0\}$, a zero-dimensional submanifold, but $D{g}_0=0$. Its kernel is all of $\mathbb{R}$, whereas its tangent space is zero. This is the phenomenon that singular defining maps can overestimate tangent spaces.

At a constrained critical point, the directional derivative of $f$ vanishes along every admissible tangent velocity:For a regular zero locus, $T_{a}M=\ker D{g}_a$. A linear functional vanishing on this kernel lies in the row space of $D{g}_a$. Equivalently there are Lagrange multipliers ${\lambda }_1,\dots ,{\lambda }_k$ such thatGeometrically, $\nabla f$ is orthogonal to the tangent space, so it lies in its orthogonal complement, spanned by the independent constraint gradients. Thus the level set of $f$ has no first-order change in a tangent direction. The equations provide candidates for constrained extrema; they must still be classified and compared, and boundary or singular cases must be treated separately. The condition is necessary for a regular constrained extremum, not a guarantee of a maximum.

The feasible set is closed and contained in the rectangleIt is therefore compact, and the continuous function $xy$ attains a global maximum there by the extreme value theorem. At the two boundary endpoints where a coordinate is zero, the product is zero. Meanwhile $(1,1)$ is feasible because the conjugate exponents satisfy $1/p+1/q=1$, and its product is one. Hence every maximizer has $x,y>0$.

On that positive region the constraint is $C^1$ with nonzero gradient. The Lagrange multiplier equations areThe first two identities imply $xy=\lambda x^p=\lambda y^q$. Since $x,y>0$, the multiplier is nonzero, so $x^p=y^q=r$. The constraint now givesThus the sole interior candidate is $(1,1)$. Existence of a maximum and the boundary comparison prove the unique global resultThis conjugate-power constrained product maximum is valid for all real $p,q>1$ with $1/p+1/q=1$; no integrality of the exponents is needed.

Use a normalization proof of Young's inequality. For arbitrary $x,y\ge 0$, putIf $A=0$, both coordinates are zero and the desired inequality is equality. If $A>0$, defineThen $X^p/p+Y^q/q=1$. Part (c) gives $XY\le 1$, and the conjugate exponent identity gives $XY=xy/A$. Therefore Young's inequality for products isThe unique equality case in part (c) says $X=Y=1$. Translated back, and including the zero case, this gives the sharp conditionConversely, if both powers equal $r\ge 0$, then $xy=r^{1/p+1/q}=r$ and the right side is $r(1/p+1/q)=r$, directly verifying equality.