Past exam of the options II course of the University of Oxford / 2010 / F / F2

Differentiability at $x_0$ means that a linear map $A:{\mathbb{R}}^n\to {\mathbb{R}}^m$ exists withThe map $A=Df(x_0)$ is the total derivative. A linear map between these finite-dimensional spaces is bounded, so $\Vert f(x_0+h)-f(x_0)\Vert \le \Vert A\Vert \Vert h\Vert +o(\Vert h\Vert )\to 0$. Thus differentiability implies continuity.

For the displayed two-variable function,Writing $r=\sqrt{x^2+y^2}$ gives $|f(x,y)|/r\le r\to 0$. Therefore it is differentiable at the origin, with derivative zero.

Finally, put $h=x-x_0$, write $f(x_0+h)=Ah+r(h)$ with $r(h)=o(\Vert h\Vert )$, and $g(x_0+h)=g(x_0)+s(h)$ with $s(h)=o(1)$. Since $f(x_0)=0$,By the Cauchy-Schwarz inequality the last two terms are $o(\Vert h\Vert )$: $A$ is bounded and $g(x_0)+s(h)$ stays bounded. HenceOnly continuity of $g$ at $x_0$ is needed, because the vanishing of $f(x_0)$ removes the term which would otherwise require differentiating $g$.

A $C^1$ embedded submanifold $M\subseteq {\mathbb{R}}^n$ of dimension $n-1$ is locally the graph of a $C^1$ function of $n-1$ coordinates after a change of coordinates. Equivalently, near each point $x_0\in M$ it is a regular level set $\{x:h(x)=0\}$ of a $C^1$ scalar function with $\nabla h(x_0)\ne 0$.

The Lagrange multiplier theorem says that if a $C^1$ function $F$ has a constrained local extremum at $x_0$ on such a level set, thenfor some real $\lambda$. Indeed the derivative of $F$ must vanish on the tangent space $T_{x_0}M=\ker Dh(x_0)$, whose normal space is the line spanned by $\nabla h(x_0)$. The nonzero-gradient condition is essential.

Work on the positive orthant, where all real powers in $g$ are defined. To justify a global maximum, enlarge the constraint set to the compact simplexIt is closed and bounded since $x_i\le 1/p_i$. The product $g$ extends continuously to $K$, is zero at every boundary point where a coordinate vanishes, and equals $1$ at $(1,\dots ,1)$. Hence a maximum exists and lies in the positive interior. For $n=1$ the constraint gives $x_1=1$ directly.

For $n\ge 2$, at an interior maximum the Lagrange multiplier equations areAll $p_i$ and $g$ are positive, so all $x_i$ are equal. The constraint and ${\sum }_i{p}_i=1$ make their common value $1$. Thus the only possible maximizer has value $1$, provingwith equality exactly when every $x_i=1$. The source writes $g$ on all of ${\mathbb{R}}^n$; for general noninteger exponents its relevant real-valued domain is the positive orthant used here.

Let $S={\sum }_i{p}_i{a}_i>0$ and set $x_i=a_i/S$. Then $x_i>0$ and ${\sum }_i{p}_i{x}_i=1$, so part (1) givesTherefore the weighted arithmetic-geometric mean inequality isEquality holds exactly when all $a_i$ are equal, by the equality case in part (1).