Past exam of the Probability, statistics and options II course of the University of Oxford / 2010 / F (Multivariable Calculus)

Differentiability at $x_0$ means that a linear map $L:{\mathbb{R}}^n\to {\mathbb{R}}^m$ exists such that $f(x_0+h)=f(x_0)+Lh+o(\Vert h\Vert )$. Since finite-dimensional linear maps are bounded, the right side tends to $f(x_0)$ as $h\to 0$, proving continuity.

For the specified scalar function, $0\le f(x,y)\le x^2\le x^2+y^2$. Dividing by $\sqrt{x^2+y^2}$ gives a bound tending to zero at the origin. Consequently $\boxed{df(0,0)=0}$ and the function is differentiable there.

For the final product assertion, write $f(x_0+h)=Lh+r(h)$, where $r(h)=o(\Vert h\Vert )$, and $g(x_0+h)=g(x_0)+s(h)$ with $s(h)=o(1)$. Since $f(x_0)=0$,Cauchy–Schwarz and boundedness of $g$ near $x_0$ justify the last estimate. Thus $\boxed{dF(x_0)h=⟨df(x_0)h,g(x_0)⟩}$; differentiability of $g$ is unnecessary because $f(x_0)=0$.

A $C^1$ embedded hypersurface $M\subseteq {\mathbb{R}}^n$ is locally carried by a $C^1$ coordinate diffeomorphism to an open subset of ${\mathbb{R}}^{n-1}\times \{0\}$. Equivalently, near each of its points it has the form $F^{-1}(0)$ for a $C^1$ function with nonzero gradient there.

The Lagrange multiplier theorem states that if a differentiable function $g$ has a local extremum on such a hypersurface at $x_0$, then $dg(x_0)$ annihilates the tangent space. In a regular local level-set description this isfor some real $\lambda$. The regularity condition $\nabla F(x_0)\ne 0$ is essential; it is what identifies the normal line and permits the local constraint chart.

Work on the positive orthant, where the real powers defining $g$ are smooth. Its closure under the constraint is the compact weighted simplex $\overline{M}=\{x_i\ge 0,\sum p_i{x}_i=1\}$, since $0\le x_i\le 1/p_i$. The product extends continuously there, is zero on its boundary when $n>1$, and is one at $(1,\dots ,1)$. Thus a maximum is attained in the positive interior.

At an interior maximizer the constraint gradient $(p_1,\dots ,p_n)$ is nonzero. The multiplier equations giveAll coordinates are equal, and the weighted constraint forces each to equal one. Therefore the maximum is one and $\boxed{g(x)\le 1}$ on $M$. For $n=1$ the constraint itself gives $x_1=1$, so the same conclusion holds directly. Equality occurs exactly at the all-ones point.

Set $A={\sum }_i{p}_i{a}_i>0$ and $x_i=a_i/A$. Then ${\sum }_i{p}_i{x}_i=1$, so part 1 gives ${\prod }_i(a_i/A{)}^{p_i}\le 1$. Since ${\sum }_i{p}_i=1$, this becomesEquality holds exactly when $x_i=1$ for all $i$, equivalently all the $a_i$ are equal. This is the weighted arithmetic–geometric mean inequality.