Proof of A Function of Class $C^2$ with Positive Semidefinite Hessian is Convex

Write $\lVert\,\cdot\,\rVert$ for the Euclidean norm, $|\,\cdot\,|$ for the absolute value, and $d_E$ for the Euclidean distance, a metric by Euclidean Distance is a Metric on $\mathbb{R}^n$; claims 2 and 5 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$ give $d_E(u,v)=\lVert u-v\rVert$ and $\lVert\mu u\rVert=|\mu|\,\lVert u\rVert$. Order arithmetic is taken from Elementary Order Arithmetic in an Ordered Field (claim 1 strict compatibility with addition, claim 2 mixed transitivity, claim 5 product of positive elements, claim 7 inverse of a positive element, claim 8 halving, claim 10 strict compatibility with multiplication by a positive element) and from Elementary Arithmetic in an Ordered Field (claim 3 translation, claim 5 multiplication by a nonnegative element); the field axioms of the field $\mathbb{R}$ and the vector space axioms of Euclidean Space $\mathbb{R}^n$ is a Real Vector Space are used for rearrangement.

Let $x,y\in U$ and let $t\in\mathbb{R}$ satisfy $0\le t$ and $t\le1$. Put $\theta=1-t$, so that $0\le\theta$ and $\theta\le1$ by translation, put $h=y-x$, and put

Step 1: $0,1\in J$. Indeed $x+0\,h=x\in U$ and $x+1\,h=x+(y-x)=y\in U$.

Step 2: $J$ is order-convex. Let $\tau_1,\tau_2\in J$ and $\tau\in\mathbb{R}$ with $\tau_1\le\tau\le\tau_2$. If $\tau_1=\tau_2$ then $\tau=\tau_1\in J$ by antisymmetry of $\le$. Otherwise $\tau_1<\tau_2$, so $0<\tau_2-\tau_1$ and

satisfies $0\le\mu$ and $\mu\le1$, by translation and multiplication by the positive element $(\tau_2-\tau_1)^{-1}$. Since $\tau_1+\mu\,(\tau_2-\tau_1)=\tau$, we get

which lies in $U$ because $U$ is convex and $x+\tau_1h,\,x+\tau_2h\in U$. Hence $\tau\in J$.

Step 3: every point of $J$ is an interior point of $J$. Let $\tau\in J$. By Euclidean Openness Agrees with Metric Openness on $\mathbb{R}^n$ the set $U$ is metric-open, so there is $r\in\mathbb{R}$ with $0<r$ such that every $z\in\mathbb{R}^n$ with $d_E(x+\tau h,z)<r$ lies in $U$. Put $\rho=r\,(\lVert h\rVert+1)^{-1}$, which is positive because $\lVert h\rVert+1$ is positive, the norm being nonnegative. If $\sigma\in\mathbb{R}$ satisfies $|\sigma-\tau|<\rho$ then

so $\sigma\in J$. Taking $u=\tau-\tfrac12\rho$ and $v=\tau+\tfrac12\rho$, which satisfy $|u-\tau|<\rho$ and $|v-\tau|<\rho$ by halving, gives $u,v\in J$ with $u<\tau<v$.

Step 4: the slice. Let $g:J\to\mathbb{R}$ and $g_1:J\to\mathbb{R}$ be given by

By step 3 every $\tau\in J$ is an interior point of $J$, so claims 1 and 2 of Derivatives of the Slice of a Function Along a Line, applied with $p=x$, show that $g$ is differentiable at $\tau$ with $g'(\tau)=g_1(\tau)$ and that $g_1$ is differentiable at $\tau$ with

where $z\cdot z'$ is the dot product and the product is the matrix-vector product.

Step 5: nonnegativity of $g_1'$. By Matrix-Vector Product, the identity $0\,a=0$, which follows from $0\,a+0\,a=(0+0)\,a=0\,a+0$ and claim 2 of Additive Cancellation and Elementary Additive Identities in a Field, and claim 3 of Properties of Finite Sums with the factor $0$, every coordinate of $0_nh$ is $0$ and hence $h\cdot(0_nh)=0$. Applying the hypothesis at the point $x+\tau h\in U$ and reading The Positive Semidefinite Ordering on Symmetric Matrices at the point $h$ gives

Step 6: conclusion. The hypotheses of A Real Function with Nonnegative Second Derivative is Convex on an Interval hold for $J$, $g$ and $g_1$, so, applied with the points $0,1\in J$ of step 1 and with $\theta$,

that is, $f(x+\theta\,h)\le(1-\theta)\,f(x)+\theta\,f(y)$. Since $\theta=1-t$ we have $1-\theta=t$ and

so $f\bigl(t\,x+(1-t)\,y\bigr)\le t\,f(x)+(1-t)\,f(y)$. As $x,y\in U$ and $t$ were arbitrary, this is the defining condition of Convex Real-Valued Function on a Convex Subset of $\mathbb{R}^n$, so $f$ is convex on $U$.