Proof of A Semiconvex Function of Class $C^2$ has Hessian Bounded Below by $-\lambda I_n$

Write $\lVert\,\cdot\,\rVert$ for the Euclidean norm, $d_E$ for the Euclidean distance, and $0_{\mathbb{R}^n}$ for the origin of $\mathbb{R}^n$. Throughout, the field axioms of the field $\mathbb{R}$ and the identity $(-a)\,b=-(a\,b)$, which follows from $a\,b+(-a)\,b=(a+(-a))\,b=0$ and claim 1 of Additive Cancellation and Elementary Additive Identities in a Field, are used for entrywise computations, together with claims 4 and 5 of that lemma. The identity $0\,a=0$ is also used; it follows from $0\,a+0\,a=(0+0)\,a=0\,a+0$ and claim 2 of the same lemma.

Step 1: the auxiliary function and its Hessian. Let $q:U\to\mathbb{R}$ be the function whose value at $y$ is $(\lambda/2)\,d_E(y,0_{\mathbb{R}^n})^2$. By claim 2 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$ and $y-0_{\mathbb{R}^n}=y$, which holds coordinatewise by claim 4 of Additive Cancellation and Elementary Additive Identities in a Field, we have $d_E(y,0_{\mathbb{R}^n})=\lVert y\rVert$, so

$$q(y)=\frac{\lambda}{2}\,\lVert y\rVert^{2}.$$

By claims 2 and 3 of A Scaled Squared Distance to a Point is of Class $C^2$, with Gradient and Hessian, applied with the point $0_{\mathbb{R}^n}$ and the constant $\lambda/2$, the function $q$ is of class $C^2$ on $U$ and $D^2q(y)=\bigl(2\,(\lambda/2)\bigr)I_n=\lambda I_n$ for every $y\in U$.

Let $g:U\to\mathbb{R}$ be the function whose value at $y$ is $f(y)+q(y)$. By Semiconvex Function on a Convex Subset of $\mathbb{R}^n$, semiconvexity of $f$ on $U$ with constant $\lambda$ says exactly that $g$ is convex on $U$.

Step 2: $g$ is of class $C^2$, with $D^2g=D^2f+\lambda I_n$. Let $k_0:U\to\mathbb{R}$ be the constant function with value $0$ and let $\tilde q=k_0-q$. By claim 2 of Differences and Constants for Functions of Class $C^2$ on a Euclidean Open Set the function $k_0$ is of class $C^2$ with vanishing Hessian, so by claim 1 of that lemma $\tilde q$ is of class $C^2$ with $D^2\tilde q(x)=0_n-\lambda I_n$, where $0_n$ is the real $n\times n$ matrix all of whose entries are $0$. Moreover $\tilde q(y)=0-q(y)=-q(y)$ by claim 4 of Additive Cancellation and Elementary Additive Identities in a Field, so

$$f(y)-\tilde q(y)=f(y)-\bigl(-q(y)\bigr)=f(y)+q(y)=g(y)$$

by claim 5 of that lemma; that is, $g=f-\tilde q$. Claim 1 of Differences and Constants for Functions of Class $C^2$ on a Euclidean Open Set therefore shows that $g$ is of class $C^2$ on $U$ and that, for every $x\in U$,

$$D^2g(x)=D^2f(x)-\bigl(0_n-\lambda I_n\bigr).$$

Computing entries with Difference of Real Matrices and Scalar Multiple of a Real Matrix, the $(i,j)$ entry of $0_n-\lambda I_n$ is $0-\lambda(I_n)_{ij}=-\bigl(\lambda(I_n)_{ij}\bigr)$, so the $(i,j)$ entry of $D^2g(x)$ is $\bigl(D^2f(x)\bigr)_{ij}+\lambda(I_n)_{ij}$. Hence, again entrywise,

$$D^2f(x)=D^2g(x)-\lambda I_n .$$

Step 3: conclusion. Fix $x\in U$ and $z\in\mathbb{R}^n$. Since $U$ is open and convex and $g$ is of class $C^2$ and convex on $U$, A Convex Function of Class $C^2$ has Positive Semidefinite Hessian gives $0_n\preceq D^2g(x)$. By Matrix-Vector Product, the identity $0\,a=0$ and claim 3 of Properties of Finite Sums with the factor $0$, the $i$th coordinate of $0_nz$ is $\sum_{j=1}^{n}0\,z_j=0$, and the same computation gives $z\cdot(0_nz)=\sum_{i=1}^{n}z_i\,0=0$; so by The Positive Semidefinite Ordering on Symmetric Matrices the displayed comparison means

$$0\le z\cdot\bigl(D^2g(x)z\bigr).$$

By claims 1 and 2 of Linearity of the Matrix-Vector Product and the Quadratic Form as a Double Sum,

$$\bigl((-\lambda)I_n\bigr)z=(-\lambda)\,(I_nz)=(-\lambda)\,z,\qquad (\lambda I_n)z=\lambda\,z,\qquad \bigl(D^2g(x)-\lambda I_n\bigr)z=D^2g(x)z-\lambda\,z ,$$

and by claim 5 of Bilinearity and Symmetry of the Dot Product on $\mathbb{R}^n$,

$$z\cdot\bigl((-\lambda)z\bigr)=(-\lambda)\,(z\cdot z),\qquad z\cdot\bigl(D^2g(x)z-\lambda z\bigr)=z\cdot\bigl(D^2g(x)z\bigr)-\lambda\,(z\cdot z).$$

Using step 2, the asserted inequality $z\cdot\bigl((-\lambda)I_nz\bigr)\le z\cdot\bigl(D^2f(x)z\bigr)$ therefore reads

$$(-\lambda)\,(z\cdot z)\le z\cdot\bigl(D^2g(x)z\bigr)-\lambda\,(z\cdot z).$$

Translating by $\lambda\,(z\cdot z)$ using claim 3 of Elementary Arithmetic in an Ordered Field, and using $(-\lambda)\,(z\cdot z)+\lambda\,(z\cdot z)=0$, this is equivalent to $0\le z\cdot\bigl(D^2g(x)z\bigr)$, which was established above. As $z\in\mathbb{R}^n$ was arbitrary, $(-\lambda)I_n\preceq D^2f(x)$.