# Optimality, convexity, and gradient descent¶

## 2 Optimality conditions¶

Course: Math 535 - Mathematical Methods in Data Science (MMiDS)
Author: Sebastien Roch, Department of Mathematics, University of Wisconsin-Madison
Updated: Oct 10, 2020

In this section, we derive optimality conditions for unconstrained continuous optimization problems.

### 2.1 Multivariate version of Taylor's theorem¶

We will make use of Taylor's Theorem, a powerful generalization of the Mean Value Theorem that provides polynomial approximations to a function around a point. We restrict ourselves to the case of a linear approximation with second-order error term, which will suffice for our purposes.

#### 2.1.1 Single-variable case¶

We begin by reviewing the single-variable case, which we will use to prove the general verison.

Theorem (Taylor): Let $f: D \to \mathbb{R}$ where $D \subseteq \mathbb{R}$. Suppose $f$ has a continuous derivative on $[a,b]$ and that its second derivative exists on $(a,b)$. Then for any $x \in [a, b]$

$$f(x) = f(a) + (x-a) f'(a) + \frac{1}{2} (x-a)^2 f''(\xi)$$

for some $a < \xi < x$.

Proof idea: The Mean Value Theorem implies that there is $a < \xi< x$ such that $f(x) = f(a) + (x - a)f'(\xi)$. One way to think of the proof of that result is the following: we constructed an affine function that agrees with $f$ at $a$ and $x$, then used Rolle to express the coefficient of the linear term using $f'$. Here we do the same with a polynomial of degree $2$. But we now have an extra degree of freedom in choosing this polynomial. Because we are looking for a good approximation close to $a$, we choose to make the first derivative at $a$ also agree. Applying Rolle twice gives the claim.

Proof: Let

$$P(t) = \alpha_0 + \alpha_1 (t-a) + \alpha_2 (t-a)^2.$$

We choose the $\alpha_i$'s so that $P(a) = f(a)$, $P'(a) = f'(a)$, and $P(x) = f(x)$. The first two lead to the conditions

$$\alpha_0 = f(a), \quad \alpha_1 = f'(a).$$

Let $\phi(t) = f(t) - P(t)$. By construction $\phi(a) = \phi(x) = 0$. By Rolle, there is a $\xi' \in (a, x)$ such that $\phi'(\xi') = 0$. Moreover, $\phi'(a) = 0$. Hence we can apply Rolle again - this time to $\phi'$ on $[a, \xi']$. It implies that there is $\xi \in (a, \xi')$ such that $\phi''(\xi) = 0$.

The second derivative of $\phi$ at $\xi$ is

$$0 = \phi''(\xi) = f''(\xi) - P''(\xi) = f''(\xi) - 2 \alpha_2$$

so $\alpha_2 = f''(\xi)/2$. Plugging into $P$ and using $\phi(x) = 0$ gives the claim. $\square$

The third term on the right-hand side of Taylor's Theorem is called the remainder. It can be seen as an error term between $f(x)$ and the linear approximation $f(a) + (x-a) f'(a)$. There are other forms for the remainder. The form we stated here is useful when one has information about the the second derivative. Here is an example.

Example: Consider $f(x) = e^x$. Then $f'(x) = f''(x) = e^x$. Suppose we are interested in approximating $f$ in the interval $[0,1]$. We take $a=0$ and $b=1$ in Taylor's Theorem. The linear term is

$$f(a) + (x-a) f'(a) = 1 + x e^0 = 1 + x.$$

Then for any $x \in [0,1]$

$$f(x) = 1 + x + \frac{1}{2}x^2 e^{\xi_x}$$

where $\xi_x \in (0,1)$ depends on $x$. We get a uniform bound on the error over $[0,1]$ by replacing $\xi_x$ with its worst possible value over $[0,1]$

$$|f(x) - (1+x)| \leq \frac{1}{2}x^2 e^{\xi_x} \leq \frac{e}{2} x^2.$$
In [1]:
# Julia version: 1.5.1
using Plots

In [2]:
x = LinRange(0,1,100)
y = exp.(x) # function f
taylor = 1 .+ x # linear approximation
err = (exp(1)/2)*x.^2 # our error bound
plot(x, y, lw=2, label="f", legend=:topleft)
plot!(x, taylor, lw=2, label="taylor")

Out[2]:

If we plot the upper and lower bounds, we see that $f$ indeed falls within them.

In [3]:
plot!(x, taylor-err, linestyle=:dot, linecolor=:green, label="lower")
plot!(x, taylor+err, linestyle=:dash, linecolor=:green, label="upper")

Out[3]:

#### 2.1.2 General case¶

In the case of several variables, we again restrict ourselves to the second order. We start with an important special case: a Multivariate Mean Value Theorem.

Theorem (Multivariate Mean Value): Let $f : D \to \mathbb{R}$ where $D \subseteq \mathbb{R}^d$. Let $\mathbf{x}_0 \in D$ and $\delta > 0$ be such that $B_\delta(\mathbf{x}_0) \subseteq D$. If $f$ is continuously differentiable on $B_\delta(\mathbf{x}_0)$, then for any $\mathbf{x} \in B_\delta(\mathbf{x}_0)$

$$f(\mathbf{x}) = f(\mathbf{x}_0) + \nabla f(\mathbf{x}_0 + \xi \mathbf{p})^T \mathbf{p}$$

for some $\xi \in (0,1)$, where $\mathbf{p} = \mathbf{x} - \mathbf{x}_0$.

Proof idea: We apply the single-variable result and the Chain Rule.

Proof: Let $\phi(t) = f(\boldsymbol{\alpha}(t))$ where $\boldsymbol{\alpha}(t) = \mathbf{x}_0 + t \mathbf{p}$. Observe that $\phi(0) = f(\mathbf{x}_0)$ and $\phi(1) = f(\mathbf{x})$. By the Chain Rule,

$$\phi'(t) = \mathbf{J}_f(\boldsymbol{\alpha}(t)) \,\mathbf{J}_{\boldsymbol{\alpha}}(t) = \nabla f(\boldsymbol{\alpha}(t))^T \mathbf{p} = \nabla f(\mathbf{x}_0 + t \mathbf{p})^T \mathbf{p}.$$

In particular, $\phi$ has a continuous first derivative on $[0,1]$. By the Mean Value Theorem in the single-variable case

$$\phi(t) = \phi(0) + t \phi'(\xi)$$

for some $\xi \in (0,t)$. Plugging in the expressions for $\phi(0)$ and $\phi'(\xi)$ and taking $t=1$ gives the claim. $\square$

We move on to the second-order result. For the more general version, see e.g. Wikipedia.

Theorem (Multivariate Taylor): Let $f : D \to \mathbb{R}$ where $D \subseteq \mathbb{R}^d$. Let $\mathbf{x}_0 \in D$ and $\delta > 0$ be such that $B_\delta(\mathbf{x}_0) \subseteq D$. If $f$ is twice continuously differentiable on $B_\delta(\mathbf{x}_0)$, then for any $\mathbf{x} \in B_\delta(\mathbf{x}_0)$

$$f(\mathbf{x}) = f(\mathbf{x}_0) + \nabla f(\mathbf{x}_0)^T \mathbf{p} + \frac{1}{2} \mathbf{p}^T \,\mathbf{H}_f(\mathbf{x}_0 + \xi \mathbf{p}) \,\mathbf{p}$$

for some $\xi \in (0,1)$, where $\mathbf{p} = \mathbf{x} - \mathbf{x}_0$.

Proof idea: We apply the single-variable result and the Chain Rule.

Proof: Let $\phi(t) = f(\boldsymbol{\alpha}(t))$ where $\boldsymbol{\alpha}(t) = \mathbf{x}_0 + t \mathbf{p}$. Observe that $\phi(0) = f(\mathbf{x}_0)$ and $\phi(1) = f(\mathbf{x})$. As observed in the proof of the Multivariate Mean Value Theorem, $\phi'(t) = \nabla f(\boldsymbol{\alpha}(t))^T \mathbf{p}$. By the Chain Rule,

$$\phi''(t) = \frac{\mathrm{d}}{\mathrm{d} t} \left[\sum_{i=1}^d \frac{\partial f(\boldsymbol{\alpha}(t))}{\partial x_i} p_i \right] = \sum_{i=1}^d \sum_{j=1}^d \frac{\partial^2 f(\boldsymbol{\alpha}(t))}{\partial x_j \partial x_i} p_j p_i = \mathbf{p}^T \,\mathbf{H}_f(\mathbf{x}_0 + t \mathbf{p}) \,\mathbf{p}.$$

In particular, $\phi$ has continuous first and second derivatives on $[0,1]$. By Taylor's Theorem in the single-variable case

$$\phi(t) = \phi(0) + t \phi'(0) + \frac{1}{2} t^2 \phi''(\xi)$$

for some $\xi \in (0,t)$. Plugging in the expressions for $\phi(0)$, $\phi'(0)$ and $\phi''(\xi)$ and taking $t=1$ gives the claim. $\square$

Example: Consider the function $f(x_1, x_2) = x_1 x_2 + x_1^2 + e^{x_1} \cos x_2$. We apply Taylor's Theorem with $\mathbf{x}_0 = (0, 0)$ and $\mathbf{x} = (x_1, x_2)$. The gradient is

$$\nabla f(x_1, x_2) = (x_2 + 2 x_1 + e^{x_1} \cos x_2, x_1 - e^{x_1} \sin x_2 )^T$$

and the Hessian is

$$\mathbf{H}_f(x_1, x_2) = \begin{pmatrix} 2 + e^{x_1} \cos x_2 & 1 - e^{x_1} \sin x_2\\ 1 - e^{x_1} \sin x_2 & - e^{x_1} \cos x_2 \end{pmatrix}.$$

So $f(0,0) = 1$ and $\nabla f(0,0) = (1, 0)^T$. Thus, by the Multivariate Taylor's Theorem, there is $\xi \in (0,1)$ such that

$$f(x_1, x_2) = 1 + x_1 + \frac{1}{2}[2 x_1^2 + 2 x_1 x_2 + (x_1^2 - x_2^2) \,e^{\xi x_1} \cos(\xi x_2) - 2 x_1 x_2 e^{\xi x_1} \sin(\xi x_2)].$$

$\lhd$

### 2.2 Unconstrained optimization¶

We will be interested in unconstrained optimization of the form:

$$\min_{\mathbf{x} \in \mathbb{R}^d} f(\mathbf{x})$$

where $f : \mathbb{R}^d \to \mathbb{R}$. In this subsection, we define several notions of solution and derive characterizations.

#### 2.2.1 Definitions¶

Ideally, we would like to find a global minimizer to the optimization problem above. Recall the definition of a global minimizer.

Definition (Global minimizer): Let $f : \mathbb{R}^d \to \mathbb{R}$. The point $\mathbf{x}^* \in \mathbb{R}^d$ is a global minimizer of $f$ over $\mathbb{R}^d$ if

$$f(\mathbf{x}) \geq f(\mathbf{x}^*), \quad \forall \mathbf{x} \in \mathbb{R}^d.$$

$\lhd$

We have observed before that, in general, finding a global minimizer and certifying that one has been found can be difficult unless some special structure is present. Therefore weaker notions of solution are needed. Recall the notion of a local minimizer.

Definition (Local minimizer): Let $f : \mathbb{R}^d \to \mathbb{R}$. The point $\mathbf{x}^* \in \mathbb{R}^d$ is a local minimizer of $f$ over $\mathbb{R}^d$ if there is $\delta > 0$ such that

$$f(\mathbf{x}) \geq f(\mathbf{x}^*), \quad \forall \mathbf{x} \in B_{\delta}(\mathbf{x}^*) \setminus \{\mathbf{x}^*\}.$$

If the inequality is strict, we say that $\mathbf{x}^*$ is a strict local minimizer. $\lhd$

In words, $\mathbf{x}^*$ is a local minimizer if there is open ball around $\mathbf{x}^*$ where it attains the minimum value. The difference between global and local minimizers is illustrated in the next figure.

(Source)

#### 2.2.2 Necessary conditions¶

Local minimizers can be characterized in terms of the gradient and Hessian of the function.

We first generalize the Descent Direction Lemma to the multivariate case. We first need to define what a descent direction is.

Definition (Descent Direction): Let $f : \mathbb{R}^d \to \mathbb{R}$. A vector $\mathbf{v}$ is a descent direction for $f$ at $\mathbf{x}_0$ if there is $\alpha^* > 0$ such that

$$f(\mathbf{x}_0 + \alpha \mathbf{v}) < f(\mathbf{x}_0), \quad \forall \alpha \in (0,\alpha^*).$$

$\lhd$

In the continuously differentiable case, the directional derivative gives a criterion for descent directions.

Lemma (Descent Direction and Directional Derivative): Let $f : \mathbb{R}^d \to \mathbb{R}$ be continuously differentiable at $\mathbf{x}_0$. A vector $\mathbf{v}$ is a descent direction for $f$ at $\mathbf{x}_0$ if

$$\frac{\partial f (\mathbf{x}_0)}{\partial \mathbf{v}} = \nabla f(\mathbf{x}_0)^T \mathbf{v} < 0$$

that is, if the directional derivative of $f$ at $\mathbf{x}_0$ in the direction $\mathbf{v}$ is negative.

Proof idea: We use the Multivariate Mean Value Theorem to show that $f$ takes smaller values in direction $\mathbf{v}$.

Proof: Suppose there is $\mathbf{v} \in \mathbb{R}^d$ such that $\nabla f(\mathbf{x}_0)^T \mathbf{v} = -\eta < 0$. For $\alpha > 0$, the Multivariate Mean Value Theorem implies that there is $\xi_\alpha \in (0,1)$ such that

$$f(\mathbf{x}_0 + \alpha \mathbf{v}) = f(\mathbf{x}_0) + \nabla f(\mathbf{x}_0 + \xi_\alpha \alpha \mathbf{v})^T(\alpha \mathbf{v}) = f(\mathbf{x}_0) + \alpha \nabla f(\mathbf{x}_0 + \xi_\alpha \alpha \mathbf{v})^T \mathbf{v}.$$

We cannot immediately apply our condition on $\mathbf{v}$ as the gradient in the previous equation is taken at $\mathbf{x}_0 + \xi_\alpha \alpha \mathbf{v}$, not $\mathbf{x}_0$. But the gradient is continuous, so there is $\delta > 0$ such that

$$\left|\nabla f(\mathbf{x})^T \mathbf{v} - \nabla f(\mathbf{x}_0)^T \mathbf{v}\right| < \eta/2$$

for all $\mathbf{x} \in B_\delta(\mathbf{x}_0)$. Hence there is $\alpha^* > 0$ small enough such that

$$\nabla f(\mathbf{x}_0 + \xi_\alpha \alpha \mathbf{v})^T \mathbf{v} < - \eta/2 < 0, \quad \forall \alpha \in (0,\alpha^*).$$

That implies

$$f(\mathbf{x}_0 + \alpha \mathbf{v}) < f(\mathbf{x}_0) - \alpha \eta/2 < f(\mathbf{x}_0), \quad \forall \alpha \in (0,\alpha^*)$$

and proves the claim. $\square$

Lemma (Descent Direction: Multivariate Version): Let $f : \mathbb{R}^d \to \mathbb{R}$ be continuously differentiable at $\mathbf{x}_0$ and assume that $\nabla f(\mathbf{x}_0) \neq 0$. Then $f$ has a descent direction at $\mathbf{x}_0$.

Proof: Take $\mathbf{v} = - \nabla f(\mathbf{x}_0)$. Then $\nabla f(\mathbf{x}_0)^T \mathbf{v} = - \|\nabla f(\mathbf{x}_0)\|^2 < 0$ since $\nabla f(\mathbf{x}_0) \neq 0$. $\square$

This leads to the following fundamental result.

Theorem (First-Order Necessary Condition): Let $f : \mathbb{R}^d \to \mathbb{R}$ be continuously differentiable on $\mathbb{R}^d$. If $\mathbf{x}_0$ is a local minimizer, then $\nabla f(\mathbf{x}_0) = 0$.

Proof idea: In a descent direction, $f$ decreases hence there cannot be one at a local minimizer.

Proof: We argue by contradiction. Suppose that $\nabla f(\mathbf{x}_0) \neq 0$. By the Descent Direction Lemma, there is a descent direction $\mathbf{v} \in \mathbb{R}^d$ at $\mathbf{x}_0$. That implies

$$f(\mathbf{x}_0 + \alpha \mathbf{v}) < f(\mathbf{x}_0), \quad \forall \alpha \in (0, \alpha^*)$$

for some $\alpha^* > 0$. So every open ball around $\mathbf{x}_0$ has a point achieving a smaller value than $f(\mathbf{x}_0)$. Thus $\mathbf{x}_0$ is not a local minimizer, a contradiction. So it must be that $\nabla f(\mathbf{x}_0) = 0$. $\square$

When $f$ is twice continuously differentiable, we also get a condition on the Hessian.

Theorem (Second-Order Necessary Condition): Let $f : \mathbb{R}^d \to \mathbb{R}$ be twice continuously differentiable on $\mathbb{R}^d$. If $\mathbf{x}_0$ is a local minimizer, then $\mathbf{H}_f(\mathbf{x}_0)$ is positive semidefinite.

Proof idea: The proof goes along the same lines as the argument leading to the First-Order Necessary Condition, but we use the Multivariate Taylor's Theorem to the second order instead.

Proof: We argue by contradiction. Suppose that $\mathbf{H}_f(\mathbf{x}_0)$ is not positive semidefinite. From the Symmetry of the Hessian Theorem and the Spectral Theorem, $\mathbf{H}_f(\mathbf{x}_0)$ has a spectral decomposition. From the Characterization of Positive Semidefiniteness, however, it follows that $\mathbf{H}_f(\mathbf{x}_0)$ must have at least one negative eigenvalue $- \eta < 0$. Let $\mathbf{v}$ be a corresponding eigenvector.

We have that $\langle \mathbf{v}, \mathbf{H}_f(\mathbf{x}_0) \,\mathbf{v} \rangle = - \eta < 0$. For $\alpha > 0$, the Multivariate Taylor's Theorem implies that there is $\xi_\alpha \in (0,1)$ such that

\begin{align*} f(\mathbf{x}_0 + \alpha \mathbf{v}) &= f(\mathbf{x}_0) + \nabla f(\mathbf{x}_0)^T(\alpha \mathbf{v}) + (\alpha \mathbf{v})^T \mathbf{H}_f(\mathbf{x}_0 + \xi_\alpha \alpha \mathbf{v}) (\alpha \mathbf{v})\\ &= f(\mathbf{x}_0) + \alpha^2 \mathbf{v}^T \mathbf{H}_f(\mathbf{x}_0 + \xi_\alpha \alpha \mathbf{v})\,\mathbf{v} \end{align*}

where we used $\nabla f(\mathbf{x}_0) = 0$ by the First-Order Necessary Condition.

Since the Hessian is continuous, there is $\delta > 0$ such that

$$\left| \mathbf{v}^T \mathbf{H}_f(\mathbf{x}) \,\mathbf{v} - \mathbf{v}^T \mathbf{H}_f(\mathbf{x}_0) \,\mathbf{v} \right| < \eta/2$$

for all $\mathbf{x} \in B_\delta(\mathbf{x}_0)$. So taking $\alpha$ small enough gives

$$\mathbf{v}^T \mathbf{H}_f(\mathbf{x}_0 + \xi_\alpha \alpha \mathbf{v}) \,\mathbf{v} < - \eta/2 < 0.$$

That implies

$$f(\mathbf{x}_0 + \alpha \mathbf{v}) < f(\mathbf{x}_0) - \alpha^2 \eta/2 < f(\mathbf{x}_0).$$

Since this holds for all sufficiently small $\alpha$, every open ball around $\mathbf{x}_0$ has a point achieving a lower value than $f(\mathbf{x}_0)$. Thus $\mathbf{x}_0$ is not a local minimizer, a contradiction. So it must be that $\mathbf{H}_f(\mathbf{x}_0) \succeq 0$. $\square$

#### 2.2.3 Sufficient conditions¶

The necessary conditions in the previous subsection are not in general sufficient, as the following example shows.

Definition (Stationary Point): Let $f : \mathbb{R}^d \to \mathbb{R}$ be continuously differentiable on $\mathbb{R}^d$. If $\nabla f(\mathbf{x}_0) = 0$, we say that $\mathbf{x}_0$ is a stationary point of $f$.$\lhd$

Example: Let $f(x) = x^3$. Then $f'(x) = 3 x^2$ and $f''(x) = 6 x$ so that $f'(0) = 0$ and $f''(0) \geq 0$. Hence $x=0$ is a stationary point. But $x=0$ is not a local minimizer. Indeed $f(0) = 0$ but, for any $\delta > 0$, $f(-\delta) < 0$.

In [4]:
f(x) = x^3
x = LinRange(-2,2, 100)
y = f.(x)
plot(x, y, lw=2, legend=false, ylim = (-5,5))

Out[4]:

$\lhd$

We give sufficient conditions for a local minimizer. We will need the following lemma, whose proof relies on the next exercises.

Exercise: Prove the following claim, which is known as the Subspace Intersection Lemma. Let $\mathcal{S}_1$ and $\mathcal{S}_2$ be linear subspaces of $\mathbb{R}^d$ and let

$$\mathcal{S}_1 + \mathcal{S}_2 = \{\mathbf{x}_1 + \mathbf{x}_2 \,:\, \forall \mathbf{x}_1 \in \mathcal{S}_1, \mathbf{x}_2 \in \mathcal{S}_2\}.$$

Then it holds that

$$\mathrm{dim}(\mathcal{S}_1 + \mathcal{S}_2) = \mathrm{dim}(\mathcal{S}_1) + \mathrm{dim}(\mathcal{S}_2) - \mathrm{dim}(\mathcal{S}_1 \cap \mathcal{S}_2).$$

[Hint: Consider a basis of $\mathcal{S}_1 \cap \mathcal{S}_2$ and complete into bases of $\mathcal{S}_1$ and $\mathcal{S}_2$. Show that the reuslting list of vectors is linear independent.] $\lhd$

Exercise: Show that, for any linear subspaces $\mathcal{S}_1, \ldots, \mathcal{S}_m$ of $\mathcal{V} = \mathbb{R}^d$, it holds that

$$\mathrm{dim}\left(\bigcap_{k=1}^m \mathcal{S}_k\right) \geq \sum_{k=1}^m \mathrm{dim}\left(\mathcal{S}_k\right) - (m-1) \,\mathrm{dim}(\mathcal{V}).$$

[Hint: Use the Subspace Intersection Lemma and induction.] $\lhd$

For a symmetric matrix $C \in \mathbb{R}^{d \times d}$, we let $\lambda_j(C)$, $j=1, \ldots, d$, be the eigenvalues of $C$ in non-increasing order with corresponding orthonormal eigenvectors $\mathbf{v}_j$, $j=1, \ldots, d$. Define the subspaces

$$\mathcal{V}_k(C) = \mathrm{span}(\mathbf{v}_1, \ldots, \mathbf{v}_k) \quad\text{and}\quad \mathcal{W}_{d-k+1}(C) = \mathrm{span}(\mathbf{v}_k, \ldots, \mathbf{v}_d).$$

The following lemma is one version of what is known as Weyl's Inequality.

Lemma (Weyl's Inequality): Let $A \in \mathbb{R}^{d \times d}$ and $B \in \mathbb{R}^{d \times d}$ be symmetric matrices. Then, for all $j=1, \ldots, d$,

$$\max_{j \in [d]} \left|\lambda_j(B) - \lambda_j(A)\right| \leq \|B- A\|_2$$

where $\|C\|_2$ is the induced $2$-norm of $C$.

Proof idea: We use the extremal characterization of the eigenvalues together with a dimension argument.

Proof: Let $H = B - A$. We prove only the upper bound. The other direction follows from interchanging the roles of $A$ and $B$. Because

$$\mathrm{dim}(\mathcal{V}_j(B)) + \mathrm{dim}(\mathcal{W}_j(A)) + \mathrm{dim}(\mathcal{W}_1(H)) = j + (d-j+1) + d = 2d+1$$

it follows from the exercise above that

$$\mathrm{dim}\left(\mathcal{V}_j(B) \cap \mathcal{W}_j(A) \cap \mathcal{W}_1(H)\right) \geq (2d+1) - 2d = 1.$$

Hence the $\mathcal{V}_j(B) \cap \mathcal{W}_j(A) \cap \mathcal{W}_1(H)$ is non-empty. Let $\mathbf{v}$ be a unit vector in that intersection.

By Courant-Fischer,

$$\lambda_j(B) \leq \langle \mathbf{v}, (A+H) \mathbf{v}\rangle = \langle \mathbf{v}, A \mathbf{v}\rangle + \langle \mathbf{v}, H \mathbf{v}\rangle \leq \lambda_j(A) + \langle \mathbf{v}, H \mathbf{v}\rangle.$$

Moreover, by Cauchy-Schwarz, since $\|\mathbf{v}\|=1$

$$\langle \mathbf{v}, H \mathbf{v}\rangle \leq \|\mathbf{v}\| \|H\mathbf{v}\| \leq \|H\|_2$$

which proves the claim. $\square$

Theorem (Second-Order Sufficient Condition): Let $f : \mathbb{R}^d \to \mathbb{R}$ be twice continuously differentiable on $\mathbb{R}^d$. If $\nabla f(\mathbf{x}_0) = \mathbf{0}$ and $\mathbf{H}_f(\mathbf{x}_0)$ is positive definite, then $\mathbf{x}_0$ is a strict local minimizer.

Proof idea: We use the Multivariate Taylor's Theorem again. This time we use the positive definiteness of the Hessian to bound the value of the funciton from below. We use Weyl's Inequality to show that the eigenvalues of the Hessian are continuous, which implies that the Hessian remains positive definite in an open ball around $\mathbf{x}_0$.

Proof: We first claim that there is $\rho > 0$ such that $\mathbf{H}_f(\mathbf{x})$ is positive definite for all $\mathbf{x} \in B_{\rho}(\mathbf{x}_0)$. That is the case when $\mathbf{x} = \mathbf{x}_0$ and let $\mu_1 > 0$ be the smallest eigenvalue $\mathbf{H}_f(\mathbf{x}_0)$. We use Weyl's Inequality to bound the eigenvalues of the Hessian from below around $\mathbf{x}_0$. For any vector $\mathbf{v} \in \mathbb{R}^d$, we have for all $j =1, \ldots, d$

$$\lambda_j(\mathbf{H}_f(\mathbf{x}_0 + \mathbf{v})) \geq \lambda_j(\mathbf{H}_f(\mathbf{x}_0)) - \|\mathbf{H}_f(\mathbf{x}_0 + \mathbf{v}) - \mathbf{H}_f(\mathbf{x}_0)\|_2$$

where we used the notation introduced above. We bound $\lambda_j(\mathbf{H}_f(\mathbf{x}_0)) \geq \lambda_1(\mathbf{H}_f(\mathbf{x}_0)) = \mu_1$ and

$$\|\mathbf{H}_f(\mathbf{x}_0 + \mathbf{v}) - \mathbf{H}_f(\mathbf{x}_0)\|_2 \leq \|\mathbf{H}_f(\mathbf{x}_0 + \mathbf{v}) - \mathbf{H}_f(\mathbf{x}_0)\|_F.$$

The Frobenius norm above is continuous in $\mathbf{v}$ as a composition of continuous functions. Moreover, we of course have at $\mathbf{v} = \mathbf{0}$ that this Frobenius is $0$. Hence, by definition of continuity, there is $\rho > 0$ such that for all $\mathbf{v} \in B_{\rho}(\mathbf{0})$ we have $\|\mathbf{H}_f(\mathbf{x}_0 + \mathbf{v}) - \mathbf{H}_f(\mathbf{x}_0)\|_F < \mu_1/2$. Plugging back above finally gives for such $\mathbf{v}$'s and all $j=1,\ldots,d$

$$\lambda_j(\mathbf{H}_f(\mathbf{x}_0 + \mathbf{v})) > \mu_1/2 > 0.$$

In particular, $\mathbf{H}_f(\mathbf{x}_0 + \mathbf{v})$ is positive definite and $\langle \mathbf{u}, \mathbf{H}_f(\mathbf{x}_0 + \mathbf{v}) \,\mathbf{u} \rangle > \frac{\mu_1}{2} \|\mathbf{u}\|^2$ for any $\mathbf{u}$ by Courant-Fischer.

By the Multivariate Taylor's Theorem, $\forall \mathbf{v} \in B_{\rho}(\mathbf{0}) \setminus \{\mathbf{0}\}$ there is $\xi \in (0,1)$

\begin{align*} f(\mathbf{x}_0 + \mathbf{v}) &= f(\mathbf{x}_0) + \nabla f(\mathbf{x}_0)^T \mathbf{v} + \mathbf{v} \,\mathbf{H}_f(\mathbf{x}_0 + \xi \mathbf{v}) \,\mathbf{v}\\ &> f(\mathbf{x}_0) + \frac{\mu_1}{2} \|\mathbf{v}\|^2\\ &> f(\mathbf{x}_0) \end{align*}

where we used that $\|\xi \mathbf{v}\| = \xi \|\mathbf{v}\| \leq \rho$. Therefore $\mathbf{x}_0$ is a strict local minimizer. $\square$