[proofplan] We first recall the finite-dimensional symmetric-matrix fact that the operator norm of $A$ is the supremum of the absolute quadratic form $|x^\top A x|$ over the unit sphere. We then approximate an arbitrary unit vector $v$ by a point $u$ in the $1/4$-net and compare the two quadratic forms. The comparison error is controlled by $\|A\|_{\mathrm{op}}|u-v|$ twice, giving an error at most $\|A\|_{\mathrm{op}}/2$, and taking the supremum over $v$ yields the desired bound. [/proofplan] [step:Identify the spectral norm with the largest absolute quadratic form] Define the function $q : S^{d-1} \to \mathbb{R}$ by $q(x) = x^\top A x$ for every $x \in S^{d-1}$. Since $A$ is symmetric, the real spectral theorem for finite-dimensional symmetric matrices gives an [orthonormal basis](/page/Orthonormal%20Basis) $e_1,\dots,e_d$ of $\mathbb{R}^d$ and [real numbers](/page/Real%20Numbers) $\lambda_1,\dots,\lambda_d \in \mathbb{R}$ such that $A e_i = \lambda_i e_i$ for each $i \in \{1,\dots,d\}$. For any $x \in S^{d-1}$, write \begin{align*} x = \sum_{i=1}^d c_i e_i, \end{align*} where $c_i \in \mathbb{R}$ and $\sum_{i=1}^d c_i^2 = 1$. Then \begin{align*} x^\top A x = \sum_{i=1}^d \lambda_i c_i^2, \end{align*} so \begin{align*} |x^\top A x| \le \max_{1 \le i \le d} |\lambda_i| \sum_{i=1}^d c_i^2 = \max_{1 \le i \le d} |\lambda_i|. \end{align*} Taking $x=e_j$ for an index $j \in \{1,\dots,d\}$ satisfying $|\lambda_j|=\max_{1 \le i \le d}|\lambda_i|$ gives equality in the supremum. Also, \begin{align*} \|A\|_{\mathrm{op}} = \sup_{x \in S^{d-1}} |Ax| = \max_{1 \le i \le d} |\lambda_i|. \end{align*} Therefore \begin{align*} \|A\|_{\mathrm{op}} = \sup_{x \in S^{d-1}} |x^\top A x|. \end{align*} [/step] [step:Compare the quadratic form at a unit vector with its net approximation] Define \begin{align*} M := \max_{u \in N} |u^\top A u|. \end{align*} The maximum exists because $N$ is finite. Let $v \in S^{d-1}$. Since $N$ is a $1/4$-net of $S^{d-1}$, choose $u \in N$ such that $|u-v| \le 1/4$. Since $u,v \in S^{d-1}$, both vectors have Euclidean norm $1$. Using bilinearity of the matrix product and symmetry only through the operator-norm estimate, we compute \begin{align*} v^\top A v - u^\top A u &= (v-u)^\top A v + u^\top A(v-u). \end{align*} Taking absolute values and applying the [Cauchy-Schwarz inequality](/theorems/432) in $\mathbb{R}^d$ to each term gives \begin{align*} |v^\top A v - u^\top A u| \le |v-u|\,|Av| + |u|\,|A(v-u)|. \end{align*} By the definition of the operator norm applied first to $v$ and then to $v-u$, \begin{align*} |v-u|\,|Av| + |u|\,|A(v-u)| \le |v-u|\,\|A\|_{\mathrm{op}}|v| + |u|\,\|A\|_{\mathrm{op}}|v-u|. \end{align*} Since $|u|=|v|=1$, this becomes \begin{align*} |v-u|\,\|A\|_{\mathrm{op}}|v| + |u|\,\|A\|_{\mathrm{op}}|v-u| = 2\|A\|_{\mathrm{op}}|v-u|. \end{align*} Using $|v-u| \le 1/4$, we obtain \begin{align*} |v^\top A v - u^\top A u| \le \frac{1}{2}\|A\|_{\mathrm{op}}. \end{align*} Therefore \begin{align*} |v^\top A v| \le |u^\top A u| + \frac{1}{2}\|A\|_{\mathrm{op}} \le M + \frac{1}{2}\|A\|_{\mathrm{op}}. \end{align*} [guided] Fix a unit vector $v \in S^{d-1}$. The role of the net is to replace $v$ by a nearby vector from the finite set $N$, where the quadratic form is one of the quantities appearing in $M$. By the $1/4$-net property, there exists $u \in N$ with $|u-v| \le 1/4$. Since $N \subset S^{d-1}$, we also have $|u|=1$, and because $v \in S^{d-1}$, we have $|v|=1$. We compare $q(v)=v^\top A v$ and $q(u)=u^\top A u$ by adding and subtracting the mixed expression $u^\top A v$: \begin{align*} v^\top A v - u^\top A u = v^\top A v - u^\top A v + u^\top A v - u^\top A u. \end{align*} Grouping the first two terms and the last two terms gives \begin{align*} v^\top A v - u^\top A u = (v-u)^\top A v + u^\top A(v-u). \end{align*} This decomposition is useful because each term contains the small vector $v-u$. Taking absolute values and using the triangle inequality gives \begin{align*} |v^\top A v - u^\top A u| \le |(v-u)^\top A v| + |u^\top A(v-u)|. \end{align*} Apply the Cauchy-Schwarz inequality in the Euclidean [inner product](/page/Inner%20Product) on $\mathbb{R}^d$ to the first term: \begin{align*} |(v-u)^\top A v| \le |v-u|\,|Av|. \end{align*} Applying the same inequality to the second term gives \begin{align*} |u^\top A(v-u)| \le |u|\,|A(v-u)|. \end{align*} By the definition of the operator norm, $|Aw| \le \|A\|_{\mathrm{op}}|w|$ for every $w \in \mathbb{R}^d$. Applying this first with $w=v$ and then with $w=v-u$ gives \begin{align*} |v^\top A v - u^\top A u| \le |v-u|\,\|A\|_{\mathrm{op}}|v| + |u|\,\|A\|_{\mathrm{op}}|v-u|. \end{align*} Because $|u|=|v|=1$, the right-hand side is \begin{align*} 2\|A\|_{\mathrm{op}}|v-u|. \end{align*} Since $|u-v|\le 1/4$, we obtain \begin{align*} |v^\top A v - u^\top A u| \le \frac{1}{2}\|A\|_{\mathrm{op}}. \end{align*} Finally, the [reverse triangle inequality](/theorems/2300) in the form $|a|\le |b|+|a-b|$ gives \begin{align*} |v^\top A v| \le |u^\top A u| + |v^\top A v-u^\top A u| \le M + \frac{1}{2}\|A\|_{\mathrm{op}}. \end{align*} [/guided] [/step] [step:Take the supremum and absorb the error term] The preceding estimate holds for every $v \in S^{d-1}$. Taking the supremum over $v \in S^{d-1}$ gives \begin{align*} \sup_{v \in S^{d-1}} |v^\top A v| \le M + \frac{1}{2}\|A\|_{\mathrm{op}}. \end{align*} Using the identity from the first step, \begin{align*} \|A\|_{\mathrm{op}} \le M + \frac{1}{2}\|A\|_{\mathrm{op}}. \end{align*} Subtracting $\frac{1}{2}\|A\|_{\mathrm{op}}$ from both sides yields \begin{align*} \frac{1}{2}\|A\|_{\mathrm{op}} \le M. \end{align*} Multiplying by $2$ gives \begin{align*} \|A\|_{\mathrm{op}} \le 2M = 2\max_{u \in N}|u^\top A u|. \end{align*} This is the desired estimate. [/step]