[proofplan] We prove each part in sequence. For Part 1, we expand $x$ in the standard basis, apply linearity of $\tau$, then use the triangle inequality followed by the Cauchy--Schwarz inequality to obtain the bound $|\tau(x)| \leq \|\tau\| \cdot |x|$. Part 2 follows by applying Part 1 twice (once for $\sigma$, once for $\tau$) and computing the Frobenius norm of the composition via basis images. Part 3 reduces Lipschitz continuity to Part 1 using linearity. [/proofplan] [step:Bound $|\tau(x)|$ by $\|\tau\| \cdot |x|$ via Cauchy--Schwarz] Write $x = \sum_{i=1}^m x_i e_i$. By linearity of $\tau$, we have $\tau(x) = \sum_{i=1}^m x_i \tau(e_i)$. Applying the triangle inequality and then the Cauchy--Schwarz inequality to the sequences $(|x_1|, \ldots, |x_m|)$ and $(|\tau(e_1)|, \ldots, |\tau(e_m)|)$: \begin{align*} |\tau(x)| = \left|\sum_{i=1}^m x_i \tau(e_i)\right| \leq \sum_{i=1}^m |x_i| |\tau(e_i)| \leq \left(\sum_{i=1}^m x_i^2\right)^{1/2} \left(\sum_{i=1}^m |\tau(e_i)|^2\right)^{1/2} = |x| \cdot \|\tau\|. \end{align*} [guided] We want to show that $\tau$ cannot stretch any vector by more than a factor of $\|\tau\|$. The strategy is to decompose $x$ into its basis components and exploit linearity. Write $x = \sum_{i=1}^m x_i e_i$ where $(e_1, \ldots, e_m)$ is the standard basis of $\mathbb{R}^m$. By linearity of $\tau \in \mathcal{L}(\mathbb{R}^m, \mathbb{R}^n)$: \begin{align*} \tau(x) = \tau\left(\sum_{i=1}^m x_i e_i\right) = \sum_{i=1}^m x_i \tau(e_i). \end{align*} Applying the triangle inequality for the norm on $\mathbb{R}^n$: \begin{align*} |\tau(x)| = \left|\sum_{i=1}^m x_i \tau(e_i)\right| \leq \sum_{i=1}^m |x_i| |\tau(e_i)|. \end{align*} Now we apply the Cauchy--Schwarz inequality in $\mathbb{R}^m$: for real sequences $(a_i)$ and $(b_i)$, $\sum_{i=1}^m a_i b_i \leq (\sum a_i^2)^{1/2}(\sum b_i^2)^{1/2}$. Setting $a_i = |x_i|$ and $b_i = |\tau(e_i)|$: \begin{align*} \sum_{i=1}^m |x_i| |\tau(e_i)| \leq \left(\sum_{i=1}^m x_i^2\right)^{1/2} \left(\sum_{i=1}^m |\tau(e_i)|^2\right)^{1/2} = |x| \cdot \|\tau\|. \end{align*} The last equality uses the definition of the Euclidean norm $|x| = (\sum x_i^2)^{1/2}$ and the Frobenius norm $\|\tau\| = (\sum |\tau(e_i)|^2)^{1/2}$. [/guided] [/step] [step:Prove submultiplicativity $\|\sigma \circ \tau\| \leq \|\sigma\| \cdot \|\tau\|$] For any $x \in \mathbb{R}^m$, applying Part 1 first to $\sigma$ and then to $\tau$: \begin{align*} |(\sigma \circ \tau)(x)| = |\sigma(\tau(x))| \leq \|\sigma\| \cdot |\tau(x)| \leq \|\sigma\| \cdot \|\tau\| \cdot |x|. \end{align*} Computing the Frobenius norm of $\sigma \circ \tau$ directly from its definition: \begin{align*} \|\sigma \circ \tau\|^2 = \sum_{i=1}^m |(\sigma \circ \tau)(e_i)|^2 = \sum_{i=1}^m |\sigma(\tau(e_i))|^2 \leq \sum_{i=1}^m \|\sigma\|^2 |\tau(e_i)|^2 = \|\sigma\|^2 \|\tau\|^2. \end{align*} Taking square roots gives $\|\sigma \circ \tau\| \leq \|\sigma\| \cdot \|\tau\|$. [/step] [step:Deduce Lipschitz continuity from the operator bound and linearity] For any $x, y \in \mathbb{R}^m$, linearity of $\tau$ gives $\tau(x) - \tau(y) = \tau(x - y)$. Applying Part 1: \begin{align*} |\tau(x) - \tau(y)| = |\tau(x - y)| \leq \|\tau\| \cdot |x - y|. \end{align*} This is precisely the [Lipschitz condition](/page/Lipschitz%20Continuity) with Lipschitz constant $\|\tau\|$. Since every Lipschitz [function](/page/Function) is [continuous](/page/Continuity), $\tau$ is continuous. [/step]