[proofplan] We use the canonical identification between $\mathcal{L}(\mathbb{R}^m, \mathcal{L}(\mathbb{R}^m, \mathbb{R}^n))$ and $\mathrm{Bil}(\mathbb{R}^m \times \mathbb{R}^m, \mathbb{R}^n)$. The forward direction evaluates the [differentiability](/page/Derivative) expansion of $Df$ at a fixed vector $k$ to extract the bilinear map $T(h, k) = [f''(a)(h)](k)$. The reverse direction defines the linear operator $\tau(h) = T(h, \cdot)$ and verifies the operator norm of the error vanishes using the Frobenius norm over basis vectors. [/proofplan] [step:Forward direction: extract the bilinear map from the second derivative] Suppose $f$ is twice [differentiable](/page/Derivative) at $a$, so $Df: V \to \mathcal{L}(\mathbb{R}^m, \mathbb{R}^n)$ is differentiable at $a$ with derivative $f''(a) \in \mathcal{L}(\mathbb{R}^m, \mathcal{L}(\mathbb{R}^m, \mathbb{R}^n))$. The differentiability expansion of $Df$ reads: \begin{align*} Df_{a + h} = Df_a + f''(a)(h) + |h|E(h), \end{align*} where $E(h) \in \mathcal{L}(\mathbb{R}^m, \mathbb{R}^n)$ and $\|E(h)\|_{\mathrm{op}} \to 0$ as $h \to \mathbf{0}$. Evaluating both sides at a fixed $k \in \mathbb{R}^m$: \begin{align*} Df_{a + h}(k) = Df_a(k) + [f''(a)(h)](k) + |h|E(h)(k). \end{align*} Define $T: \mathbb{R}^m \times \mathbb{R}^m \to \mathbb{R}^n$ by $T(h, k) = [f''(a)(h)](k)$. This is bilinear: linear in $h$ because $f''(a)$ is a [linear map](/page/Linear%20Map), and linear in $k$ because each $f''(a)(h) \in \mathcal{L}(\mathbb{R}^m, \mathbb{R}^n)$ is linear. The error satisfies $|\varepsilon(h, k)| = |E(h)(k)| \leq \|E(h)\|_{\mathrm{op}} \cdot |k| \to 0$ as $h \to \mathbf{0}$ for each fixed $k$. [/step] [step:Reverse direction: reconstruct the second derivative from the bilinear expansion] Suppose the bilinear expansion holds with $T \in \mathrm{Bil}(\mathbb{R}^m \times \mathbb{R}^m, \mathbb{R}^n)$. For each $h$, the map $k \mapsto T(h, k)$ is linear, so define $\tau(h) \in \mathcal{L}(\mathbb{R}^m, \mathbb{R}^n)$ by $\tau(h)(k) = T(h, k)$. By bilinearity in $h$, the map $h \mapsto \tau(h)$ is linear, so $\tau \in \mathcal{L}(\mathbb{R}^m, \mathcal{L}(\mathbb{R}^m, \mathbb{R}^n))$. The expansion becomes: \begin{align*} Df_{a + h} = Df_a + \tau(h) + |h|E(h), \end{align*} where $E(h)$ is the operator $k \mapsto \varepsilon(h, k)$. We must verify $\|E(h)\|_{\mathrm{op}} \to 0$. Computing via the Frobenius norm (which dominates the operator norm): \begin{align*} \|E(h)\|^2 = \sum_{j=1}^m |E(h)(e_j)|^2 = \sum_{j=1}^m |\varepsilon(h, e_j)|^2 \to 0, \end{align*} since $\varepsilon(h, e_j) \to \mathbf{0}$ for each basis vector $e_j$. Therefore $Df$ is [differentiable](/page/Derivative) at $a$ with $f''(a) = \tau$, and $T(h, k) = f''(a)(h, k)$. [guided] The key insight is the canonical identification between iterated linear maps and bilinear maps. An element $\tau \in \mathcal{L}(\mathbb{R}^m, \mathcal{L}(\mathbb{R}^m, \mathbb{R}^n))$ sends $h$ to a linear operator $\tau(h)$, which then acts on $k$. The result $[\tau(h)](k)$ depends on both variables and is bilinear: linear in $h$ because $\tau$ is linear, and linear in $k$ because each $\tau(h)$ is a linear map. The forward direction is straightforward: evaluate the differentiability expansion at a fixed $k$ to extract the bilinear map $T(h, k) = [f''(a)(h)](k)$. The reverse direction is more subtle. We know $\varepsilon(h, k) \to \mathbf{0}$ for each fixed $k$, but we need $\|E(h)\|_{\mathrm{op}} \to 0$ where $E(h)$ is the operator $k \mapsto \varepsilon(h, k)$. How do we pass from pointwise to norm convergence? In finite dimensions, it suffices to check convergence on the standard basis $e_1, \ldots, e_m$. The Frobenius norm satisfies $\|E(h)\|_{\mathrm{op}} \leq \|E(h)\|_F = \bigl(\sum_{j=1}^m |E(h)(e_j)|^2\bigr)^{1/2}$. Each term $|E(h)(e_j)|^2 = |\varepsilon(h, e_j)|^2 \to 0$ as $h \to \mathbf{0}$. The finite sum of $m$ terms each tending to $0$ also tends to $0$. This is where finite-dimensionality is essential: in infinite dimensions, pointwise convergence on a basis does not imply norm convergence (the sum could be infinite). [/guided] [/step]