[proofplan] We reduce to the scalar case ($n = 1$) by [Componentwise Differentiability](/theorems/324). For scalar $f$, we analyse the second-order difference $\Delta(s,t) = f(a + se_i + te_j) - f(a + se_i) - f(a + te_j) + f(a)$ over a small rectangle. Applying the one-dimensional [Mean Value Theorem](/theorems/186) twice -- first in $e_i$ then $e_j$ -- expresses $\Delta(s,t)$ in terms of $f''$ at an intermediate point. Applying it in the reverse order gives the same $\Delta(s,t)$ with swapped arguments. [Continuity](/page/Continuity) of $f''$ at $a$ then forces equality of the two orderings in the limit. [/proofplan] [step:Reduce to the scalar case and set up the second-difference quotient] By [Componentwise Differentiability](/theorems/324), it suffices to prove symmetry for scalar-valued [functions](/page/Function) ($n = 1$). By bilinearity, it suffices to show $f''(a)(e_i, e_j) = f''(a)(e_j, e_i)$ for all $i \neq j$ (the case $i = j$ is automatic). For small $s, t > 0$, define the second difference: \begin{align*} \Delta(s, t) = f(a + se_i + te_j) - f(a + se_i) - f(a + te_j) + f(a). \end{align*} [/step] [step:Apply the Mean Value Theorem twice: first in $e_i$, then in $e_j$] Define $\varphi(u) = f(a + ue_i + te_j) - f(a + ue_i)$, so $\Delta(s,t) = \varphi(s) - \varphi(0)$. By the [Mean Value Theorem](/theorems/186), there exists $\sigma \in (0, s)$ with \begin{align*} \Delta(s, t) = s \, \varphi'(\sigma) = s \bigl[D_i f(a + \sigmae_i + te_j) - D_i f(a + \sigmae_i)\bigr]. \end{align*} Define $\psi(v) = D_i f(a + \sigmae_i + ve_j)$, so the bracket is $\psi(t) - \psi(0)$. Applying the Mean Value Theorem again, there exists $\tau \in (0, t)$ with \begin{align*} \Delta(s, t) = st \, D_j D_i f(a + \sigmae_i + \taue_j). \end{align*} By the [Second Derivative and Iterated Partial Derivatives](/theorems/331) theorem, $D_j D_i f = f''(\cdot)(e_j, e_i)$, so \begin{align*} \Delta(s, t) = st \, f''(a + \sigmae_i + \taue_j)(e_j, e_i). \end{align*} [/step] [step:Apply the Mean Value Theorem in the reverse order: first in $e_j$, then in $e_i$] Repeating the argument with the roles of $i$ and $j$ swapped -- first differencing in $e_j$, then differencing the result in $e_i$ -- the same quantity $\Delta(s,t)$ equals \begin{align*} \Delta(s, t) = st \, f''(a + \sigma'e_i + \tau'e_j)(e_i, e_j) \end{align*} for some $\sigma' \in (0, s)$, $\tau' \in (0, t)$. [/step] [step:Take the limit $s, t \to 0^+$ using continuity of $f''$] From the two expressions for $\Delta(s,t)$: \begin{align*} f''(a + \sigmae_i + \taue_j)(e_j, e_i) = f''(a + \sigma'e_i + \tau'e_j)(e_i, e_j). \end{align*} As $s, t \to 0^+$, all intermediate points converge to $a$ (since $0 < \sigma, \sigma' < s$ and $0 < \tau, \tau' < t$). By [continuity](/page/Continuity) of $f''$ at $a$: \begin{align*} f''(a)(e_j, e_i) = f''(a)(e_i, e_j). \end{align*} [guided] The second difference $\Delta(s,t) = f(a + se_i + te_j) - f(a + se_i) - f(a + te_j) + f(a)$ is a single quantity that can be computed by differencing in either order. This is the key: regardless of whether we first vary $e_i$ then $e_j$, or vice versa, we get the same number $\Delta(s,t)$. **Order 1 ($e_i$ then $e_j$):** Two applications of the [Mean Value Theorem](/theorems/186) give $\Delta(s,t) = st \cdot f''(a + \sigmae_i + \taue_j)(e_j, e_i)$ for some $\sigma \in (0,s)$, $\tau \in (0,t)$. **Order 2 ($e_j$ then $e_i$):** The same argument with swapped roles gives $\Delta(s,t) = st \cdot f''(a + \sigma'e_i + \tau'e_j)(e_i, e_j)$ for some $\sigma' \in (0,s)$, $\tau' \in (0,t)$. Equating and dividing by $st > 0$: $f''(a + \sigmae_i + \taue_j)(e_j, e_i) = f''(a + \sigma'e_i + \tau'e_j)(e_i, e_j)$. Both intermediate points lie in the rectangle $a + [0,s]e_i + [0,t]e_j$, which shrinks to $\{a\}$ as $s, t \to 0^+$. The [continuity](/page/Continuity) hypothesis on $f''$ is consumed precisely here: it ensures that $f''$ at any point in the shrinking rectangle converges to $f''(a)$. Taking $s, t \to 0^+$ gives $f''(a)(e_j, e_i) = f''(a)(e_i, e_j)$. Without continuity of $f''$, the intermediate points might approach $a$ along different paths, and $f''$ at these points need not have the same limit. This is why mere existence of second derivatives does not suffice for symmetry. [/guided] [/step]