Let $f: U \subset \mathbb{R}^m \rightarrow \mathbb{R}^n$ twice differentiable at the point $a \in U$. suppose that $\lim_{k \rightarrow \infty} t_k=0$ in $\mathbb{R}$ and that $\lim_{k \rightarrow \infty} v_k=v$; $\lim_{k \rightarrow \infty} w_k=w$ prove that $$f''(a) vw= \lim_{k \rightarrow \infty} \frac{f(a+t_k(v_k+w_k))-f(a+t_kv_k)-f(a+t_kw_k)+f(a)}{t_k^2}$$

We know that since $f$ is differentiable there is a linear transformation $$T_a: \mathbb{R}^m \rightarrow \mathbb{R }^n$$ such that $f(a+h)-f(a)=T_a h + r(h)$ where $\lim_{h \rightarrow \infty} \frac{\|r(h)\|}{\|h\|}=0$, I don't know how to proceed, any idea please.

  • $\begingroup$ Note that you need to use of definition of differentiable function in $(f')$. $\endgroup$ – Александр Пальма Aug 25 at 23:31

Taylor gives $f(x+h) =f(x) + f'(x)h + {1 \over 2} h^T f''(x)h + r(h)$, where $\lim_{h \to 0} {r(h) \over \|h\|^2} = 0$.

Then $f(a+t_k(v_k+w_k)) = f(a) + f'(a) t_k(v_k+w_k) +{1 \over 2} t_k^2 (v_k+w_k)^T f''(x) (v_k+w_k) + r(t_k(v_k+w_k)) $

$f(a+t_k v_k ) = f(a) + f'(a) t_k v_k +{1 \over 2} t_k^2 v_k^T f''(x) v_k + r(t_k v_k) $

$f(a+t_k w_k ) = f(a) + f'(a) t_k w_k +{1 \over 2} t_k^2 w_k^T f''(x) w_k + r(t_k w_k) $

and so ${f(a+t_k(v_k+w_k)) - f(a+t_k v_k ) - f(a+t_k w_k ) + f(a) \over t_k^2 } = v_k^T f''(a) w_k + { r(t_k(v_k+w_k)) - r(t_k v_k) - r(t_k w_k) \over t_k^2}$, and taking limits gives $v^T f''(a)w$.

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