Conjecture 1: Let $f : \mathbb{R}^n\to \mathbb{R}$ be a continuously differentiable function and $a, b\in\mathbb{R}^n$ such that $a_i \leq b_i$ for each $i\in \{1, 2, \ldots, n\}$. Define $v = \nabla f(a)$. If $f(a) \leq f(b)$ and $v_k < 0$ for some $k\in \{1, 2, \ldots, n\}$ is it true that $f$ obtains a minimum on $(a,b)$?
My idea for a proof:
Assume for contradiction that $f$ does not obtain a minimum on $(a,b)$. By the Extreme Value Theorem $f$ obtains a minimum on $[a,b]$ implying that $a$ is the minimizer of $f$ over $[a,b]$ (or both $a$ and $b$ are minimzers if $f(a) = f(b)$). Since $v_k < 0$ and $\nabla f$ is continuous then there exists an $\epsilon > 0$ such that $a+\epsilon e_k\in [a,b]$ and $f(a+\epsilon e_k) < f(a)$ where $e_k$ is the $k$th standard basis vector of $\mathbb{R}^n.$ Implying that $a$ is not a minizer of $f$ over $[a,b]$, which is a contradiction. Therefore, $f$ obtains a minimum on $(a,b)$.
Are there any errors in this proof I may have overlooked?
Edited: Conjecture 1 is true for $n = 1$ (that is $f'(a) < 0, f(a)\leq f(b)$, and continuous derivative imply $f(x)$ will obtain a local min on $(a,b)$) so the attempt was to expand this idea to higher dimensions. But, Conjecture 1 is certainly not true for all $n > 1$. As a counter example consider $f(x_1, x_2, \ldots, x_n) = x_1^2-x_1^2-\cdots-x_n^2$. Then \begin{align*} \nabla f &= \begin{bmatrix} 2x_1 & -2x_2 & \cdots & -2x_n \end{bmatrix}^T \\ H_f &= \text{diag}(2,-2,\ldots,-2). \end{align*} Let $a = (-1, 1, \ldots, 1)$ and $b = (1,0,\ldots,0)$. Then $f(a) = -(n-2)<1 = f(b)$ and $[\nabla f(a)]_k = -2 < 0$, but the origin is the only critical point of $f$ and it is a saddle point. So, $f$ does not obtain a minimum on $(a,b)$.
Thanks to a comment by Doug M, I believe a better extension of the 1d idea is the following:
Conjecture 2: Let $f : \mathbb{R}^n\to \mathbb{R}$ be a continuously differentiable function and $a, b\in\mathbb{R}^n$ such that $a_i \leq b_i$ for each $i\in \{1, 2, \ldots, n\}$. Let $\mathcal{A}_i = \{x\in [a,b]: x_i = a_i\}$ and $\mathcal{B}_i = \{x\in[a,b]: x_i = b_i\}$. Assume for each $i\in \{1, 2, \ldots, n\}$ that $\underset{x\in \mathcal{A}_i}{\min} f(x) \leq \underset{x\in \mathcal{B}_i}{\min} f(x)$ and $[\nabla f(x)]_i < 0$ for every $x\in \mathcal{A}_i$. Then $f$ obtains a minimum on $(a,b)$.
Proof. Assume for contradiction that $f$ does not obtain a minimum on $(a,b)$. By the Extreme Value Theorem, $f$ obtains a minimum on $[a,b]$ implying that a minimum occurs on the boundary. Hence, there exists a $k$ such that a minimum of $f$ occurs at $x_{\text{min}}\in\mathcal{A}_k$. Since $[\nabla f(x_{\min})]_k<0$ and $\nabla f$ is continuous then there exists $\epsilon > 0$ such that $x_{\min}+\epsilon e_k\in [a,b]$ and $f(x_{\min}+\epsilon e_k) < f(x_{\min})$ where $e_k$ is the $k$th standard basis vector of $\mathbb{R}^n.$ Implying that $x_{\min}$ is not a minizer of $f$ over $[a,b]$, which is a contradiction. Therefore, $f$ obtains a minimum on $(a,b)$.
