For a positive function, can the length of gradient get arbitrarily close to 0? Consider a differentiable function $f:R^d\rightarrow R_+$. Let $a=\inf\limits_{x\in R^d} f(x)$. It's evident that $a\ge0$.
Now if there exists $x\in R^d$ such that $f(x)=a$, then $f'(x)=0$.
If there doesn't exist $x\in R^d$ such that $f(x)=a$, can we prove that $\forall \epsilon>0, \exists x\in R^d$ such that $|f'(x)|<\epsilon$? Is there any relevant theorem?
 A: Here is a roundabout way to prove this assuming that $f$ is $C^1$.
Suppose $\|\nabla f(x)\| \ge \delta >0$ for all $x$.
Pick some $x_0$ and apply steepest descent to $f$ starting at $x_0$. We will obtain a contradiction.
Let $\lambda_k$ be the largest $\lambda \in \{1,{1 \over 2}, { 1\over 4},.. \}$ such that
$f(x_k-\lambda \nabla f(x_k)) \le f(x_k) - {1 \over 2} \lambda \|\nabla f(x_k)\|^2$
(cf. the Armijo step size rule).
Since $df(x_k, -\nabla f(x_k)) = -\|\nabla f(x_k)\|^2$, we see that $\lambda_k$ is
well defined as long as $\nabla f(x_k) \neq 0$. Let
$x_{k+1} = x_k - \lambda_k \nabla f(x_k)$ and note that 
$f(x_{k+1}) -f(x_k) \le -{1 \over 2}\lambda_k \|\nabla f(x_k)\|^2$, hence
non increasing.
Summing, we get $f(x_n)-f(x_0) \le -{1 \over 2} \sum_i \lambda_{k=0}^{n-1} \|\nabla f(x_k)\|^2$
Since $f$ is bounded below, we see that $\sum_k \lambda_k \|\nabla f(x_k)\|^2 $ is bounded, and since $\|\nabla f(x_k)\| \ge \delta$, we see that
$\sum_k \lambda_k \|\nabla f(x_k)\| $ is bounded and hence $x_k \to x^*$ for
some $x^*$.
Since we have assumed that $\nabla f(x^*) \neq 0$ and $\nabla f$ is continuous, it is not hard to show that we must have $\lambda_k \ge \lambda^* >0$ for some $\lambda^*$ and for $x_k$ sufficiently close to $x^*$. In particular, this gives
$f(x_{k+1}) \le f(x_k) - {1 \over 2} \lambda^* \|\nabla f(x_k) \|^2 \le f(x_k) - {1 \over 4} \lambda^* \|\nabla f(x^*) \|^2$ for sufficiently large $k$. However, this
gives $f(x_k) \downarrow - \infty$ which contradicts continuity of $f$ at $x^*$.
Hence $\inf_x \| \nabla f(x) \| = 0$.
