Let
- $f\in C^1(\mathbb R^n)$
- $x\in\mathbb R^n$ and $d\in\mathbb R^n$ with $$\langle\nabla f(x),d\rangle<0\tag{1}$$
Then, $t>0$ is said to satisfy the Wolfe-Powell conditions with parameters $\sigma\in (0,1/2)$ and $\rho\in [\sigma,1)$ $:\Leftrightarrow$ $$\varphi(t):=f(x+td)\le f(x)+\sigma t\langle\nabla f(x),d\rangle=:\varphi_0(t)\tag{2}$$ and $$\varphi'(t)\ge\rho\varphi'(0)\tag{3}\;.$$
Why do we choose $\sigma<1/2$ instead of $\sigma <1$?
I've read, that we do this, because otherwise we would exclude the exact minimizer of a quadratic $\varphi$. So, I tried to take a look at what happens if $f(x+\;\cdot\;d)$ is indeed described by a quadratic function $$\varphi(t)=at^2+bt+c\tag{4}$$ and it turned out to be a contradiction to $(1)$, since $$\langle\nabla f(x),d\rangle\stackrel{(2)}=\varphi'(0)\stackrel{(4)}=\left.2at+b\right|_{t=0}=b\;,$$ unless $b<0$.
It's not a contradiction to $(1)$. First, $t=0$ is excluded by $(3)$. Second, the quadratic $\varphi$ attains its minimum at $$t_*=-\frac b{2a}\;.$$ It's easy to show, that $t_*$ satisfies the Armijo condition $(2)$, if $$(1-2\sigma)b^2\ge 0\;.$$ Since we can't assume $b=0$, we need $\sigma<1/2$, if we don't want to exclude the exact minimizer $t_*$.