So let us assume our manifold is complete, simply connected, and has nonpositive sectional curvature. If we assume that the displacement function $f(x)=d(x,\phi(x))$, for an isometry $\phi:M\rightarrow M$, is bounded from below by a positive value, show that there exists a geodesic axis. That is there exists a geodesic $\gamma$ such that $\phi\circ \gamma(t)=\gamma(t+t_0)$ assuming that $f(x)=\inf(f)$ for some $x\in M$.
So here is what I have so far. It seems that I should select the geodesic that goes from $x$ to $\phi(x)$ where $f(x)=\inf(f)$. I had two ideas for showing that this is an axis. Namely we could show that $\phi\circ\gamma(-t_0)=\gamma(0)$, where $\gamma(0)=x$ and $\gamma(t_0)=\phi(x)$, then we could use that geodesics are unique in this type of manifold, or we could show that $f(\gamma)$ is constant which could also establish our result, but I'm not sure how to go about with either. Any suggestions? Thanks.
Your first idea (using the uniqueness of geodesics) is good. The key formula is the derivative of the distance function: we are given a minimum $x_0$ of $f$, so we'd like to extract some information from the critical point condition $Df_{x_0} = 0$. The hypotheses on the manifold guarantee the distance function will be smooth away from the diagonal, so this is definitely true.
If we let $\gamma$ be an arclength-parametrized minimizing geodesic joining $x$ to $y$ and $L$ be its length, then you should be able to derive from the first variation formula that
$$ (Dd_{(x,y)})(u,v) =\langle\dot \gamma(L),v\rangle - \langle \dot \gamma(0), u\rangle.$$
Applying this with $x = x_0, y = \phi(x_0), v = D\phi(u)$ we get
$$ Df_{x_0}(u) = (Dd_{(x_0,\phi(x_0)))})(u,D\phi(u))=\langle\dot \gamma(L), D\phi(u)\rangle - \langle \dot \gamma(0), u\rangle=0$$
for all vectors $u \in T_{x_0} M$. Since $\phi$ is an isometry, we can rewrite this as $$\langle\dot \gamma(L) - D\phi(\dot \gamma(0)), D\phi(u)\rangle=0$$ for all $u$, and thus the fact that $D\phi_{x_0}$ is bijective tells us that $\dot \gamma(L) = D\phi(\dot \gamma(0))$. Since $\phi$ is an isometry we know $\phi\circ \gamma$ is geodesic, so the uniqueness of geodesics extends this equality of velocities at a point to equalities of the entire geodesics; i.e. $\phi(\gamma(t)) = \gamma(t + L)$.