Bonnet myers

Suppose some pair exceeds the proposed bound, and choose a minimizing segment. [F1, given] Assume, toward a contradiction, that there are points $p,q\in M$ with $l:=d_g(p,q)>\pi \mathrm{/}\sqrt{k}$; by [F1] there is a minimizing geodesic segment $\gamma :[0,l]\to M$ from $p$ to $q$, which after affine reparametrization is unit speed, with $\gamma (0)=p$, $\gamma (l)=q$ and $L(\gamma )=l=d_g(p,q)$.

The minimizing segment minimizes energy as well as length. [F2, given] For any piecewise smooth competitor $c:[0,l]\to M$ with $c(0)=p$ and $c(l)=q$, [F2] gives $2lE(c)\ge L(c{)}^2\ge d_g(p,q{)}^2=l^2,$ so $E(c)\ge l\mathrm{/}2$; since $\gamma$ is unit speed, $E(\gamma )=\frac{1}{2}{\int }_0^{l}1dt=l\mathrm{/}2$. Hence $E(\gamma )\le E(c)$ for every such $c$, and the endpoint-fixed energy functional attains its minimum at $\gamma$.

Orthonormal normal frame, sine test fields, and nonnegativity of their index forms. [F3, F5, F6, F7, step 1.2] By [F5] choose an orthonormal parallel frame $E_1,\dots ,E_{n-1}$ along $\gamma$ whose initial vectors form an orthonormal basis of the orthogonal complement of $\mathbb{R}\dot{\gamma }(0)$; then each $E_i(t)$ is normal to $\dot{\gamma }(t)$ and $(E_1(t),\dots ,E_{n-1}(t),\dot{\gamma }(t))$ is an orthonormal basis of $T_{\gamma (t)}M$. Put $\phi (t):=\sin \text{\hspace{0.17em}⁣}\left(\frac{\pi t}{l}\right),V_i(t):=\phi (t)E_i(t),t\in [0,l]\mathrm{.}$ By [F7] $\phi (0)=\phi (l)=\sin 0=\sin \pi =0$, and $\phi$ and the $E_i$ are smooth, so $V_i\in {\mathcal{X}}_0(\gamma )$ for every $i$. For fixed $i$ and small $s$, define ${\alpha }_i(s,t):={\exp }_{\gamma (t)}(s\phi (t)E_i(t))$; by [F6] this is smooth in $(s,t)$, its central curve is the geodesic $\gamma$, and ${\alpha }_i(s,0)=\gamma (0)$, ${\alpha }_i(s,l)=\gamma (l)$ for every $s$ because $\phi$ vanishes at the endpoints. Its variation field at $s=0$ is $V_i$, so [F3] applied to the two-parameter family $F(s,r,t):={\alpha }_i(s+r,t)$ — whose two variation fields both equal $V_i$ — identifies ${\mathrm{\partial }}_s{\mathrm{\partial }}_{r}E(F)$ at $(0,0)$ with $I_{\gamma }(V_i,V_i)$. That second derivative is the second derivative of $s\mapsto E({\alpha }_i(s,\cdot ))$ at its minimum $s=0$ from step 1.2, hence is $\ge 0$; therefore $I_{\gamma }(V_i,V_i)\ge 0\text{for every }i=1,\dots ,n-1.$

Computing the index forms and summing the Ricci trace. [F4, F7, step 2.1] Since $E_i$ is parallel, $D_t{V}_i={\phi }^{\mathrm{\prime }}{E}_i$ and $g(D_t{V}_i,D_t{V}_i)={\phi }^{\mathrm{\prime }2}$; the curvature term is $g(R(V_i,\dot{\gamma })\dot{\gamma },V_i)={\phi }^{2}g(R(E_i,\dot{\gamma })\dot{\gamma },E_i)={\phi }^2{K}_i$ with $K_i(t):=K(\mathrm{span}\{E_i(t),\dot{\gamma }(t)\})=\mathrm{Rm}(E_i,\dot{\gamma },\dot{\gamma },E_i),$ by [F4] and the orthonormality of the pair. Hence the definition of the index form in [F3] gives $I_{\gamma }(V_i,V_i)={\int }_0^l({\phi }^{\mathrm{\prime }}(t{)}^2-K_i(t)\phi (t{)}^2)dt\mathrm{.}$ On the interval $[0,l]$ the elementary integrals are ${\int }_0^l{\phi }^2={\int }_0^l\frac{1-\cos (2\pi t\mathrm{/}l)}{2}dt=\frac{l}{2},$ since $\frac{d}{dt}\frac{l}{4\pi }\sin (2\pi t\mathrm{/}l)=\frac{1}{2}\cos (2\pi t\mathrm{/}l)$ and $\sin (2\pi )=\sin 0=0$ by [F7], and likewise ${\int }_0^l{\phi }^{\mathrm{\prime }2}=\frac{{\pi }^2}{l^2}{\int }_0^l{\cos }^2\text{\hspace{0.17em}⁣}\left(\frac{\pi t}{l}\right)dt=\frac{{\pi }^2}{l^2}\cdot \frac{l}{2}=\frac{{\pi }^2}{2l},$ using ${\cos }^2=(1+\cos (2\cdot ))\mathrm{/}2$ from [F7]. Therefore ${\sum }_{i=1}^{n-1}{I}_{\gamma }(V_i,V_i)=(n-1)\frac{{\pi }^2}{2l}-{\int }_0^l({\sum }_{i=1}^{n-1}{K}_i(t))\phi (t{)}^{2}dt\mathrm{.}$ By [F4] the frame $(E_1,\dots ,E_{n-1},\dot{\gamma })$ is orthonormal, so ${\sum }_{i=1}^{n-1}{K}_i(t)={\sum }_{i=1}^{n-1}\mathrm{Rm}(E_i,\dot{\gamma },\dot{\gamma },E_i)={\mathrm{Ric}}_{\gamma (t)}(\dot{\gamma }(t),\dot{\gamma }(t))-\mathrm{Rm}(\dot{\gamma },\dot{\gamma },\dot{\gamma },\dot{\gamma })=\mathrm{Ric}(\dot{\gamma },\dot{\gamma }),$ the last curvature term vanishing by the skew-symmetry in [F4] applied to its last two slots. The hypothesis $\mathrm{Ric}\ge (n-1)kg$ and $g(\dot{\gamma },\dot{\gamma })=1$ thus give ${\sum }_i{K}_i(t)\ge (n-1)k$ for every $t$, and since $l>0$, ${\sum }_{i=1}^{n-1}{I}_{\gamma }(V_i,V_i)\le (n-1)\frac{{\pi }^2}{2l}-(n-1)\frac{kl}{2}=\frac{(n-1)l}{2}(\frac{{\pi }^2}{l^2}-k)<0,$ because $l>\pi \mathrm{/}\sqrt{k}$ is equivalent to ${\pi }^2\mathrm{/}{l}^2<k$.

The contradiction gives the diameter bound. [step 2.1, step 3.1] Step 2.1 gives $I_{\gamma }(V_i,V_i)\ge 0$ for every $i$, hence ${\sum }_i{I}_{\gamma }(V_i,V_i)\ge 0$, while step 3.1 gives ${\sum }_i{I}_{\gamma }(V_i,V_i)<0$. This contradiction shows that no pair can have distance greater than $\pi \mathrm{/}\sqrt{k}$. Fix $p_0\in M$, possible by nonemptiness. Then $M\subset B(p_0,\pi \mathrm{/}\sqrt{k}+1)$, so $M$ is bounded. Its diameter is now defined by [F1] as the supremum of pairwise distances and is at most $\pi \mathrm{/}\sqrt{k}$.

Compactness and boundary conventions. [F1, step 4.1] By step 4.1 the manifold is bounded: $d_g(x,y)\le \pi \mathrm{/}\sqrt{k}$ for all $x,y$. It is closed in itself, so [F1] makes $M$ compact. In dimension $n=1$ the normal space is zero, there are no test fields $V_i$, and the theorem has no content — consistently, a complete one-dimensional manifold is $\mathbb{R}$ or a circle and the bound is stated only for $n\ge 2$. The strict positivity $k>0$ is used in the strict inequality ${\pi }^2\mathrm{/}{l}^2<k$; for $k=0$ the statement is false as a diameter bound (Euclidean space is complete with nonnegative Ricci and is unbounded; its diameter is undefined under Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space). The case $l=\pi \mathrm{/}\sqrt{k}$ is not excluded by the argument and is attained by the round sphere of curvature $k$, so the bound is sharp. All frames and test fields were chosen as finite explicit families from the initial orthonormal frame, so nothing beyond the inherited [A1] is selected.