We start by the definition of the second fundamental form
$$\|h\|^{2}=\bigl\| h\bigl(\mathcal{D}^{\theta}, \mathcal{D}^{\theta}\bigr)\bigr\| ^{2}+\bigl\| h\bigl(\mathcal {D}^{\perp}, \mathcal{D}^{\perp} \bigr)\bigr\| ^{2}+2\bigl\| h\bigl(\mathcal{D}^{\theta}, \mathcal{D}^{\perp} \bigr)\bigr\| ^{2}. $$
Since
M is a mixed totally geodesic, we get
$$ \|h\|^{2}=\bigl\| h\bigl(\mathcal{D}^{\perp}, \mathcal{D}^{\perp} \bigr)\bigr\| ^{2}+\bigl\| h\bigl(\mathcal {D}^{\theta}, \mathcal{D}^{\theta}\bigr)\bigr\| ^{2}. $$
(4.6)
Thus by (
2.15), we obtain
$$\|h\|^{2}\geq\sum_{l=m+1}^{2n+1}\sum _{r, k=1}^{\alpha}g\bigl(h(e_{r}, e_{k}), e_{l}\bigr)^{2}. $$
The above equation can be expressed in the components of
\(\phi\mathcal {D}^{\perp}\),
\(F\mathcal{D}^{\theta}\), and
ν as
$$ \begin{aligned}[b] \|h\|^{2}\geq{}&\sum_{l,r,k=1}^{\alpha} g\bigl(h(e_{r}, e_{k}), \bar{e}_{l}\bigr)^{2}+\sum _{l=\alpha+1}^{2\beta+\alpha}\sum_{r, k=1}^{\alpha }g \bigl(h(e_{r}, e_{k}), \bar{e}_{l} \bigr)^{2}\\ &{}+\sum_{l=m}^{2(n-m+1)}\sum _{r, k=1}^{\alpha}g\bigl(h(e_{r}, e_{k}), \bar {e}_{l}\bigr)^{2}. \end{aligned} $$
(4.7)
We shall leave all the terms, except the second term, and we get
$$\|h\|^{2}\geq\sum_{l=1}^{2\beta}\sum _{r, k=1}^{\alpha}g\bigl(h(e_{r}, e_{k}), \tilde{e}_{l}\bigr)^{2}. $$
Thus using the adapted frame for
\(F\mathcal{D}^{\theta}\), we derive
$$\|h\|^{2}\geq \operatorname{csc}^{2}\theta\sum _{j=1}^{\beta}\sum_{r=1}^{\alpha}g \bigl(h(e_{r}, e_{r}), Fe^{*}_{j} \bigr)^{2}+\operatorname{csc}^{2}\theta \operatorname{sec}^{2} \theta\sum_{j=1}^{\beta}\sum _{r=1}^{\alpha}g\bigl(h(e_{r}, e_{r}), FPe^{*}_{j}\bigr)^{2}. $$
Using Lemma
3.4 for a mixed totally geodesic warped product submanifold and the fact that
\(\eta(e_{j})=0\),
\(1\leq j\leq d_{1}-1\) for an orthonormal frame, we arrive at
$$\|h\|^{2}\geq \operatorname{csc}^{2}\theta\sum _{j=1}^{\beta}\sum_{r=1}^{\alpha} \bigl(Pe^{*}_{j}\ln f\bigr)^{2}g(e_{r}, e_{r})+\operatorname{csc}^{2}\theta \cos^{2}\theta\sum _{j=1}^{\beta}\sum_{r=1}^{\alpha} \bigl(e^{*}_{j}\ln f\bigr)^{2}g(e_{r}, e_{r}). $$
Hence by hypothesis, we obtain
$$\|h\|^{2}\geq\alpha \operatorname{csc}^{2}\theta\sum _{j=1}^{\beta}\bigl(Pe^{*}_{j}\ln f \bigr)^{2}+\alpha \operatorname{cot}^{2}\theta\sum _{j=1}^{\beta}\bigl(e^{*}_{j}\ln f \bigr)^{2}. $$
By the property of trigonometric functions, we arrive at
$$\|h\|^{2}\geq\alpha \operatorname{csc}^{2}\theta\sum _{j=1}^{2\beta}\bigl(e^{*}_{j}\ln f \bigr)^{2}-\alpha \sum_{j=1}^{\beta} \bigl(e^{*}_{j}\ln f\bigr)^{2}. $$
Then by adding and subtracting the same terms
\(\xi\ln f\) in the above equation, we get
$$\|h\|^{2}\geq\alpha \operatorname{csc}^{2}\theta\sum _{j=1}^{2\beta+1}\bigl(e^{*}_{j}\ln f \bigr)^{2}-\alpha \sum_{j=1}^{\beta} \bigl(e^{*}_{j}\ln f\bigr)^{2}-\alpha \operatorname{csc}^{2}\theta( \xi\ln f)^{2}. $$
\(\xi\ln f=1\) [
13] for a warped product submanifold of a nearly Kenmotsu manifold. Thus the above equation gives
$$\|h\|^{2}\geq\alpha\Biggl[\operatorname{csc}^{2}{\theta}\bigl\{ \bigl\| \nabla^{\theta}\ln f\bigr\| ^{2}-1\bigr\} -\sum _{j=1}^{\beta}\bigl(e^{*}_{j}\ln f \bigr)^{2}\Biggr]. $$
If the equality holds, from the terms left in (
4.6) and (
4.7), we obtain the following conditions from the first and third terms:
$$\bigl\| h(\mathcal{D},\mathcal{D})\bigr\| ^{2}=0,\qquad g\bigl(h\bigl( \mathcal{D}^{\perp}, \mathcal{D}^{\perp}\bigr), \phi \mathcal{D}^{\perp}\bigr)=0, $$
and
$$g\bigl(h\bigl(\mathcal{D}^{\perp}, \mathcal{D}^{\perp}\bigr), \nu \bigr)=0, $$
where
\(\mathcal{D}=\mathcal{D}^{\theta}\oplus\xi\), this means that
\(M_{\theta}\) is a totally geodesic in
M̃ and
\(h(\mathcal {D}^{\perp}, \mathcal{D}^{\perp})\subseteq F\mathcal{D}^{\theta}\). Now from Lemma
3.4 for a mixed totally geodesic submanifold we find
$$g\bigl(h(Z, W), FX\bigr)=-(PX\ln f)g(Z, W), $$
for any
\(Z, W\in\Gamma(TM_{\perp})\) and
\(X\in\Gamma(TM_{\theta})\). The above equations imply that
\(M_{\perp}\) is totally umbilical in
M̃, so the equality holds. This completes the proof of the theorem. □