[proofplan] The proof identifies both analytic and optimal-transport curvature-dimension conditions with the same pointwise weighted Ricci lower bound. First, the weighted Bochner formula turns the Bakry-Emery condition into an algebraic inequality involving the Hessian trace and the weighted Ricci tensor. The implication from the tensor bound to $BE(K,N)$ follows from a sharp trace-dimension inequality, while the reverse implication is obtained by testing $BE(K,N)$ on compactly supported functions with prescribed first and second jets at an arbitrary point. Finally, the smooth weighted Lott-Sturm-Villani consistency theorem identifies $CD(K,N)$ with the same tensor inequality, so the three assertions are equivalent. [/proofplan]

[step:Recall the Bochner identity and the two curvature-dimension conventions] For $f\in C_c^\infty(M)$, the definitions of $\Gamma$ and $\Gamma_2$ give \begin{align*} \Gamma(f)=|\nabla f|^2. \end{align*} By the [Bochner Formula for the Witten Laplacian][citetheorem:9592], applied to the weighted manifold $(M,g,m)$ with $m=e^{-V}\operatorname{vol}_g$, one has the pointwise identity \begin{align*} \Gamma_2(f)=|\nabla^2f|_{\mathrm{HS}}^2+\operatorname{Ric}_{V,\infty}(\nabla f,\nabla f). \end{align*} Here $|\nabla^2f|_{\mathrm{HS}}$ denotes the Hilbert-Schmidt norm of the symmetric $2$-tensor $\nabla^2f$ with respect to $g$, and \begin{align*} \operatorname{Ric}_{V,\infty}=\operatorname{Ric}+\nabla^2V. \end{align*} When $N=\infty$, the Bakry-Emery condition is therefore \begin{align*} |\nabla^2f|_{\mathrm{HS}}^2+\operatorname{Ric}_{V,\infty}(\nabla f,\nabla f)\ge K|\nabla f|^2. \end{align*} When $N\in(n,\infty)$, it is \begin{align*} |\nabla^2f|_{\mathrm{HS}}^2+\operatorname{Ric}_{V,\infty}(\nabla f,\nabla f)\ge K|\nabla f|^2+\frac{1}{N}(Lf)^2. \end{align*} [/step]

[step:Derive $BE(K,N)$ from the weighted Ricci tensor lower bound]Assume first that $\operatorname{Ric}_{V,N}\ge Kg$. If $N=\infty$, then for every $f\in C_c^\infty(M)$, \begin{align*} \Gamma_2(f)=|\nabla^2f|_{\mathrm{HS}}^2+\operatorname{Ric}_{V,\infty}(\nabla f,\nabla f)\ge K|\nabla f|^2=K\Gamma(f), \end{align*} because $|\nabla^2f|_{\mathrm{HS}}^2\ge0$. Thus $BE(K,\infty)$ holds. Now let $N\in(n,\infty)$. Fix $p\in M$ and $f\in C_c^\infty(M)$. Define the real numbers \begin{align*} a:=(\Delta f)(p) \end{align*} and \begin{align*} b:=g_p(\nabla V(p),\nabla f(p)). \end{align*} Then \begin{align*} (Lf)(p)=a-b. \end{align*} The tensor bound gives \begin{align*} \operatorname{Ric}_{V,\infty}(\nabla f,\nabla f)(p)\ge K|\nabla f(p)|^2+\frac{b^2}{N-n}. \end{align*} The Hilbert-Schmidt trace inequality for the symmetric endomorphism corresponding to $\nabla^2f(p)$ gives \begin{align*} |\nabla^2f(p)|_{\mathrm{HS}}^2\ge \frac{a^2}{n}. \end{align*} Combining these estimates with the Bochner identity gives \begin{align*} \Gamma_2(f)(p)\ge K|\nabla f(p)|^2+\frac{a^2}{n}+\frac{b^2}{N-n}. \end{align*} By the weighted Cauchy-Schwarz inequality in $\mathbb R^2$, \begin{align*} \frac{a^2}{n}+\frac{b^2}{N-n}\ge \frac{(a-b)^2}{N}. \end{align*} Hence \begin{align*} \Gamma_2(f)(p)\ge K\Gamma(f)(p)+\frac{1}{N}(Lf(p))^2. \end{align*} Since $p\in M$ and $f\in C_c^\infty(M)$ were arbitrary, $BE(K,N)$ holds.[/step]

[guided]The goal is to turn the tensor inequality into the scalar Bakry-Emery inequality for every test function. The Bochner identity already expresses $\Gamma_2(f)$ as the sum of a nonnegative Hessian term and the weighted Ricci tensor applied to $\nabla f$: \begin{align*} \Gamma_2(f)=|\nabla^2f|_{\mathrm{HS}}^2+\operatorname{Ric}_{V,\infty}(\nabla f,\nabla f). \end{align*} For $N=\infty$, there is no dimensional correction term. The assumption is exactly \begin{align*} \operatorname{Ric}_{V,\infty}\ge Kg. \end{align*} Therefore, for every $f\in C_c^\infty(M)$, \begin{align*} \Gamma_2(f)=|\nabla^2f|_{\mathrm{HS}}^2+\operatorname{Ric}_{V,\infty}(\nabla f,\nabla f)\ge K|\nabla f|^2=K\Gamma(f). \end{align*} This is precisely $BE(K,\infty)$. For finite $N$, the only extra work is to produce the term $(Lf)^2/N$. Fix a point $p\in M$ and a function $f\in C_c^\infty(M)$. Define \begin{align*} a:=(\Delta f)(p) \end{align*} and \begin{align*} b:=g_p(\nabla V(p),\nabla f(p)). \end{align*} These are chosen because the weighted Laplacian at $p$ is their difference: \begin{align*} (Lf)(p)=a-b. \end{align*} The finite-dimensional weighted Ricci tensor is \begin{align*} \operatorname{Ric}_{V,N}=\operatorname{Ric}_{V,\infty}-\frac{1}{N-n}\nabla V\otimes\nabla V. \end{align*} Thus the assumption $\operatorname{Ric}_{V,N}\ge Kg$, evaluated on the vector $\nabla f(p)\in T_pM$, says \begin{align*} \operatorname{Ric}_{V,\infty}(\nabla f,\nabla f)(p)\ge K|\nabla f(p)|^2+\frac{b^2}{N-n}. \end{align*} The Hessian term supplies the trace contribution. Since $\nabla^2f(p)$ is a symmetric bilinear form on the $n$-dimensional inner product space $T_pM$, the Hilbert-Schmidt trace inequality gives \begin{align*} |\nabla^2f(p)|_{\mathrm{HS}}^2\ge \frac{(\operatorname{tr}_g\nabla^2f(p))^2}{n}=\frac{a^2}{n}. \end{align*} Substituting these two estimates into the Bochner formula yields \begin{align*} \Gamma_2(f)(p)\ge K|\nabla f(p)|^2+\frac{a^2}{n}+\frac{b^2}{N-n}. \end{align*} It remains to compare the last two terms with $(Lf(p))^2/N=(a-b)^2/N$. This is exactly Cauchy-Schwarz with weights $n$ and $N-n$: \begin{align*} (a-b)^2=\left(\sqrt n\,\frac{a}{\sqrt n}-\sqrt{N-n}\,\frac{b}{\sqrt{N-n}}\right)^2\le N\left(\frac{a^2}{n}+\frac{b^2}{N-n}\right). \end{align*} Dividing by $N$ gives \begin{align*} \frac{a^2}{n}+\frac{b^2}{N-n}\ge \frac{(a-b)^2}{N}. \end{align*} Therefore \begin{align*} \Gamma_2(f)(p)\ge K\Gamma(f)(p)+\frac{1}{N}(Lf(p))^2. \end{align*} Because both the point $p$ and the test function $f$ were arbitrary, this proves $BE(K,N)$.[/guided]

[step:Recover the weighted Ricci tensor lower bound from $BE(K,N)$] Assume $BE(K,N)$. Fix $p\in M$ and $X\in T_pM$. Let $A:T_pM\times T_pM\to\mathbb R$ be any symmetric bilinear form. By the compactly supported jet-realization lemma in a smooth manifold, there exists $f\in C_c^\infty(M)$ such that \begin{align*} \nabla f(p)=X \end{align*} and \begin{align*} \nabla^2f(p)=A. \end{align*} This lemma follows by taking normal coordinates on a relatively compact coordinate ball around $p$, prescribing the degree-two Taylor polynomial there, and multiplying by a smooth cutoff equal to $1$ near $p$. For $N=\infty$, choose $A=0$. The Bochner identity and $BE(K,\infty)$ give \begin{align*} \operatorname{Ric}_{V,\infty}(X,X)=\Gamma_2(f)(p)\ge K\Gamma(f)(p)=K|X|^2. \end{align*} Since $p$ and $X$ were arbitrary, $\operatorname{Ric}_{V,\infty}\ge Kg$. Let $N\in(n,\infty)$. Define \begin{align*} b:=g_p(\nabla V(p),X). \end{align*} For each $a\in\mathbb R$, choose $A_a:T_pM\times T_pM\to\mathbb R$ to be \begin{align*} A_a:=\frac{a}{n}\,g_p. \end{align*} Then $\operatorname{tr}_{g_p}A_a=a$ and $|A_a|_{\mathrm{HS}}^2=a^2/n$. Applying $BE(K,N)$ to a compactly supported function with $\nabla f(p)=X$ and $\nabla^2f(p)=A_a$ gives \begin{align*} \frac{a^2}{n}+\operatorname{Ric}_{V,\infty}(X,X)\ge K|X|^2+\frac{(a-b)^2}{N}. \end{align*} Rearranging, \begin{align*} \operatorname{Ric}_{V,\infty}(X,X)-K|X|^2\ge \frac{(a-b)^2}{N}-\frac{a^2}{n}. \end{align*} The right-hand side is maximized over $a\in\mathbb R$ at \begin{align*} a=-\frac{nb}{N-n}, \end{align*} and the maximum value is \begin{align*} \frac{b^2}{N-n}. \end{align*} Therefore \begin{align*} \operatorname{Ric}_{V,\infty}(X,X)-K|X|^2\ge \frac{b^2}{N-n}. \end{align*} Equivalently, \begin{align*} \operatorname{Ric}_{V,N}(X,X)\ge K|X|^2. \end{align*} Since $p\in M$ and $X\in T_pM$ were arbitrary, $\operatorname{Ric}_{V,N}\ge Kg$. [/step]

[step:Identify the Lott-Sturm-Villani condition with the same tensor inequality] For $N=\infty$, the smooth weighted Riemannian consistency theorem for entropy convexity states that, on a complete smooth weighted manifold with locally finite full-support measure $m=e^{-V}\operatorname{vol}_g$, entropy $K$-displacement convexity on finite-entropy endpoints is equivalent to \begin{align*} \operatorname{Ric}_{V,\infty}\ge Kg. \end{align*} This is exactly the weighted entropy convexity criterion [Weighted Entropy Convexity from Bakry-Emery Ricci Curvature Lower Bounds][citetheorem:9590] together with its smooth converse. For $N\in(n,\infty)$, the smooth weighted Lott-Sturm-Villani consistency theorem states that the weak condition $CD(K,N)$ for bounded-support absolutely continuous endpoint measures is equivalent to \begin{align*} \operatorname{Ric}_{V,N}\ge Kg. \end{align*} This is the smooth weighted curvature-dimension criterion [Weighted Riemannian Curvature-Dimension Criterion][citetheorem:9591] together with its smooth converse. These consistency theorems use the same normalization of the generator $L=\Delta-g(\nabla V,\nabla\cdot)$, the same weighted measure $m=e^{-V}\operatorname{vol}_g$, and the same finite-dimensional correction term in $\operatorname{Ric}_{V,N}$ as in the statement. [/step]

[step:Combine the two equivalences] The previous two analytic steps prove \begin{align*} BE(K,N)\iff \operatorname{Ric}_{V,N}\ge Kg. \end{align*} The smooth weighted Lott-Sturm-Villani consistency theorem gives \begin{align*} CD(K,N)\iff \operatorname{Ric}_{V,N}\ge Kg. \end{align*} Hence \begin{align*} CD(K,N)\iff BE(K,N)\iff \operatorname{Ric}_{V,N}\ge Kg. \end{align*} This proves the equivalence of the three assertions in both the finite-dimensional case $N\in(n,\infty)$ and the infinite-dimensional case $N=\infty$. [/step]