[proofplan] The finite-trace hypothesis implies that the restriction of $\operatorname{Tr}$ to the positive cone of $pNp$ is finite, so normalization by $\operatorname{Tr}(p)$ gives a state after linear extension. We verify faithfulness, normality, and the trace identity directly from the corresponding properties of $\operatorname{Tr}$. Factoriality is obtained from central support: since $N$ is a factor and $p\ne0$, the central support of $p$ is $1$, and the standard corner-center correspondence forces $Z(pNp)=\mathbb C p$. Finally, any minimal projection in $pNp$ would be a minimal projection in $N$, contradicting the Type $II$ hypothesis. [/proofplan]

[step:Normalize the finite restriction of the trace to the corner]The algebra $pNp=\{pxp:x\in N\}$ is a von Neumann algebra with identity $p$. Let $(pNp)_+$ denote its positive cone. For $x\in(pNp)_+$, let $\|x\|_{\mathrm{op}}$ denote the operator norm of $x$ in the unital $C^*$-algebra $pNp$. Positivity in $pNp$ gives \begin{align*} 0\le x\le \|x\|_{\mathrm{op}}p. \end{align*} Since $\operatorname{Tr}$ is positive and $\operatorname{Tr}(p)<\infty$, it follows that \begin{align*} 0\le \operatorname{Tr}(x)\le \|x\|_{\mathrm{op}}\operatorname{Tr}(p)<\infty. \end{align*} Because $p\ne0$ and $\operatorname{Tr}$ is faithful, $\operatorname{Tr}(p)>0$. Hence the map \begin{align*} \tau_p:(pNp)_+&\to[0,\infty) \end{align*} \begin{align*} x&\mapsto \frac{\operatorname{Tr}(x)}{\operatorname{Tr}(p)} \end{align*} is well-defined and satisfies \begin{align*} \tau_p(p)=1. \end{align*}[/step]

[guided]The first point is that the formula for $\tau_p$ must be finite on every positive element of the corner. The identity of $pNp$ is $p$, so if $x\in(pNp)_+$, then the usual order estimate for a positive element in a unital $C^*$-algebra gives \begin{align*} 0\le x\le \|x\|_{\mathrm{op}}p. \end{align*} Applying positivity of $\operatorname{Tr}$ to this order inequality gives \begin{align*} 0\le \operatorname{Tr}(x)\le \operatorname{Tr}(\|x\|_{\mathrm{op}}p). \end{align*} Since $\operatorname{Tr}$ is homogeneous on positive elements, the right-hand side is \begin{align*} \operatorname{Tr}(\|x\|_{\mathrm{op}}p)=\|x\|_{\mathrm{op}}\operatorname{Tr}(p). \end{align*} The hypothesis $\operatorname{Tr}(p)<\infty$ therefore implies \begin{align*} 0\le \operatorname{Tr}(x)\le \|x\|_{\mathrm{op}}\operatorname{Tr}(p)<\infty. \end{align*} We also need the denominator to be nonzero. Since $p$ is a nonzero positive element and $\operatorname{Tr}$ is faithful, $\operatorname{Tr}(p)=0$ is impossible. Thus $\operatorname{Tr}(p)>0$, and the normalization \begin{align*} \tau_p(x)=\frac{\operatorname{Tr}(x)}{\operatorname{Tr}(p)} \end{align*} defines a finite positive functional on $(pNp)_+$. Finally, \begin{align*} \tau_p(p)=\frac{\operatorname{Tr}(p)}{\operatorname{Tr}(p)}=1, \end{align*} so the normalized functional has total mass one on the identity of the corner.[/guided]

[step:Extend the normalized positive trace linearly] Let $(pNp)_{\mathrm{sa}}$ denote the self-adjoint part of $pNp$. For $a\in(pNp)_{\mathrm{sa}}$, let $a_+\in(pNp)_+$ and $a_-\in(pNp)_+$ be the positive and negative parts of $a$ obtained by continuous functional calculus in $pNp$, so that \begin{align*} a=a_+-a_-. \end{align*} Define \begin{align*} \tau_p(a)=\tau_p(a_+)-\tau_p(a_-). \end{align*} This is well-defined because $a_+$ and $a_-$ are uniquely determined by $a$. For an arbitrary $x\in pNp$, define the self-adjoint elements \begin{align*} \operatorname{Re}(x)=\frac{x+x^*}{2} \end{align*} and \begin{align*} \operatorname{Im}(x)=\frac{x-x^*}{2i}. \end{align*} Then $x=\operatorname{Re}(x)+i\operatorname{Im}(x)$, and we set \begin{align*} \tau_p(x)=\tau_p(\operatorname{Re}(x))+i\tau_p(\operatorname{Im}(x)). \end{align*} The finite-trace restriction $\operatorname{Tr}|_{(pNp)_+}$ is additive and homogeneous on the positive cone because $\operatorname{Tr}$ is a trace. Since it is finite on $(pNp)_+$, the standard ordered-vector-space [extension theorem](/theorems/59) for positive additive homogeneous functionals on the positive cone of a unital $C^*$-algebra gives a unique positive linear functional on $pNp$ whose restriction to $(pNp)_+$ is the formula above. The preceding Jordan and complex decompositions describe this extension explicitly, so $\tau_p:pNp\to\mathbb C$ is complex-linear. [/step]

[step:Verify that the extension is a faithful normal tracial state] The extension is positive by construction and satisfies $\tau_p(p)=1$, so it is a state. If $x\in(pNp)_+$ and $\tau_p(x)=0$, then $\operatorname{Tr}(x)=0$, and faithfulness of $\operatorname{Tr}$ gives $x=0$; hence $\tau_p$ is faithful. Let $(x_i)_{i\in I}$ be an increasing bounded net in $(pNp)_+$ with supremum $x\in(pNp)_+$. Since $pNp\subset N$ as a von Neumann subalgebra and the order supremum is the same in the corner, normality of $\operatorname{Tr}$ gives \begin{align*} \tau_p(x)=\frac{\operatorname{Tr}(x)}{\operatorname{Tr}(p)} =\sup_{i\in I}\frac{\operatorname{Tr}(x_i)}{\operatorname{Tr}(p)} =\sup_{i\in I}\tau_p(x_i). \end{align*} Thus $\tau_p$ is normal. Let $\widetilde{\operatorname{Tr}}_p:pNp\to\mathbb C$ denote the finite linear extension of $\operatorname{Tr}|_{(pNp)_+}$ obtained by the same ordered-vector-space extension procedure. Thus $\tau_p=\widetilde{\operatorname{Tr}}_p/\operatorname{Tr}(p)$ on all of $pNp$. If $x,y\in pNp$, then $xy,yx\in pNp$, and the tracial property of this finite restriction gives \begin{align*} \tau_p(xy)=\frac{\widetilde{\operatorname{Tr}}_p(xy)}{\operatorname{Tr}(p)} =\frac{\widetilde{\operatorname{Tr}}_p(yx)}{\operatorname{Tr}(p)} =\tau_p(yx). \end{align*} Therefore $\tau_p:pNp\to\mathbb C$ is a faithful normal tracial state. [/step]

[step:Use central support to show the corner is a factor] Let $c_N(p)\in Z(N)$ denote the central support of $p$ in $N$, namely the smallest central projection in $N$ that dominates $p$. Since $N$ is a factor, its center is $Z(N)=\mathbb C1$. Therefore the only central projections in $N$ are $0$ and $1$. Because $p\ne0$, the central support cannot be $0$, so \begin{align*} c_N(p)=1. \end{align*} We use the standard corner-center correspondence for central support: for a projection $p$ in a von Neumann algebra $N$, the center $Z(pNp)$ is naturally identified with $Z(N)c_N(p)$, with the unit of the corner corresponding to $c_N(p)$. Applying the [[Central Carrier Theorem](/theorems/9286)][citetheorem:9286] to the projection $p$ gives the required central carrier object, and the correspondence applies because $p$ is a projection in the von Neumann algebra $N$. Since $Z(N)=\mathbb C1$ and $c_N(p)=1$, this gives \begin{align*} Z(pNp)=\mathbb C p. \end{align*} Thus $pNp$ is a factor. [/step]

[step:Exclude minimal projections in the corner] Suppose, toward a contradiction, that $e\in pNp$ is a minimal projection in $pNp$. Since the identity of $pNp$ is $p$, we have $e\le p$. For every projection $f\in N$ satisfying $f\le e$, the inequalities $f\le e\le p$ imply $f=pfp$, hence $f\in pNp$. Since $e$ is minimal in $pNp$, every projection $f\in N$ with $f\le e$ is either $0$ or $e$. Therefore $e$ is a minimal projection in $N$. But $N$ is Type $II_\infty$, so by definition it has no nonzero minimal projections. This contradicts $e\ne0$. Hence $pNp$ has no nonzero minimal projections. [/step]

[step:Conclude that the corner has Type $II_1$] We have shown that $pNp$ is a factor and that it has no nonzero minimal projections. The functional $\tau_p$ is a faithful normal tracial state on $pNp$, so the identity $p$ is finite and the factor is finite. Therefore $pNp$ is a finite non-atomic factor, hence a Type $II_1$ factor. The normalized trace on its positive cone is exactly \begin{align*} \tau_p(x)=\frac{\operatorname{Tr}(x)}{\operatorname{Tr}(p)} \end{align*} for $x\in(pNp)_+$, and the preceding construction gives its unique linear extension to a faithful normal tracial state on $pNp$. [/step]