= Solution
Write $T_v=T(v)$. We prove that $T$ has a closed graph. Suppose
$$
v_n\longrightarrow v\quad\text{in }V,
\qquad
T_{v_n}\longrightarrow f\quad\text{in }V^*.
$$
Put $u_n=v_n-v$ and $g=f-T_v$. Then $u_n\to0$ and $T_{u_n}\to g$ in $V^*$.
Fix $w\in V$ and $t\in\mathbb R$. Positivity and linearity give
$$
\begin{aligned}
0&\leq T_{u_n+tw}(u_n+tw)\\
&=T_{u_n}(u_n)
+t\bigl(T_{u_n}(w)+T_w(u_n)\bigr)
+t^2T_w(w).
\end{aligned}
$$
Because $(T_{u_n})$ converges in the <continuous dual space>, its norms are bounded, so
$$
|T_{u_n}(u_n)|\leq\lVert T_{u_n}\rVert\lVert u_n\rVert\longrightarrow0.
$$
Also $T_{u_n}(w)\to g(w)$, while $T_w(u_n)\to0$ because the fixed functional $T_w$ is continuous. Passing to the limit yields
$$
0\leq t g(w)+t^2T_w(w)
$$
for every real $t$. If $g(w)\ne0$, a sufficiently small $t$ of the opposite sign makes the right-hand side negative. Hence $g(w)=0$ for every $w$, so $g=0$ and $f=T_v$.
The graph is therefore closed. The <closed graph theorem> now proves that $T:V\to V^*$ is continuous. This is precisely <continuity of a positive linear map into a dual space>.
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