Path Poincare inequality on an anchored finite metric graph

If $E=\mathrm{\varnothing }$, then $H=\{0\}$ and the inequality is $0\le 0$. Otherwise fix an edge $e$ in a component containing an anchor. If $e$ is a loop, deleting its interior leaves the component connected because any path that traverses the loop can omit that closed excursion. If its endpoints are distinct, every original path that uses $e$ reaches one endpoint before traversing it; after deleting $e$, vertices are therefore connected to one of its two endpoints. Thus removing $e$ either leaves the component connected or separates it into two components, one containing each endpoint. In the connected case choose either endpoint; in the separated case choose the endpoint in the component containing an anchor. By [F3] there is a finite path from an anchor to that endpoint avoiding $e$. Deleting the portions between repeated vertices gives a simple path, so no edge is traversed twice. For a loop choose its tail incidence; otherwise use the incidence at the chosen endpoint. For any $x$ on $e$, append the segment from that incidence to $x$. The total length is at most $L_{\mathrm{\Gamma }}$, and the common vertex traces agree along the path by [F1]. Since $w=0$ at the anchor, [F1] and [F2] imply $\mathrm{\mid }{w}_e(x){\mathrm{\mid }}^2\le L_{\mathrm{\Gamma }}{\sum }_{a\text{ on the concatenated path}}{\int }_a\mathrm{\mid }{w}_a^{\mathrm{\prime }}{\mathrm{\mid }}^2\le L_{\mathrm{\Gamma }}{\sum }_{a\in E}\mathrm{\parallel }{w}_a^{\mathrm{\prime }}{\mathrm{\parallel }}_{L^2(a)}^2\mathrm{.}$

Integrate the pointwise bound of step 1.1 over every edge and sum. The total measure of the disjoint edge interiors is ${\sum }_e{\mathrm{\ell }}_e=L_{\mathrm{\Gamma }}$, so $\mathrm{\parallel }w{\mathrm{\parallel }}_{L^2(\mathrm{\Gamma })}^2\le L_{\mathrm{\Gamma }}^2{\sum }_{e\in E}\mathrm{\parallel }{w}_e^{\mathrm{\prime }}{\mathrm{\parallel }}_{L^2(e)}^2\mathrm{.}$ Taking nonnegative square roots proves the statement.