Under the usual walk convention allowing repeated edges, the stated hypothesis is insufficient. Take the connected graph consisting of one zero-weight edge $xy$. It has no simple cycles, so the cycle condition holds. Nevertheless $Q=x,y,x,y$ is a minimum-weight $xy$-walk of weight zero and is not a path.

If the course uses “walk” to mean an edge-nonrepeating trail, the intended argument is valid. If such a minimum trail repeats a vertex, take its first repeated-vertex segment. Its interior vertices are distinct and no edge repeats, so this segment is a simple cycle $C$. Removing it leaves an $xy$-trail of weight $w(Q)-w(C)<w(Q)$, a contradiction. Hence $Q$ is a path. Since every $xy$-path is a competing trail, it is a shortest path.

For unrestricted walks, the same proof works if every nontrivial closed subwalk has positive weight, in particular if every edge has positive weight. This is cycle erasure in minimum-weight walks. The exact 2017 course convention was not available to check; the counterexample and the trail interpretation distinguish the two meanings explicitly.

Use a triangle with vertices $v,a,b$ and weights $w(va)=w(vb)=2$, $w(ab)=-1$. The unique shortest $va$-path is $v,b,a$, of weight one, while the unique shortest $vb$-path is $v,a,b$, also of weight one. A shortest-path tree would therefore need all three edges of the triangle. This is impossible for a tree.

Thus negative edge weights can prevent a shortest-path tree, even though the shortest simple path to each individual vertex exists. This is the example in signed edge weights can prevent a shortest-path tree.

Use the squaring map from a quadrant to the upper half-plane:By conformal invariance of harmonicity, the answer isOn the horizontal axis it gives $f(x^2,0)=e^{-x^2}$, and on the vertical axis it gives $f(-y^2,0)=e^{-y^2}$. Also $|(x+iy{)}^2|=x^2+y^2$, so spatial infinity maps to spatial infinity and the required decay follows. The two boundary values agree at the origin.

For integrable functions $u,v$, the convolution theorem in the chosen convention isLet $F(s,y)$ be the Fourier transform of $f(\cdot ,y)$ in the horizontal coordinate. Transforming Laplace's equation givesFor $s\ne 0$, exclude the growing exponential and obtain $F(s,y)=2e^{-|s|y}/(1+s^2)$. At $s=0$, continuity in frequency gives $F(0,y)=2$; a constant zero-frequency transform is compatible with spatial decay of $f$.

By the transform pair in part (a), the inverse transform of $e^{-|s|y}$ is the Poisson kernel for the upper half-plane, $P_y(x)=y/[\pi (x^2+y^2)]$. ThusThe Poisson extension of exponential boundary data is harmonic, approaches the continuous boundary data and decays at spatial infinity. For uniqueness, apply the maximum principle for harmonic functions to the difference of two solutions on increasingly large half-discs: it is zero on the real boundary and uniformly small on the distant semicircle.