A sparse host has many leaf extensions, few smaller copies, or a sparse pair

Suppose outcomes 2 and 3 fail. If $yn\le 1$, choose a singleton $S$; because $\mathrm{\mid }{H}^{\mathrm{\prime }}\mathrm{\mid }\ge 2$, it has no copy of $H^{\mathrm{\prime }}$, giving outcome 2. If $y^{a}n\le 1\le yn$, take any vertex as $A$ and all its nonneighbors outside $A$ as $B$. Then $\mathrm{\mid }A\mathrm{\mid }\ge y^{a}n$ and $\mathrm{\mid }B\mathrm{\mid }\ge n-1-yn\ge (1-hy)n$, since $(h-1)yn\ge 1$; moreover $B$ is $x$-sparse to $A$. Thus failure of both outcomes implies $y^{a}n>1$ and $yn>1$. In particular $y^{2}n\ge y^{a}n>1$, so $yn>2$.

Choose any set $S$ of $\lceil yn\rceil$ vertices. For each induced embedding $\varphi :J\hookrightarrow G[S]$, let $I_{\varphi }$ be the embeddings of $H^{\mathrm{\prime }}$ into $G[S]$ restricting to $\varphi$; the image of $u$ distinguishes their members. Let $T$ be the $\varphi$ for which $\mathrm{\mid }{I}_{\varphi }\mathrm{\mid }\ge y^{a}n$. Failure of outcome 2 gives ${\mathrm{ind}}_{H^{\mathrm{\prime }}}(G[S])>y^{a-2}\mathrm{\mid }S{\mathrm{\mid }}^{h-1}\mathrm{.}$ There are at most $\mathrm{\mid }S{\mathrm{\mid }}^{h-2}$ maps $J\to S$, while each $\varphi \mathrm{\notin }T$ has fewer than $y^{a}n\le y^{a-1}\mathrm{\mid }S\mathrm{\mid }$ extensions. As $y\le 1\mathrm{/}2$, the embeddings over $T$ therefore number more than $(y^{a-2}-y^{a-1})\mathrm{\mid }S{\mathrm{\mid }}^{h-1}\ge y^{a-1}\mathrm{\mid }S{\mathrm{\mid }}^{h-1}$. Each $\varphi$ has at most $\mathrm{\mid }S\mathrm{\mid }$ extensions, so |T|>y^{a-1}|S|^{h-2}.\tag{1}

Fix $\varphi \in T$ and let $P=\varphi (V(J))$. Let $A^{\mathrm{\prime }}$ be the images of $u$ in $I_{\varphi }$, so $\mathrm{\mid }{A}^{\mathrm{\prime }}\mathrm{\mid }\ge y^{a}n$, and choose $A\subseteq A^{\mathrm{\prime }}$ of size $\lceil y^{a}n\rceil$. Let $B$ consist of vertices outside $S$ with no neighbor in $P$. At most $(h-2)yn$ vertices are excluded by adjacency to $P$, whence |B|\ge n-\lceil yn\rceil-(h-2)yn \ge(1-(h-1)y)n-1.\tag{2} Because outcome 3 fails, fewer than $(1-hy)n$ vertices of $B$ have at most $x\mathrm{\mid }A\mathrm{\mid }$ neighbors in $A$; otherwise those vertices, with $A$, would give that outcome. By (2), more than $yn-1\ge y^{2}n$ vertices of $B$ have more than $x\mathrm{\mid }A\mathrm{\mid }$ neighbors in $A$. The last inequality follows from $y^{2}n>1$ and $y\le 1\mathrm{/}2$. Hence there are more than xy^2n|A|\ge xy^{a+2}n^2\ge x^{a+3}n^2\tag{3} edges from $B$ to $A$.

Every edge in (3), together with $\varphi$, yields a distinct induced embedding of $H$: its endpoint in $A$ supplies $u$, its endpoint in $B$ supplies $v$, and the latter has no neighbor in $P$. Different $\varphi$ give different embeddings because restriction to $J$ recovers $\varphi$. Using (1), (3), $\mathrm{\mid }S\mathrm{\mid }\ge yn$, and $x\le y$, we get ${\mathrm{ind}}_H(G)>y^{a-1}\mathrm{\mid }S{\mathrm{\mid }}^{h-2}{x}^{a+3}{n}^2\ge x^{a+3}{y}^{a+h-3}{n}^h\ge x^{2a+h}{n}^h\mathrm{.}$ This is outcome 1.