[proofplan]
The adjointness identity is exactly the defining property of $f^\perp$ as the adjoint of multiplication by $f$. The degree assertion follows from the grading and from orthogonality of distinct homogeneous degrees: if $f$ has degree $r$ and $g$ has degree $m$, then $fg$ has degree $m+r$, so it can pair nontrivially with a homogeneous element $h$ of degree $n$ only when $m+r=n$. Nondegeneracy of the Hall [inner product](/page/Inner%20Product) on each homogeneous component then forces $f^\perp h$ to have only degree $n-r$, and no component at all if $n<r$.
[/proofplan]
[step:Identify the adjointness identity from the definition of $f^\perp$]
Fix $f \in \operatorname{Sym}_{\mathbb Q}$. By definition, $f^\perp:\operatorname{Sym}_{\mathbb Q}\to\operatorname{Sym}_{\mathbb Q}$ is the adjoint of the multiplication map
\begin{align*}
M_f:\operatorname{Sym}_{\mathbb Q} &\to \operatorname{Sym}_{\mathbb Q} \\
g &\mapsto fg.
\end{align*}
Therefore, for every $g,h \in \operatorname{Sym}_{\mathbb Q}$,
\begin{align*}
\langle M_f g,h\rangle=\langle g,f^\perp h\rangle.
\end{align*}
Since $M_f g=fg$, this gives
\begin{align*}
\langle fg,h\rangle=\langle g,f^\perp h\rangle.
\end{align*}
[/step]
[step:Use orthogonality of homogeneous degrees to locate $f^\perp h$]
Assume now that $f \in \operatorname{Sym}_{\mathbb Q,r}$ is homogeneous of degree $r$. Fix $n \geq 0$ and $h \in \operatorname{Sym}_{\mathbb Q,n}$. Write the finite homogeneous decomposition of $f^\perp h$ as
\begin{align*}
f^\perp h=\sum_{k \geq 0} u_k,
\end{align*}
where each $u_k \in \operatorname{Sym}_{\mathbb Q,k}$ and all but finitely many $u_k$ are zero.
Let $m \geq 0$ and let $g \in \operatorname{Sym}_{\mathbb Q,m}$. Since $f$ has degree $r$, the product $fg$ lies in $\operatorname{Sym}_{\mathbb Q,m+r}$. By orthogonality of distinct homogeneous components,
\begin{align*}
\langle fg,h\rangle=0
\end{align*}
unless $m+r=n$. Using the adjointness identity already proved,
\begin{align*}
\langle g,f^\perp h\rangle=\langle fg,h\rangle=0
\end{align*}
for every $g \in \operatorname{Sym}_{\mathbb Q,m}$ whenever $m+r \neq n$.
Because $g \in \operatorname{Sym}_{\mathbb Q,m}$ pairs orthogonally with every $u_k$ for $k \neq m$, we have
\begin{align*}
\langle g,f^\perp h\rangle=\langle g,u_m\rangle.
\end{align*}
Thus, for every $m \geq 0$ with $m+r \neq n$,
\begin{align*}
\langle g,u_m\rangle=0
\end{align*}
for all $g \in \operatorname{Sym}_{\mathbb Q,m}$. The Hall inner product is nondegenerate on $\operatorname{Sym}_{\mathbb Q,m}$, so $u_m=0$ for every $m$ with $m+r \neq n$.
[guided]
We want to prove that $f^\perp h$ has a specific homogeneous degree. Since the Hall inner product is graded, the way to detect a homogeneous component is to pair against arbitrary homogeneous test elements of the same degree.
Fix $m \geq 0$ and choose an arbitrary $g \in \operatorname{Sym}_{\mathbb Q,m}$. Since $f \in \operatorname{Sym}_{\mathbb Q,r}$, multiplication in the graded ring gives
\begin{align*}
fg \in \operatorname{Sym}_{\mathbb Q,m+r}.
\end{align*}
The element $h$ lies in $\operatorname{Sym}_{\mathbb Q,n}$. Distinct homogeneous components are orthogonal for the Hall inner product, so
\begin{align*}
\langle fg,h\rangle=0
\end{align*}
whenever $m+r \neq n$.
Now apply the adjointness identity from the previous step:
\begin{align*}
\langle g,f^\perp h\rangle=\langle fg,h\rangle.
\end{align*}
Therefore, whenever $m+r \neq n$,
\begin{align*}
\langle g,f^\perp h\rangle=0
\end{align*}
for every $g \in \operatorname{Sym}_{\mathbb Q,m}$.
Write the homogeneous decomposition of $f^\perp h$ as
\begin{align*}
f^\perp h=\sum_{k \geq 0} u_k,
\end{align*}
where $u_k \in \operatorname{Sym}_{\mathbb Q,k}$ and only finitely many $u_k$ are nonzero. Since $g$ has degree $m$, orthogonality of homogeneous components gives
\begin{align*}
\langle g,f^\perp h\rangle
=
\left\langle g,\sum_{k \geq 0}u_k\right\rangle
=
\langle g,u_m\rangle.
\end{align*}
Thus, for each $m$ with $m+r \neq n$,
\begin{align*}
\langle g,u_m\rangle=0
\end{align*}
for all $g \in \operatorname{Sym}_{\mathbb Q,m}$. Nondegeneracy of the Hall inner product on $\operatorname{Sym}_{\mathbb Q,m}$ then forces $u_m=0$. Hence every homogeneous component of $f^\perp h$ vanishes except possibly the one of degree $n-r$.
[/guided]
[/step]
[step:Conclude the degree shift and the vanishing below degree $r$]
From the previous step, the only possible nonzero homogeneous component of $f^\perp h$ has degree $m=n-r$. If $n \geq r$, this gives
\begin{align*}
f^\perp h \in \operatorname{Sym}_{\mathbb Q,n-r}.
\end{align*}
If $n<r$, then there is no integer $m \geq 0$ satisfying $m+r=n$, so every homogeneous component $u_m$ of $f^\perp h$ is zero. Hence
\begin{align*}
f^\perp h=0.
\end{align*}
Since $h \in \operatorname{Sym}_{\mathbb Q,n}$ was arbitrary, this proves
\begin{align*}
f^\perp\left(\operatorname{Sym}_{\mathbb Q,n}\right)\subseteq \operatorname{Sym}_{\mathbb Q,n-r},
\end{align*}
with $\operatorname{Sym}_{\mathbb Q,n-r}=\{0\}$ when $n<r$.
[/step]
