Is the value of $\max_{f:V\rightarrow [\frac{-1}{2},\frac{1}{2}], \\ \sum_{v}{f(v)}=0} \frac{f^T L_G f}{n-f^Tf}$ polynomially computable? For a graph $G$ on $n$ vertices, what is the value of following ratio: 

$$\max_{f:V\rightarrow [\frac{-1}{2},\frac{1}{2}], \\ \sum_{v}{f(v)}=0} \frac{f^T L_G f}{n-f^Tf} ,$$ 

where $L_G=D_G-A_G$ is the laplacian matrix of $G$? 

Is this parameter related to the spectrum of $G$? 

Is this parameter polynomially computable? 

Remark: Note that we have 
$$\max_{f:V\rightarrow [\frac{-1}{2},\frac{1}{2}], \\ \sum_{v}{f(v)}=0} \frac{f^T L_G f}{f^Tf} = \lambda_n(G),$$ 
where $\lambda_n(G)$ is the largest eigenvalue of $L_G$.
Without any further constraints, this expression will in general be unbounded, so the maximum won't exist.

Let $V$ be $\{v_1,\ldots,v_n\}$ with $n\ge 2$. Pick $i\neq j$ such that $v_i,v_j$ are not both isolated. The submatrix of the Laplacian for $v_i,v_j$ has the form $\begin{pmatrix}d_i &amp; -a_{ij} \\ -a_{ij} &amp; d_j\end{pmatrix}$ with $a_{ij}\in\{0,1\}$ and $d_i,d_j\ge a_{ij}$, and $d_i+d_j&gt;0$.

For any $\varepsilon&gt;0$, define a function $f_\varepsilon:V\to\mathbb{R}$ by $f_\varepsilon(v_i) = \sqrt{n}-\varepsilon, f_\varepsilon(v_j) = \varepsilon - \sqrt{n}$, and $f_\varepsilon(v) = 0$ for all other $v\in V$. With this, your expression becomes

$(d_i + d_j + 2a_{ij})\frac{n-2\varepsilon\sqrt{n}+\varepsilon^2}{2\varepsilon\sqrt{n}-\varepsilon^2}$.

$(d_i+d_j-2a_{ij}) \ge 1$ by the above, and for $\varepsilon\to 0$, the numerator and denominator of the fraction converge to $n$ and $0$ (from above); therefore, the expression goes to $+\infty$.