I found the following statement in a paper (here (see 4th line of page 1940) or here (towards the middle of page 10)):
The $L^2$ operator norm of a symmetric, square matrix is bounded by its $L^\infty$ operator norm.
The $L^2$ operator norm of a matrix $A$ is defined as: $$||A||_2 := \sup_{||x|| =1}||Ax||$$ and the $L^\infty$ operator norm of $A$ is defined as: $$||A||_\infty := \max_i \sum_{j}|A_{i,j}|~.$$ So, the paper basically claims that for a symmetric, square matrix $A$, $||A||_2 \le ||A||_\infty$. Is the statement true at all? I cannot find the statement anywhere else, and the closest I found, is that for a general $m\times n$ matrix $A$, $||A||_2 \le \sqrt{m}||A||_\infty.$
See the very last inequality on this wikipedia page and note that for a symmetric $A$ the norms $\lVert A \rVert_1$ and $\lVert A \rVert_{\infty}$ are equal.