How to bound the $\|\operatorname{diag}(A^TA)\|_2$ by $\|A\|_2$? In the question, $A\in R^{n\times d}$ is a matrix, $\operatorname{diag}(A^TA)\in R^{d\times d}$ is a square diagonal matrix that keeps the diagonal entries of $A^TA$ in its diagonal, and $\|\cdot\|_2$ is matrix spectral norm. 
Thanks.

The bound you really want is $\|diag(A^TA)\| \leq \|A^TA\| = \|A\|^2$.  Your bound won't work in general; to see that it won't, it suffices to consider diagonal matrices $A$.
In any case, it suffices to note that
$$
{\|A\|_2}^2 = \max_{\|x\| = 1} {\|Ax\|_2}^2 = 
\max_{\|x\| = 1} (x^TA^T)(Ax) \leq 
\max_{i=1,\dots,n} e_i(A^TA)e_i
$$
where $e_i$ are the canonical basis vectors (so $e_1 = (1,0,\dots,0)$).  From there, it suffices that this upper bound is exactly the same as $\|diag(A^TA)\|_2$. 
