Equivalent Definitions of the Spectral Norm There are many equivalent definitions of the spectral norm $\|A\|_2$ for when $A$ is a symmetric matrix, the most common ones being
$$\sup_{\|x\|_{2} = 1}{\|Ax\|_{2}} = \sup_{\|x\|_{2}=1}|{\langle Ax,x \rangle|} = \text{largest eigenvalue of $A$ in absolute value}$$ 
Recently, while going through a paper on compressed sensing (http://statweb.stanford.edu/~candes/papers/PartialMeasurements.pdf), I was met with the following definition of the spectral norm(search "spectral" in the paper): 
$$\|Y\|_2 = \displaystyle\sup_{\|f_1\|_{2} = \|f_1\|_{2} = 1 } \langle f_1, Yf_2 \rangle$$
where $f_1, f_2$ are unit norm vectors. After going through some naive calculations, I could not find out why this norm is equivalent to the ones I defined above, nor have I found another source that defines it this way. I was wondering if someone can clear this up for me as to why the definitions are equivalent.
Thanks all beforehand! 
 A: On one hand,
\begin{align}
   \sup_{\|f_1\|=\|f_2\|=1} |  \langle f_1, Yf_2\rangle|
   &\ge \sup_{\|f_2\|=1,Yf_2 \ne 0} |\langle \frac{1}{\|Yf_2\|}Yf_2,Yf_2\rangle| \\
   & =\sup_{\|f_2\|=1,Yf_2\ne 0} \|Yf_2\| \\
   & = \sup_{\|f_2\|=1}\|Yf_2\| =\|Y\|
\end{align}
On the other hand,
\begin{align}
     \sup_{\|f_1\|=\|f_2\|=1} |\langle f_1,Yf_2\rangle| &\le \sup_{\|f_1\|=\|f_2\|=1} \|f_1\|\|Yf_2\| \\
   &= \sup_{\|f_1\|=\|f_2\|=1}\|Yf_2\| \\
   &=\sup_{\|f_2\|=1}\|Yf_2\| = \|Y\|.
\end{align}
A: Clearly $\sup_{\|x\|_{2}=1}|{\langle Ax,x \rangle|} \leq \sup_{\|f_1\|_{2} = \|f_1\|_{2} = 1 } \langle f_1, Af_2 \rangle$.  The reverse inequality can be proven in several ways.  For instance, by the spectral theorem, we may assume that $A$ is diagonal.  If every eigenvalue of $A$ has absolute value $\leq C$ and we have two vectors $f_1=(f_1^1,\dots,f_1^n)$ and $f_2=(f_2^1,\dots,f_2^n)$, we have $$\langle f_1,Af_2\rangle=\sum_i f_1^i A_{ii}f_2^i\leq C\sum_i |f_1^if_2^i|\leq C\|f_1\|_2\|f_2\|_2.$$ We can conclude that $\sup_{\|f_1\|_{2} = \|f_1\|_{2} = 1 } \langle f_1, Af_2 \rangle\leq C$.
