I am sorry for the lack of specificity in the title. I am having a hard time comprehending a portion of proof that simply asserts an inequality, where I fail to see why that statement should hold true.
To give bit of context.
Let $\gamma$ denote an $N\times{1}$ column vector such that $\gamma\in\mathbb{R}^n$.
Moreover, $\varepsilon$ is a $T\times{N}$ random matrix such that for a sample space $\Omega$ it follows that $\varepsilon:\Omega\to\mathbb{R}^{T\times{N}}$. I
suppose, however, that the random nature of $\varepsilon$ has no implications for my question. So $\varepsilon$ might be thought of as a simple real $T\times{N}$ matrix.
Starting from
$$N^{-2}T^{-1}\,\,\gamma^\prime\varepsilon^\prime\varepsilon^{\phantom{\prime}}\gamma\,=\,N^{-2}T^{-1}\,\,\sum_j\sum_i\sum_t\,\gamma_j\,\gamma_i\,\varepsilon_{jt}\,\varepsilon_{it}$$ it follows by distributivity that. $$N^{-2}T^{-1}\,\,\gamma^\prime\varepsilon^\prime\varepsilon^{\phantom{\prime}}\gamma\,=\,N^{-2}\,\,\sum_j\sum_i\,\gamma_j\,\gamma_i\,\left[T^{-1}\sum_t\,\varepsilon_{jt}\,\varepsilon_{it}\right]$$ So far that's absolutely fine. I am struggling with the next part: $$N^{-2}\,\,\sum_j\sum_i\,\gamma_j\,\gamma_i\,\left[T^{-1}\sum_t\,\varepsilon_{jt}\,\varepsilon_{it}\right]\leq\left[N^{-2}\,\,\sum_j\sum_i\,\gamma^2_j\,\gamma^2_i\,\right]^{1/2}\times\,\left[N^{-2}\,\,\sum_j\sum_i\,\Big(T^{-1}\sum_t\,\varepsilon_{jt}\,\varepsilon_{it}\Big)^2\,\right]^{1/2}$$
I might be at a loss here, but how do I see that this statement is true. The prove moves on the consider the supremum of the above expression over all $\,\gamma:N^{-1}\gamma^\prime\gamma=1\,$. But - according to the proof - the statement above should hold without any restrictions being placed on $\gamma$.
I feel that I could easily construct scenarios where e.g. $$\sum_j\sum_i\,\gamma_j\,\gamma_i>\left[\sum_j\sum_i\,\gamma^2_j\,\gamma^2_i\,\right]^{1/2}.$$
Thank you so very much for taking time.
Best wishes,
Jon
