Question on LSI system
Given an impulse response of 
$$h(t) = 
\begin{cases} \frac{1}{t^2} &amp;,&amp; \text{ for } t &lt; 1/2 \\ 0 &amp;,&amp; \text{ otherwise } \\ \end{cases}
$$

Its associated CTFT (frequency response) is 
$$H(\Omega) = \int_{-\infty}^{\infty} h(t) e^{-j \Omega t} dt $$

Now look at its value at $\Omega = 0$ which is $H(0)$; 
$$H(0) = \int_{-\infty}^{\infty} h(t) dt = \int_{-\infty}^{\infty} \frac{1}{t^2} dt = 2 $$

Conclusion: since $H(0)$ is $2$ , then $1$ or $1/2$ or any number $x$ less than $2$; i.e., $x &lt; 2$, cannot be an upper bound for $H(\Omega)$ as it already passes that value...