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Consider the function $u(t) \in C_c\mathbb(R) = $ { $f(t)\in C\mathbb(R) | f(t) = 0, |t| \geq T$, where $T \geq 0 $ }.

Then $u(t)$ bounded and continuous. Can we say that $u(t)$ $\in$ $L^2$ as well?

ofir_13
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  • The square of the absolute value of the function will also be continuous, and it will also be zero outside a compact. –  Jul 24 '18 at 09:46

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$\int_{-\infty } ^{\infty } |u(t)|^{2} \, dt=\int_{-T } ^{T } |u(t)|^{2}\, dt \leq 2CT$ where $C=\sup \{|u(x)|:|x|\leq T\}$, so $u \in L^{2}$.