Discussion
Let \(T : L^{2}[-1,1] \to L^{2}[-1,1]\) be defined by \(Tf = \tilde{f}\), where \(\tilde{f}(x) = f(-x)\) almost everywhere. If M is the kernel of I - T, then the distance between the function \(\phi(t) = e^{t}\) and M is
Answer: D
The distance is \(\frac{1}{2}\sqrt{e^{2} - e^{-2} - 4}\).
The kernel of I - T is the set of f with f(-x) = f(x), the even functions. Its orthogonal complement in \(L^{2}[-1, 1]\) is the odd functions. Split \(e^{t} = \cosh t + \sinh t\): the even part lies in M, so the distance from \(e^{t}\) to M is the norm of the odd part, sinh t:
\[ \begin{aligned} \int_{-1}^{1} \sinh^{2} t\,dt &= \int_{-1}^{1} \frac{\cosh 2t - 1}{2}\,dt \\ &= \frac{\sinh 2}{2} - 1 \\ &= \frac{e^{2} - e^{-2} - 4}{4} \end{aligned} \]
Taking the square root gives \(\frac{1}{2}\sqrt{e^{2} - e^{-2} - 4}\). The option with + 4 has the sign of the -1 term reversed, the one with - 2 halves it, and \(\frac{1}{2}\sqrt{e^{2} - 4}\) drops the \(e^{-2}\) term.
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