Exercise 8.2.6

Question

Which of these functions from $C[0,1]$ to $\mathbf{R}$ (with the usual metric) are continuous?
\begin{enumerate}[label=(\alph*)] 
\item $g(f) = \int^1_0 fk$, where $k$ is some fixed function in $C[0,1]$.
\item $g(f) = f(1/2)$.
\item $g(f) = f(1/2)$, but this time with respect to the metric on $C[0,1]$ from Exercise 8.2.2 (c).
 \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] \item Continuous. Let $M$ be a bound on $|k|$, and for any $\epsilon > 0$, let $\delta = \epsilon / M$. Then for any function $h$ satisfying $d(f,h) < \delta$, we have $$\left|\int^1_0 fk - \int^1_0 hk\right| = \left|\int^1_0 (f-h)k\right| \leq \int^1_0 \left|(f-h)\right|\left|k\right| < \int^1_0 \delta M = \epsilon$$ \item Continuous. Let $\delta = \epsilon$, and note that $\left|f(1/2) - h(1/2)\right| \leq d(f,h) < \delta = \epsilon$ \item Not continuous. Let $f = 0$, and for any $\delta > 0$, define $$h_\delta(x) = \begin{cases} 1/2 & x \in V_\delta(1/2) \ 0 & \text{otherwise} \end{cases} $$ Clearly for any $\delta$, $d(h_\delta,f) = \delta$. But $(h-f)(1/2) = 1/2$, so any $\epsilon < 1/2$ cannot be achieved. \end{enumerate}

Exercise 8.2.6

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Exercise 8.2.6

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