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I have the following problem: Consider the identity map $id: C_{max} \rightarrow C_{int}$ where $C_{max}$ is the metric space $C([a,b],\mathbb{R})$ of continuous real valued function defined on $[a,b]$ equipped with the metric $d_{max}(f,g) = \max|f(x)-g(x)|$, and $C_{int}$ is $C([a,b],\mathbb{R})$ equipped with the integral metric, $d_{int}(f,g) = \int_{a}^b |f(x) - g(x)| dx$. Show that $id$ is a continuous linear bijection (an isomorphism) but its inverse is not continuous.

I proved continuity in the following manner: Note that it is trivially clear that the identity map $id$ is a bijection. We must then show that $id$ is continuous. For any given $\epsilon > 0$, suppose that $d_{max}(f,g) = \max|f - g| < \delta$ for $\delta < \frac{\epsilon}{b-a}$. Then we notice that

\begin{align} d_{int}(f,g) &= \int^a_b|f - g|dx \\ &< \int^a_b\delta \\ &= \delta(b - a) \\ &< \epsilon. \end{align}

However, I'm absolutely stuck on showing that the inverse of ${id}^{-1}$ is not continuous. My intuition says to us proof by contradiction, but I don't know what to do here. How would I begin going about proving that fact that ${id}^{-1}$ is not continuous?

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  • $\begingroup$ hint: $f_n(x) = x^n$ on $[0,1]$ $\endgroup$ – user251257 Sep 30 '15 at 1:42
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The sequence of functions $$f_n(x) := \max(1 - n|x|,0)$$ $C_{\text{int}}([-1,1])$-converge to $0$, but $‖ f_n - 0 ‖_{C_{\text{max}}} = 1 \not\to0$.

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