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?

  • $\begingroup$ hint: $f_n(x) = x^n$ on $[0,1]$ $\endgroup$ – user251257 Sep 30 '15 at 1:42

The sequence of functions $$f_n(x) := \max(1 - n|x|,0)$$ $C_{\text{int}}([-1,1])$-converge to $0$, but $‖ f_n - 0 ‖_{C_{\text{max}}} = 1 \not\to0$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.