Here is Prob. 9 in the Problems after Sec. 3.2 in Introductory Functional Analysis With Applications by Erwine Kreyszig:

Let $V$ be the vector space of all continuous complex-valued functions on $J = [a,b]$. Let $X_1 = \left( V, \Vert \cdot \Vert_\infty \right)$, where $\Vert x \Vert_\infty = \max_{t \in J} \vert x(t) \vert$; and let $X_2 = \left( V, \Vert \cdot \Vert_2 \right)$, where $$\Vert x \Vert_2 = \langle x, x \rangle^{1/2}, \ \ \ \ \langle x, y \rangle \ = \ \int_a^b \ x(t) \overline{y(t)} \ \mathrm{d} t. $$ Show that the identity mapping $x \mapsto x$ of $X_1$ onto $X_2$ is continuous. ...

My Solution:

For any $x, y \in V$, we have $$ \begin{align*} \Vert x-y \Vert_2^2 \ &= \ \int_a^b \ \vert x(t) - y(t) \vert^2 \ \mathrm{d} t \\ &\leq \ \int_a^b \ \max_{s \in J} \vert x(s) - y(s) \vert^2 \ \mathrm{d} t \\ &= \ \int_a^b \ \left( \max_{s \in J} \vert x(s) - y(s) \vert \right)^2 \ \mathrm{d} t \ \ \ \mbox{ [Is this step justified, I wonder?]} \\ &= \ \left( \max_{s \in J} \vert x(s) - y(s) \vert \right)^2 \ \int_a^b \ \mathrm{d} t \\ &= \Vert x-y \Vert_\infty^2 (b-a); \end{align*} $$ so $$ \Vert x-y \Vert_2 \leq \Vert x-y \Vert_\infty \sqrt{b-a}. $$ So, given $\epsilon > 0$, if we choose $\delta$ such that $0 < \delta < \frac{\epsilon}{\sqrt{b-a}}$, then $$ \Vert x-y \Vert_2 < \epsilon $$ for all $x, y \in V$ such that $$ \Vert x-y \Vert_\infty < \delta, $$ showing that the identity map is even uniformly continuous.

Is the above proof correct?

If so, then how to show directly if the inverse map is also continuous or not?

  • 1
    $\begingroup$ That intermediate step is redundant, as you suspected. The proof is correct. The statement essentially tells you that if the sequence $f_n$ tends uniformly to $f$, then the integrals of $f_n$ will tend to the integral of $f$. $\endgroup$ – Alex M. Apr 22 '15 at 7:10
  • $\begingroup$ If you instead note that $|x-y| \le \|x-y\|_{\infty}$, then the steps become more obvious. The max operator can be confusing. $\endgroup$ – DisintegratingByParts Apr 22 '15 at 8:02
  • $\begingroup$ your proof is correct but notice one thing that proceeding the same way as you did you can show that $\|x\|_2\leq \sqrt{b-a}\|x\|_{\infty}$ for all $x \in X_1$. So you are done here as this shows that the identity operator is bounded and hence continuous. $\endgroup$ – Urban PENDU Apr 22 '15 at 8:55

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