# continuity of metric space maps

1.Let $(X_i, d_i)$, $i = 1, 2, 3$, be the metric spaces where $X_1 = X_2 = X_3 =C[0, 1]$ and \begin{align} d_1(f, g) & = \sup_{t \in [0,1]} |f(t) − g(t)| \\ d_2(f, g) & =\int_0^1|f(x)-g(x)|~dx \\ d_3(f, g) &= \left( \int_0^1|f(x)-g(x)|^2~ dx \right)^\frac12 \end{align} Let $\textrm{id}$ be the identity map of $C[0, 1]$ onto itself. Pick out the true statements.

a. $\textrm{id} : X_1 → X_2$ is continuous.
b. $\textrm{id} : X_2 → X_1$ is continuous.
c. $\textrm{id} : X_3 → X_2$ is continuous.

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I don't understand some of this. It will help if you knew some tex commands. –  Taylor Sep 12 '12 at 4:34
∫_0^1 means integration 0 to 1 . sorry i cant wright it properly. –  poton Sep 12 '12 at 4:43
See meta.math.stackexchange.com/questions/5020/… for tips on writing math. I would edit your question myself, but I can't figure out what that supt is supposed to be. –  Gerry Myerson Sep 12 '12 at 4:46
Perhaps adding link to the original source of the problem (probably this one) would be a good idea, especially since you've originally posted the question with some typos and without using LaTeX (therefore the pdf-file is better readable). –  Martin Sleziak Sep 12 '12 at 5:20

Some hints:

(a) Does $f\approx g$ in the sense of $d_1$ imply $f\approx g$ in the sense of $d_2$ ? More precisely: Given $\varepsilon>0$, can you find $\delta>0$ such that $\sup_{t\in[0,1]}|f(t)-g(t)|<\delta$ implies $\int_0^1|f(x)-g(x)|dx<\varepsilon$? Actually, this one is easy as you should observe that the simple choice $\delta=\varepsilon$ works.

(b) Consider $f(x)=0$ and $g_n(x)=x^n$. What is $d_1(f,g_n)$ and $d_2(f,g_n)$? Thus given $\varepsilon=\frac12$, can there exist $\delta>0$ such that ...?

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To answer a and b you can use an answer I posted recently:

The topology $X_1$ contains less open sets than $X_2$ so that a and b are easy to answer.

For c you can use that if the measure space has finite measure and $p \leq q$ then $\|\cdot\|_{L^q} \leq \mu(X)^{1/q - 1/p} \|\cdot\|_{L^p}$ so that $\|f\|_{L^1} \leq \|f\|_{L^2}$ for all $f$ and hence $X_2 \subset X_3$, that is, every open ball in $X_2$ is also open in $X_3$.

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$\newcommand{\norm}[1]{\lVert{#1}\rVert}$If you have already learned about linear normed spaces - in particular if you know that a linear map between normed space is continuous if and only if it is bounded - it might help you with this problem.

Let us recall that a function $\varphi \colon X\to Y$ between two normed spaces is called bounded if there exists a constant $C$ such that $$\norm{\varphi(x)}_Y \le C\norm{x}_X$$ for each $x\in X$.

Notice that all 3 metrics are derived from norms on $C[0,1]$: $$\norm{f}_1=\sup_{x\in[0,1]} |f(x)|\\ \norm{f}_2=\int_0^1 |f(x)| \, dx\\ \norm{f}_3=\left(\int_0^1 |f(x)|^2 \,dx\right)^{1/2}.$$ (I have kept notation corresponding to $d_{1,2,3}$ from your post, but usually the notation $\norm{\cdot}_p$ is used differently, see Wikipedia.)

a. It is relatively easy to show that $\norm{f}_2 \le \norm{f}_1$, which means that $id \colon X_1 \to X_2$ is a bounded linear map, hence it is continuous.

b. The map $id \colon X_1 \to X_2$ is not bounded, you can find functions such that $\int_0^1 f(x)\,dx =1$ but $\sup|f(x)|$ is arbitrarily large. Imagine a function with a high peak on a very short interval.

c. From Cauchy-Schwarz inequality (or by Jensen's inequality applied to the convex function $\varphi(x)=x^2$, as suggested by Matt in comments) you get $$\left(\int |f(x)| \, dx\right)^2 \le \int |f(x)|^2 \,dx$$ which is equivalent to $$\int |f(x)| \, dx \le \left(\int |f(x)|^2 \,dx\right)^{1/2},$$ which is, in your notation $$\norm{f}_2 \le \norm{f}_3.$$ Hence $id \colon X_3 \to X_2$ is bounded (and thus continuous).

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And I'm a bit confused. I can see how to get the first inequality by applying Jensen's inequality. But I can't seem to manage using Cauchy Schwarz. Could you show me how? Thanks a lot! –  Matt N. Sep 24 '12 at 11:17
@Matt Thanks for your comments, I've corrected the link. I meant the version of Cauchy-Schwarz mentioned in Wikipedia article: $\left|\int f(x) \overline{g(x)}\,dx\right|^2\leq\int \left|f(x)\right|^2\,dx \cdot \int\left|g(x)\right|^2\,dx$. If I put $g(x)=1$, I have the required inequality. –  Martin Sleziak Sep 24 '12 at 17:02
Right, of course : ) –  Matt N. Oct 18 '12 at 13:24