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Let $ \ C[0,1] \ $ stands for the real vector space of continuous functions $ \ [0,1] \to [0,1] \ $ on the unit interval with the usual subspace topology from $\mathbb{R}$. Let $$\lVert f \rVert_1 = \int_0^1 |f(x)| \ dx \qquad \text{ and } \qquad \lVert f \rVert_{\infty} = \max_{x \in [0,1]} |f(x)|$$ be the usual norms defined on that space. Let $ \ \Delta : C[0,1] \to C[0,1] \ $ be the diagonal function, ie, $ \ \Delta f=f \ $, $\forall f \in C[0,1]$. Then $$ \Delta = \big\{ (f,g) \in C[0,1] \times C[0,1] \ : \ g=f \ \big\} \ . $$ My questions are

(1) $ \ \ $ Is $ \ \Delta \ $ a closed set of $ \ C[0,1] \times C[0,1] \ $, with respect to the product topology induced by these norms?

(2) $ \ \ $ Is $ \ \Delta : (C[0,1], \lVert \cdot \rVert_1) \to (C[0,1], \lVert \cdot \rVert_{\infty}) \ $ continuous?

(3) $ \ \ $ Does $ \ \Delta : (C[0,1], \lVert \cdot \rVert_1) \to (C[0,1], \lVert \cdot \rVert_{\infty}) \ $ maps closed sets of $ \ (C[0,1], \lVert \cdot \rVert_1) \ $ onto closed sets of $ \ (C[0,1], \lVert \cdot \rVert_{\infty}) \ $?

(4) $ \ \ $ Is $ \ \Delta : (C[0,1], \lVert \cdot \rVert_{\infty}) \to (C[0,1], \lVert \cdot \rVert_1) \ $ continuous?

(5) $ \ \ $ Does $ \ \Delta : (C[0,1], \lVert \cdot \rVert_{\infty}) \to (C[0,1], \lVert \cdot \rVert_1) \ $ maps closed sets of $ \ (C[0,1], \lVert \cdot \rVert_{\infty}) \ $ onto closed sets of $ \ (C[0,1], \lVert \cdot \rVert_1) \ $?

Now about some terminology, when I say that "$\Delta \ $ is closed", or that "$\Delta \ $ is a closed map" or that "$\Delta \ $ is a closed operator"?

Thanks in advance.

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Gustavo
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    I think you'd better ask them separately in several questions. Also please show what you have tried. – John Feb 08 '17 at 14:08
  • $\Delta$ function would be better named $i$ or $\operatorname{id}$. Also, the Diagonal set is closed in $Y \times Y$ iff $Y$ is Hausdorff. – Henno Brandsma Feb 08 '17 at 17:32
  • @JohnZHANG Sorry, I am now writing my efforts. – Gustavo Feb 08 '17 at 18:18
  • I did some research and found a funny terminology: a closed map is one that maps closed sets onto closed sets and a closed operator is a map that is closed w.r.t. the product topology. This is weird to me: it is like an operator could be something which is not a function... – Gustavo Feb 09 '17 at 02:23

3 Answers3

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Only for (1). Clearly $(x^n,x^n)\in\Delta$ for each $n$. Note $x^n\to 0$ if $\in[0,1)$ and $x^n\to 1$ if $x=1$. Let $$ f=\left\{\begin{array}{rl} 0&\text{ if }x\in[0,1)\\ 1&\text{ if }x=1 \end{array}\right. $$ It is easy to see that $(x^n,x^n)\to (f,f)$ with norm $\|\cdot\|_1$. However $(f,f)\not\in\Delta$. Thus $\Delta$ is not closed with the product topology induced by $\|\cdot\|_1$.

Note if $u_n\to u$ with norm $\|\cdot\|_{\infty}$, then $u\in C[0,1]$ and hence $(u,u)\in\Delta$. It easy to see $(u_n,u_n)\to (u,u)$ with the product topology induced by $\|\cdot\|_{\infty}$. So $\Delta$ is closed with the product topology induced by $\|\cdot\|_{\infty}$.

xpaul
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  • ${(y,y); y \in Y}$ is closed in $Y \times Y$ in the product topology iff $Y$ is Hausdorff. Both norm topologies on $C([0,1])$ are metric so Hausdorff. – Henno Brandsma Feb 08 '17 at 17:58
  • Your example with the $x^n$ does not count. It just shows that $C([0,1])$ is not closed in $L^1([0,1])$, but that's not the question. – Henno Brandsma Feb 08 '17 at 17:59
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As to 1), this is just general topology: $\{(x,x): x \in X\}$ is closed in $X \times X$ in the product topology iff $X$ is Hausdorff. All metric topologies are Hausdorff. So yes, $\Delta$ is closed in both product topologies. See these answers. This assumes that both copies of $C([0,1])$ have the same topology. The question is unclear on this point.

As to (2), consider the spike functions $f_n(x): f_n(x) = 0$ for $x \in [\frac{1}{n}, 1]$ and $f_n(x)$ is linear from $1$ to $0$ between $0$ and $\frac{1}{n}$.

Then $f_n(x) \rightarrow 0$ (the all zero-function) in the $L^1$ norm but $\left||\Delta(f_n)\right||_\infty = 1$ for all $n$, so does not converge to $\Delta(0) = 0$ in the $L^\infty$ norm.

If $f_n \rightarrow f$ in the sup norm, $||f_n -f||_1 = \int_0^1 |f_n -f|dx \le ||f_n - f||_\infty$ and this shows $\Delta$ in (4) is even Lipschitz.

I'll have to think about the closedness a bit.

As to closedness $i_1: (X, \mathcal{T}) \rightarrow (X,\mathcal{T}') ,i_1(x) =x$ is continuous iff $i_2: (X, \mathcal{T}') \rightarrow (X,\mathcal{T}), i_2(x) =x$ maps closed sets to closed sets: suppose the first $i_1$ is continuous, if $C$ be closed in $\mathcal{T}'$ , then $i_2[C] = (i_1)^{-1}[C]$ is closed in $\mathcal{T}$. The other direction is similar.

This should settle the closed-map questions.

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Henno Brandsma
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  • I also got it in a different (but similar) way

    (4) $ \ \ $ Let $ \ f_0 \in C[0,1] , $. Let $ \ \varepsilon > 0 , $. Choose $ \ \delta = \varepsilon > 0 \ $ and we have that, for all $ \ f \in C[0,1] , $, if $ \ \lVert f - f_0 \rVert_{\infty} < \delta , $, then $$\lVert \Delta f - \Delta f_0 \rVert_1 = \lVert f - f_0 \rVert_1 \leq \lVert f - f_0 \rVert_{\infty} < \delta = \varepsilon \ . $$ Then $ \ \Delta : (C[0,1] , \lVert \cdot \rVert_{\infty}) \to (C[0,1] , \lVert \cdot \rVert_{\infty}) \ $ is continuous at $f_0 \ $.

     – Gustavo Feb 08 '17 at 20:47
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    I did not know that the theorem about closedness of the diagonal referent to the Hausdorffness of the space works if the factors have different topologies. However, in my case, we have that $ \ \Delta_{11} \ $ is closed in $ \ C_1 \times C_1 \ $ and $$\Delta_{(4)} \times \Delta_1 : C_{\infty} \times C_1 \to C_1 \times C_1$$ is continuous. So, $ \ \Delta_{\infty 1} = (\Delta_{(4)} \times \Delta_1)^{-1} [\Delta_{11}] \ $ is also closed in $ \ C_{\infty} \times C_1 \ $. – Gustavo Feb 08 '17 at 21:01
  • @Gustavo you're giving the proof that Lipschitz functions are continuous. It's also what I use. – Henno Brandsma Feb 08 '17 at 21:02
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    I disagree that "the question is unclear". The question (1) was "Is $ \ \Delta \ $ a closed set of $ \ C[0,1] \times C[0,1] \ $, with respect to the product topology induced by these norms?" I was talking about (these) two norms, not just one alone. Sorry, but I think that sounds pretty obvious in the body of the question. However, my last comment has fixed this detail. – Gustavo Feb 08 '17 at 21:32
  • @Gustavo I concur. $\Delta_{\infty 1} $ is indeed closed in $C_\infty \times C_1$. – Henno Brandsma Feb 09 '17 at 01:21
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    @Gustavo You could interpret "these norms" as meaning two cases: one where both have L1 norm, and one where both have $L^\infty$ norm. It is ambiguous in this sense. – mathematician Feb 09 '17 at 01:34
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My edition took too many lines, so I decided to write it as an answer.

(1) $ \ \ $ The spaces $ \ (C[0,1], \lVert \underline{} - \underline{} \rVert_1) \ $ and $ \ (C[0,1], \lVert \underline{} - \underline{} \rVert_{\infty}) \ $ are metric spaces. Thus the product topology is induced by, say, the max-product metric $$d : (C[0,1] \times C[0,1]) \times (C[0,1] \times C[0,1]) \to \mathbb{R}$$ given by $ \ \ d \big( (f,g) , (h,k) \big) = \max \{ \lVert f-g \rVert_1 \, , \lVert h-k \rVert_{\infty} \} \ $, $\forall f,g,h,k \in C[0,1]$.

Let $ \ \bar{\Delta} = \mathcal{C} \ell ( \Delta) \ $ be the topological closure of $\Delta$ with respect to the product topology. It is obvious that $ \ \Delta \subset \bar{\Delta} \subset C[0,1] \times C[0,1] \ $. Let $ \ v \in \bar{\Delta} \ $. By countable choice, there exists $ \ (v_n)_{n \in \mathbb{N}^*} \in \Delta^{\mathbb{N}^*} \ $ such that $ \displaystyle \ \lim_{n \to \infty} v_n = v \ $ with respect to the product topology. So, for each $ \ n \in \mathbb{N}^*$, there exists $ \ f_n \in C[0,1] \ $ such that $ \ v_n = (f_n , f_n) \in \Delta \, $. Furthermore, there exists $ \ \varphi , \psi \in C[0,1] \ $ such that $$\lim_{n \to \infty} (f_n , f_n) = \lim_{n \to \infty} v_n = v = (\varphi , \psi) \in C[0,1] \times C[0,1] \ . $$ Therefore, for all $ \ \varepsilon > 0 \, $, there exists $ \ N \in \mathbb{N}^* \ $ such that, $\forall n \in \mathbb{N}^*$, if $ \ n>N$, then $$\max \{ \lVert f_n - \varphi \rVert_1 \, , \lVert f_n - \psi \rVert_{\infty} \} = d \big( (f_n , f_n) , (\varphi , \psi) \big) = d(v_n,v) < \varepsilon \ . $$ Thus $ \ \lim f_n = \varphi \ \ $ in $ \ (C[0,1], \lVert \cdot \rVert_1) \ $ and $ \ \lim f_n = \psi \ \ $ in $ \ (C[0,1], \lVert \cdot \rVert_{\infty}) \ $.

For each $ \ n \in \mathbb{N}^*$ we have that

\begin{eqnarray*} \lVert \varphi - \psi \rVert_1 & \leq & \lVert \varphi - f_n \rVert_1 + \lVert f_n - \psi \rVert_1 \\ & = & \lVert \varphi - f_n \rVert_1 + \int_0^1 |f_n (x) - \psi(x)| \ dx \\ & \leq & \lVert \varphi - f_n \rVert_1 + \int_0^1 \sup_{x \in [0,1]} |(f_n - \psi)(x)| \ dx \\ & = & \lVert \varphi - f_n \rVert_1 + \lVert f_n - \psi \rVert_{\infty} \ \ \ . \end{eqnarray*}

Let $ \ \varepsilon > 0 \, $. There exists $ \ N_1 , N_2 \in \mathbb{N}^*$ such that, for all $ \ n \in \mathbb{N}^*$, $$n>N_1 \ \Rightarrow \ \lVert \varphi - f_n \rVert_1 < \frac{\varepsilon}{2}$$ and $$n>N_2 \ \Rightarrow \ \lVert f_n - \psi \rVert_{\infty} < \frac{\varepsilon}{2} \ \ . $$ It follows that, $\forall n \in \mathbb{N}^*$, if $ \ n > N = \max \{ N_1 \, , N_2 \}$, then $$\lVert \varphi - \psi \rVert_1 \leq \lVert \varphi - f_n \rVert_1 + \lVert f_n - \psi \rVert_{\infty} < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon \ . $$ We are left with $ \ 0 \leq \lVert \varphi - \psi \rVert_1 < \varepsilon \ $, for all $ \ \varepsilon > 0 \, $. Therefore, $\lVert \varphi - \psi \rVert_1 = 0 \ \Rightarrow \ \varphi = \psi \ $. Thus $ \ (\varphi , \psi) = v \in \Delta \, $. As $v$ was arbitrarily chosen, we have that $ \ \bar{\Delta} \subset \Delta \, $. Hence $ \ \bar{\Delta} = \Delta \ $ and we conclude that $\Delta$ is closed w.r.t. the product topology.

Gustavo
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