Is the diagonal $ \ \Delta : C[0,1] \to C[0,1] \ $ closed?

Question

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.

Answer 1

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.

Answer 2

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 $\|\cdot\|_{\infty}$.

Answer 3

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.