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Earlier today I was reading over some analysis notes, and I noticed something interesting and unintuitive. The metric $d_1: \mathcal{C}([0,1],\mathbb{C}) \times \mathcal{C}([0,1],\mathbb{C}) \to \mathbb{R}$ was defined via $d_1(f,g) = \int_0^1|f(x)-g(x)|\mathrm{d}x$, where $\mathcal{C}([0,1],\mathbb{C})$ is the space of continuous functions from $[0,1]$ to $\mathbb{C}$. There were some other simpler theorems proven, but right at the end it said:

Interestingly, one may notice that with $x$ fixed in $[0,1]$, the evaluation map $f \mapsto f(x)$ from $\mathcal{C}([0,1],\mathbb{C})$ to $\mathbb{C}$ is not continuous with respect to $d_1$.

There was no explanation for this. I was trying to think of how I could show this, but nothing I think of seems like it would work. Is there a simple explanation for this?

Charlie Fairfax
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2 Answers2

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I think what might help is a sort of geometric intuition for what is going on here. This doesn't always give you the answer when more complicated spaces are involved, but it's often a good way to try to reason about problems like this.

Consider this: your metric determines the distance between functions based on the area between the graphs of their absolute values. However, the distance between the evaluation of two functions $f$ and $g$ at $x$ is a length, i.e. the straight line distance between $f(x)$ and $g(x)$ in $\mathbb{C}$. Do you see where I am going with this?

EDIT: Check out this graph, and see if you can take it from here.

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This is a linear operator $$ev_x \colon (\mathcal{C}[0, \, 1], \vert \vert \cdot \vert\vert_1) \to (\mathbb{C},\, \vert \cdot \vert), \quad ev_x(f):=f(x)$$ between normed spaces. Now, such an operator is continuous if and only if it is bounded, i.e., there exists a constant $M$ such that $$|ev_x(f)| \leq M \quad \textrm{for all}\,\, f\,\, \mathrm{with}\,\, \vert \vert f \vert \vert_1=1.$$ But clearly you can construct functions $f$ satisfying $\int_0^1 |f(x)|dx=1$ and such that their value in $x$ is arbitrarily large, so the evaluation map $ev_x$ is not bounded.

Francesco Polizzi
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