Disprove continuity by e-d criterium

Question

The image

$d\colon C([a,b])\times C([a,b])\to \mathbb{R}_{\ge 0}\\ (f,g)\mapsto \int_{[a,b]}|f(t)-g(t)|\mathrm{d}t$

defines a metric on the space of continuous functions.

I want to show that

$\Phi_x\colon C([a,b])\to \mathbb{R}\\ f\mapsto f(x)\\ x\in[a,b]$

is NOT (edit: sry, I forgot) continuous.

Thus I have to find one $\epsilon >0$ for which every $\delta > 0$ such that $\int_{[a,b]}|f(t)-g(t)|\mathrm{d}t\ge \epsilon$ for all points for which $|f(x)-g(x)|<\delta$.

Since $f(x)-g(x)\le |f(x)-g(x)|<\delta$ and $\int_{[a,b]}|f(t)-g(t)|\mathrm{d}t\ge \left | \int_{[a,b]}(f(t)-g(t))\mathrm{d}t\right |$ I can choose $\epsilon = \left | \int_{[a,b]}(f(t)-g(t))\mathrm{d}t\right |$ and since $x\in[a,b]$ the inequality

$\int_{[a,b]}|f(t)-g(t)|\mathrm{d}t\ge \epsilon$ holds for every $\delta >0$

Is this correct?

Answer 1

So take the standard counterexample that ZHANG used. $f_n: [0,1] \rightarrow \mathbb{R}$ is given by $f_n(0) =1, f_n(x) = 0$ for $x \ge \frac{1}{n}$ and on $[0, \frac{1}{n}]$ linear down from 1 to 0. All $f_n \in C([0,1])$, and $g(x) =0$ for all $x$ is also in $C[0,1]$.

Then (claim): $\Phi_0$ is not continuous at $g$.

Continuity of $\Phi_0$ at $g$ means in terms of the metrics $d$ and $|.|$ and $\varepsilon,\delta$:

$$\forall \varepsilon > 0 \exists \delta >0: \forall f \in C([0,1]): d(f,g) < \delta \rightarrow \left|\Phi_0(f)-\Phi_0(g)\right| < \varepsilon$$

where $d(f,g) = \int_0^1 \left| f(x) - g(x)\right|dx $is the $L^1$-metric on $C([0,1])$.

So its negation (by simple logic) is:

$$ \exists \varepsilon > 0 \forall \delta >0: \exists f \in C([0,1]): (d(f,g) < \delta) \land (\left|\Phi_0(f)-\Phi_0(g)\right| \ge \varepsilon)$$

Pick $\varepsilon = 1 > 0$ and let $\delta > 0$, then pick $n$ such that $\frac{1}{2n} < \delta$. Then pick $f =f_n$ and note that $d(f_n, g) = \int_0^1 |f_n(x) | dx = \frac{1}{2n} < \delta$ and $|\Phi_0(g) - \Phi_0(f_n)| =|0 -1| = 1 \ge \varepsilon$. So the negation of continuity of $\Phi_0$ at $g$ has been shown.

Also, $x=0$ is not special, we can construct such sequences of spikes at every $x \in [0,1]$, if we like.

Note that this idea works whenever we have a sequence $f_n$ converging to $g$ such that its values do not converge to the value of $g$. So both views of conitnuity at a point, sequences and $\varepsilon, \delta$ are quite equivalent, as you should know.

Answer 2

$\Phi_x$ is not continuous under this metric $d$. You can find a counterexample by considering $g=0$ on $[0,1]$, $f_n$ be a sequence of functions such that $f_n(0)=1$ and $f_n(x)=0$ for $x\in [\frac{1}{n},1]$ and $f_n$ is linear on $ [0,\frac{1}{n}]$ and $\Phi_0$.

Using this counterexample, you can show that there exists $\epsilon>0$ such that for all $\delta>0$, there exists $f,g$ such that $d(f,g)<\delta$ but $|\Phi_0 f-\Phi_0 g|\ge \epsilon$, which proves the discontinuity of $\Phi_0$ by the definition.