Proving continuity with two different metrics

Question

Problem statement:

Let $X$ be the set of all continuous functions $f:[a,b]\rightarrow \Bbb R$, and define the metric $d^*(f,g)$ on $X$ by $$d^*(f,g) = \int_{a}^{b} |f(t) - g(t)|dt$$

Now, for each $f \in X$, $I(f) = \int_a^b f(t)dt$

Prove that $I: (X,d^*) \rightarrow (\Bbb R, d)$ is continuous, where $d(a,b) = |a-b|$.

My attempt:

I honestly feel really slow, this was the first problem at the end of the chapter and I have no idea where to start. I feel like it is very valuable that the set $X$ contains only continuous functions, but I'm not sure how to utilize it. I'm fairly sure I need to prove that $d^*(f(t),f(a)) < \delta \implies d(t,a) < \epsilon$, so I need to work backwards and find $\delta(\epsilon)$, but I am thoroughly confused...

If anyone can help, and perhaps give some tips on constructing $\delta - \epsilon$ proofs in a somewhat intuitive manner, I would be extremely greatful.

Answer 1

A note:

• The points of $X$ are functions. Thus you have to think of functions as individual objects. In school functions are introduced as being processes which change numbers (for example $x\mapsto x^2$). Now you have to take the whole function as an object.

Some tips:

1) Write down the $\epsilon$-$\delta$ definition for the current problem:

$$\forall f \in X \forall \epsilon > 0 \exists \delta > 0 \forall g \in X: d^*(f,g) < \delta \Rightarrow d(I(f),I(g))< \epsilon$$

2) Plug in the specific definitions:

$$\begin{array}{l}\forall f \in X \forall \epsilon > 0 \exists \delta > 0 \forall g \in X: \\ \quad\quad\int_a^b|f(t)-g(t)|dt < \delta \Rightarrow |\int_a^b f(t)dt-\int_a^b f(t)dt|< \epsilon \end{array}$$

So given a $f\in X$ and an $\epsilon > 0$ you have to find a $\delta(\epsilon, f) > 0$ such that

$$\int_a^b|f(t)-g(t)|dt < \delta \Rightarrow |\int_a^b f(t)dt-\int_a^b f(t)dt|< \epsilon$$

regardless of $g\in X$.

Hint: You might use $|\int_a^b h(t) dt|< \int_a^b |h(t)| dt$ for any function $h$.

Answer 2

Let $f_n \to f$ in the topology of $d^*$. Then $d(I(f_n), I(f)) =|I(f_n) - I(f)| = |I(f_n-f)| \leq I(|f_n - f|) = d^* (f_n,f)$. (I have used that $|\int f| \leq \int |f|$.) Now pick some arbitrary $\varepsilon >0$ and take $\delta = \varepsilon$ and this is all you need.

Answer 3

Since other answers have covered the direct approach, I'll give an alternate one. $I$ is a linear functional. Note that $|I(f)| \leq \|f\|$. In particular, $I$ is a bounded functional, hence continuous.