$L: C[a,b] \to \mathbb{R}$, $L(f) = \int_a^b f$ then $L$ is continuous on $C[a,b]$

Question

Given $L: C[a,b] \to \mathbb{R}$, a function on $C[a,b]$ defined by:

$\displaystyle L(f) = \int_a^b f$ where $f \in C[a,b]$.

Prove $L$ is continuous on $C[a,b]$.

C[a,b] is a metric space which consists of functions continuous on [a,b]. The metric $\Vert \cdot \Vert$ is defined by $\Vert f - g \Vert$ $\displaystyle =\max_{a \leq x \leq b} |f(x) - g(x)|$.

My attempt ($\epsilon-\delta)$ proof:

Let $\epsilon \gt 0$ be given. Choose $\delta = \frac{\epsilon}{b-a}$.

Let $f_0 \in C[a,b]$ and let $\Vert f-f_0 \Vert \lt \delta$ , where $f$ is any function in $C[a,b]$.

Now

$\displaystyle \int_a^bf \leq \left(\max_{a\leq x \leq b} f(x)\right)(b-a)$, and $\displaystyle \int_a^b f_0 \leq \left(\max_{a\leq x \leq b} f_0(x)\right)(b-a)$.

So

$\displaystyle \left| \int_a^bf \right| \leq \left|\max_{a\leq x \leq b} f(x)\right|(b-a)$, and $\displaystyle \left| \int_a^b f_0 \right| \leq \left|\max_{a\leq x \leq b} f_0(x)\right|(b-a)$

Then

$\begin{split} \displaystyle \left| L(f) - L(f_0) \right| & = \left| \int_a^bf - \int_a^b f_0 \right| \\ & \leq \left( \left| \max_{a \leq x \leq b} f(x) \right| - \left| \max_{a \leq x \leq b} f_0(x) \right| \right) \cdot (b-a) \\ & \leq \left| \max_{a \leq x \leq b} f(x) - \max_{a \leq x \leq b} f_0(x) \right| \cdot (b-a) \\ & \leq \left| \max_{a \leq x \leq b} (f(x) -f_0(x)) \right| \cdot (b-a) \\ & = \Vert f-f_0 \Vert \cdot (b-a) \\ & \lt \delta \cdot (b-a) \\ & = \epsilon \end{split}$

If I made any mistake, I would like it to be pointed out and corrected. I would also like some suggestions to improve my proof, and if possible, an alternative proof as well.

Thanks.

Answer 1

Let $f_n\to g$ in the norm specified and let $\epsilon>0$. Then there exists some $N\in\mathbb{N}$ such that $||f_k-g||<\frac{\epsilon}{b-a}$ for all $k>N$. Then

$$ |L(f_k)-L(g)|\leq\int_a^b||f_k-g||dx< (b-a)\cdot\frac{\epsilon}{b-a}=\epsilon. $$

Answer 2

The line that \begin{align*} \left|\int f-\int f_{0}\right|\leq\left(|\max f|-|\max f_{0}|\right)(b-a) \end{align*} is not correct, somehow it is like distributing the absolute value to the functions: $|u-v|\leq|u|-|v|$ is not valid anyway.

The line that \begin{align*} |\max f-\max f_{0}|=|\max(f-f_{0})| \end{align*} is not correct. We have $-\max f_{0}=\min(-f)$ actually.

Anyway, Mnifldz has provided a correct answer.

Another way to prove it is to realize that $L$ is a linear operator, it is equivalent to show that $|L(f)|\leq C\|f\|$ for some constant $C>0$ independent of $f$, in this way, one may take $C=b-a$.