Prove openess or closure of $\int_{a}^{b} f(x) \partial x$

Question

Let the space $C([a,b],\mathbb{R})$ with the metric $d_{∞}$.

a) Prove that the following set $U=(f \in C([a,b],\mathbb{R}) / \int_{a}^{b} f(x) \partial x >0 )$ is open in $C([a,b],\mathbb{R})$

b) Prove that the following set $U=(f \in C([a,b],\mathbb{R}) / \int_{a}^{b} f(x) \partial x \geq 0 )$ is closed in $C([a,b],\mathbb{R})$

My attempt

For a)

Let $f \in U$ some fixed functio. We have to prove that $f$ is an interior point of $U.

Let's take $\epsilon=min_{a\leq x \leq b}\int_{a}^{b} f(x) \partial x >0$, for which $B_{∞}(f,\epsilon) \subset U$. Thus $g\in B_{∞}(f,\epsilon)$, then $d_{∞}(f,g)<\epsilon$. Then for every $x \in [a,b]$, we have $|\int_{a}^{b} f(x) \partial x-\int_{a}^{b} g(x) \partial x| \leq max_{a\leq x \leq b} |\int_{a}^{b} f(x) \partial x-\int_{a}^{b} g(x) \partial x| < min_{a\leq x \leq b}\int_{a}^{b} f(x) \partial x \leq \int_{a}^{b} f(x) \partial x$. Then $|\int_{a}^{b} f(x) \partial x-\int_{a}^{b} g(x) \partial x| < \int_{a}^{b} f(x) \partial x$. That is \int_{a}^{b} g(x) \partial x >0, and this means $g \in U$.

For b)

We have a sequence $(f_{n})_{n \in \mathbb{N}}$ in $C$ with $\lim_{n \rightarrow ∞} f_{n} = f$ with $d_{∞}$. We have to prove that $f \in C$.

By property we have that for every $x \in [a,b]$, $\lim_{n \rightarrow ∞} \int_{a}^{b} f_{n}(x) \partial x = \int_{a}^{b} f(x) \partial x$.

Given that $f_{n} \in C$ for every $n$, we have that $\int_{a}^{b} f_{n}(x) \partial x \geq$ for every $x \in[a,b]$, and that is $\int_{a}^{b} f(x) \partial x \geq$ for every $x \in[a,b]$. Finally, this implies $f \in C$.

Answer 1

To handle both parts together, it is useful to consider the (linear) map $\varphi \colon C[0,1] \rightarrow \mathbb{R}$ given by

$$ \varphi(f) = \int_0^1 f(x) \, dx $$

and show that it is a continuous map. Then $f^{-1}((0,\infty))$ will be open and $f^{-1}([0,\infty))$ will be closed. To see that $\varphi$ is continuous, it is enough to show that it is bounded which follows from

$$ |\varphi(f)| = \left| \int_0^1 f(x) \, dx \right| \leq \int_0^1 |f(x)| \, dx \leq \int_0^1 ||f||_{\infty} = ||f||_{\infty}.$$

Answer 2

Let $E$ be your space, and consider

$$g\; \; :E \to \mathbb R$$

$$\;\;\;\;\;f \mapsto \int_a^b f(x)dx$$

$g$ is a linear application continuous at $0$ since if $(f_n)$ converges uniformly to $0$, then $g(f_n)$ converges to $g(0)=0$.

For the first case,

$U=g^{-1}((0,+\infty))$ .

but

$(0,+\infty)$ is an open in $\mathbb R$

and

$g$ continuous at $E$

$\implies U$ is an open in $E$.

for the second, it is closed.