Let $X = C([0,1];\mathbb{R} )$ be space of continuous real valued fns defined on [0,1] equipped with sup metric $d$ given by : $$d(f,g) = sup_{x \in [0,1]} |f(x) - g(x) |$$
Consider map $I: X \rightarrow \mathbb{R}$ defined by $$I(f) = 1 + \frac{1}{2} \int_{0}^{1} f(x) dx \ \ \ \forall f \in X$$
Prove map $I$ defined FROM sup metric onto $\mathbb{R}$ with usual metric is $\textbf{continuous}$
The way i started was through the definition, so here choose a, continuous at $a$ if given any $\epsilon \ \exists \delta>0 \ s.t. $ $$d_y(I(f) , I(a) ) < \epsilon \ \text{whenever}\ \ d_x(x,a) < \delta$$
So just putting these in we just get $$\left |1 + \frac{1}{2} \int_0^1 f(x)dx - (1 + \frac{1}{2} \int_0^1 f(a)dx\right| < \epsilon$$ Simplifying this, $$\left |\frac{1}{2} \int_0^1 f(x) - f(a)dx\right| \ < \ \epsilon \ \text{whenever} \ d_x(x,a) < \epsilon$$
But here, $$d_x(x,a) = sup_{x \in [0,1]} |x - a |, \ \ \ \text{I think. This is where I'm confused}$$
Simplifying further, we get $$\frac{1}{2} \int_0^1 |f(x)-f(a)| dx < \epsilon$$ whenever the above holds.
Now the part where my question comes in - how do I complete this fully? Is it just choose $$\delta \leq 2\epsilon$$ and the above will all hold, or is that too vague or wrong?
If someone could help that would be really helpful thank you!
What you are supposed to prove is that, if $f\in C([0,1];\Bbb R)$ and $\varepsilon>0$, then there is some $\delta>0$ such that $d(g,f)<\delta\implies\bigl|I(g)-I(f)\bigr|<\varepsilon$. Just take $\delta=\frac\varepsilon2$. Then, if $d(g,f)<\delta$ (that is, if $\sup_{x\in[0,1]}\bigl|g(x)-f(x)\bigr|<\delta$), you have\begin{align}\bigl|I(g)-I(f)\bigr|&=\left|\left(1+\frac12\int_0^1g(x)\,\mathrm dx\right)-\left(1+\frac12\int_0^1f(x)\,\mathrm dx\right)\right|\\&\leqslant\frac12\int_0^1\bigl|g(x)-f(x)\bigr|\,\mathrm dx\\&<\frac12\int_0^1\delta\,\mathrm dx\\&=\frac\delta2\\&=\varepsilon.\end{align}