Continuity in a Banach Space

Question

This is a past exam question:

On the Banach space $\bigg(C\bigg( \bigg[-\dfrac{1}{2}, \dfrac{1}{2} \bigg], ||\cdot||_\infty\bigg) $, consider the operator given by

$$T(f)(x)= x+ \displaystyle\int^{1/2}_0 f\bigg(\dfrac{x+t}{2}\bigg)dt$$

Prove that $Tf$ is continuous for $f \in C\bigg(\bigg[-\dfrac{1}{2},\dfrac{1}{2}\bigg]\bigg) $

Please help me in the right direction. Similar questions seem to come up every year and I'm not really sure how to tackle it.

Answer 1

Let $f,g\in C\left(\left[-\frac{1}{2},\frac{1}{2}\right],\|\cdot\|_\infty\right)$, then $$|Tf(x)-Tg(x)| = \left|\int_0^\frac{1}{2} f\left(\frac{x+t}{2}\right)-g\left(\frac{x+t}{2}\right)\,\mathrm{d}t\right| $$ $$\leq \int_0^\frac{1}{2}\left|f\left(\frac{x+t}{2}\right)-g\left(\frac{x+t}{2}\right)\right|\,\mathrm{d}t \leq \frac{1}{2}\|f-g\|_\infty.$$ Hence it follows that $\|Tf-Tg\|_\infty\leq\frac{1}{2}\|f-g\|_\infty$ and $T$ is continuous.

Answer 2

In order to solve this, we have to show that the transformation is continuous at every point, meaning:

$\forall f\in C[-\frac{1}{2}, \frac{1}{2}],\forall \varepsilon>0, \exists \delta>0: \forall g\in B(f,\delta), ||T(f)-T(g)||_{\infty}<\varepsilon $

Let's assume that $max_{x \in [-\frac{1}{2},\frac{1}{2}]} |f(x)-g(x)|=||f-g||_{\infty}<2 \varepsilon$

Let's look at: $||T(f(x))-T(g(x))||_{\infty}=||\int_0^{\frac{1}{2}} (f(\frac{x+t}{2})-g(\frac{x+t}{2}))dt||_{\infty}$

Let's note that according to our assumption, $ f(\frac{x+t}{2})-g(\frac{x+t}{2})\leq 2\varepsilon $

and therefore: $ ||\int_0^{\frac{1}{2}} (f(\frac{x+t}{2})-g(\frac{x+t}{2}))dt||_{\infty}<\frac{1}{2}\cdot 2\varepsilon=\varepsilon $

Therefore, T is continuous.