# $\varphi_f(x) = \int_0^1 k(x,y) f(y) \, dy$, and $T(f) = \varphi_f$. Show that $T$ is continuous.

(Exercise 7.5.9 Introduction to Real Analysis by Jiri Lebl): Take the metric space of continuous functions $$C([0,1], \mathbb{R})$$. Let $$k : [0,1] \times [0,1] \to \mathbb{R}$$ be a continuous function. Given $$f \in C([0,1], \mathbb{R})$$ define $$\varphi_f(x) = \int_0^1 k(x,y) f(y) \, dy.$$

a) Show that $$T(f) = \varphi_f$$ defines a function $$T:C([0,1], \mathbb{R}) \to C([0,1], \mathbb{R})$$.

Is it enough to say that since $$\varphi_f$$ is well defined, $$T$$ is well defined?

b) Show that $$T$$ is continuous.

I know that $$\varphi_f(x)$$ is continuous on $$[0,1]$$, but I do not know how to prove $$T$$ is continuous on $$C([0,1], \mathbb{R})$$.

Thanks in advance.

## 1 Answer

For (a), you need to show that if $$f$$ is continuous, then $$Tf$$ is continuous (that is, $$T$$ actually sends $$C([0,1])$$ to $$C([0,1])$$). This follows from the (uniform) continuity of $$k$$. For (b), you need to show that the map $$f\mapsto Tf$$ is continuous. Indeed, note that if one fixes $$\epsilon>0$$ and $$f,g\in C([0,1]),$$ then $$|Tf(x)-Tg(x)|\leq\int\limits_0^1 |k(x,y)||f(y)-g(y)|\, dy\leq \max|k| \|f-g\|_\infty.$$ So, choosing $$\delta=\epsilon/\max |k|$$ will establish continuity. I'll let you fill in the details (they're very straightforward).