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(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.

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1 Answer 1

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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).

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