Prove that an operator is bounded.

Question

Let $X = (C([0,1]), \| \cdot \|_{\infty})$. Define an operator $K$ by

$$ (Kf)(x) := \int_{0}^{1} k(x,y) f(y) dy. $$ Where

$k:[0,1] \times [0,1] \to \mathbb{R}$ continuous.

My try

$$ \|Kf\|_X = \sup_{x \in [0,1]} \left| \int_{n \to 1} k(x,y) f(y) dy \right| \leq \sup_{x \in [0,1]} \left| \int_{0}^{1} M f(y) dy \right|= \\ = |M| \sup_{x \in [0,1]} | \left| \int_{0}^{1} f(y) dy \right| = |M| \cdot \|f \|_X $$

Where $M = \sup_{x,y \in [0,1] \times [0,1]} k(x,y)$ which exists since $k$ continuous on compact interval.

Answer 1

Almost correct. The idea is fine, but your calculation a flaw. When you estimate an absolute value of an integral $$ \def\abs#1{\left|#1\right|}\abs{\int_0^1 k(x,y)f(y)\, dy} $$ you cannot just make the integrand "larger" to make the integral larger, because of signs. This only works for positive integrands. You have to start with the triangle inequality, $$ \abs{\int_0^1 k(x,y)f(y)\, dy} \le \int_0^1 \abs{k(x,y)f(y)}\, dy $$ Then you can go on $$ \int_0^1 \abs{k(x,y)f(y)}\, dy \le M \int_0^1 \abs{f(y)}\, dy $$ with $M := \sup_{x,y} \abs{k(x,y)}$ (note the absolute value here, it is needed!).