# Bounded integral operators in Functional analysis

Let $K: [0,1] \times \mathbb{R}^n \to \mathbb{C}$ have the properties:

1. $K(x,\cdot) \in L^2(\mathbb{R}^n)$ for all $x\in[0,1]$
2. For every $f\in L^2(\mathbb{R}^n)$ the function $$x\mapsto \int_{\mathbb{R}^n} K(x,y)f(y)\,dy$$ is continuous on $[0,1].$ Prove that the intergral operator ${\bf K}$ defined by $${\bf K}f(x) = \int_{\mathbb{R}^n} K(x,y)f(y)\,dy$$ is bounded from $L^2(\mathbb{R}^n)$ to $C([0,1])$.

I know that $L^2(\mathbb{R}^n)$ and $C([0,1])$ are Banach spaces. So maybe we could apply the closed graph theorem? Then closed $\implies$ continuous $\implies$ bounded. How could I prove it? Or should I do it some other way?

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Hint: Use Cauchy-Schwarz inequality

$$|\langle f,g\rangle|\leq ||f||_2 ||g||_2.$$ Here is how you start $$|{\bf K}f(x)| \leq \int_{\mathbb{R}^n} |K(x,y)||f(y)|dy \leq \sqrt{ \int_{\mathbb{R}^n}|k(x,y)|^2 dy} \sqrt{ \int_{\mathbb{R}^n}|f(y)|^2 dy}\dots$$

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I think this is wrong since then: $$|{\bf K}f(x)| \leq \|M(x,.)\|\|f|\|$$ And the bound would depend on $x$? – Johan Dec 3 '12 at 8:51
@Johan: You need to prove $||\bf{K}||\leq M$, so use the definition of the norm of an operator. Note that $\bf{K}f(x)\in C[0,1]$, so you have to know what norm should be used. – Mhenni Benghorbal Dec 3 '12 at 9:25