Suppose that k:[-1,1] →ℝ is a measurable function and that ∫-1^1|k(s)| d s<∞. Define K by (K ϕ)(

Step 1: u000AFirst, we need to show that $K$ is a bounded linear map from $L_2(0,1)$ to itself. To s

Suppose that k:[-1,1] →ℝ is a measurable function and that...

Suppose that $k:[-1,1] \rightarrow \mathbb{R}$ is a measurable function and that $\int_{-1}^1|k(s)| \mathrm{d} s<\infty$. Define $K$ by $$ (K \phi)(x)=\int_0^1 k(x-t) \phi(t) \mathrm{d} t \quad(0 \leqslant x \leqslant 1) $$ and show that $K$ is a bounded linear map from $L_2(0,1)$ to itself with $\|K\| \leqslant \int_{-1}^1|k(s)|$ ds. (Hint: the kernel is a Schur kernel.) Now define $k_n$ by $k_n(s)=k(s)$ if $|k(s)| \leqslant n$ and $k_n(s)=0$ if $|k(s)|>n$, so that $k_n$ is a bounded function and the operator $K_n$ defined by $\left(K_n \phi\right)(x)=\int_0^1 k_n(x-t) \phi(t) \mathrm{d} t(0 \leqslant x \leqslant 1)$ is compact. Deduce that $\left\|K_n-K\right\| \rightarrow 0$ and thus that $K$ is compact.

Suppose that $k:[-1,1] \rightarrow \mathbb{R}$ is a measurable function and that $\int_{-1}^1|k(s)| \mathrm{d} s<\infty$. Define $K$ by
$$
(K \phi)(x)=\int_0^1 k(x-t) \phi(t) \mathrm{d} t \quad(0 \leqslant x \leqslant 1)
$$
and show that $K$ is a bounded linear map from $L_2(0,1)$ to itself with $\|K\| \leqslant \int_{-1}^1|k(s)|$ ds. (Hint: the kernel is a Schur kernel.) Now define $k_n$ by $k_n(s)=k(s)$ if $|k(s)| \leqslant n$ and $k_n(s)=0$ if $|k(s)|>n$, so that $k_n$ is a bounded function and the operator $K_n$ defined by $\left(K_n \phi\right)(x)=\int_0^1 k_n(x-t) \phi(t) \mathrm{d} t(0 \leqslant x \leqslant 1)$ is compact. Deduce that $\left\|K_n-K\right\| \rightarrow 0$ and thus that $K$ is compact.

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