$Exercise \oplus Problem 17$

你好，这里是我的个人网站数学分析的每周一题栏目(数学分析每周一题，其中数学分析指的是数学中的分析学, 主要包括微积分，实分析，复分析) ——————Alina Lagrange

Prove that

lim_{δ \to 0} ‖ f * K_{δ} - f ‖ = 0 .

where

K_{δ}

is a family of good kernels.

$P r o o f .$

\begin{aligned} ‖ f * K_{δ} - f ‖ & = \int_{R^{n}} | f * K_{δ} (x) - f (x) | d x \\ = \int_{R^{n}} | \int_{R^{n}} f (x - y) K_{δ} (y) d y - f (x) | d x \\ = \int_{R^{n}} | \int_{R^{n}} (f (x - y) - f (x)) K_{δ} (y) d y | d x \\ \leq \int_{R^{n}} \int_{R^{n}} | f (x - y) - f (x) | | K_{δ} (y) | d y d x \\ = \int_{R^{n}} \int_{R^{n}} | f (x - y) - f (x) | | K_{δ} (y) | d x d y \\ = \int_{| y | \leq ℓ} \int_{R^{n}} | f (x - y) - f (x) | | K_{δ} (y) | d x d y \\ + \int_{| y | > ℓ} \int_{R^{n}} | f (x - y) - f (x) | \\ = I_{1} + I_{2} . \end{aligned}

Then for

I_{1}

\begin{aligned} I_{1} & = \int_{| y | \leq ℓ} | K_{δ} (y) | \int_{R^{n}} | f (x - y) - f (x) | d x d y \\ = \int_{| y | \leq ℓ} | K_{δ} (y) | ‖ f (x - y) - f (x) ‖_{1} d x d y \end{aligned}

where

ℓ > 0

. By the continuity of the Lebesgue integral i.e.

\forall ε > 0, \exists ζ > 0,

when

| y | < ζ, ‖ f (x - y) - f (x) ‖_{1} < ε .

Hence

I_{1} \leq ε \int_{| y | \leq ℓ} | K_{δ} (y) | d y \leq M ε .

Then for

I_{2}

\begin{aligned} I_{2} = & \int_{| y | > ℓ} \int_{R^{n}} | f (x - y) - f (x) | | K_{δ} | d x d y \\ \leq & 2 ‖ f ‖_{L^{1} (R^{n})} \int_{| y | > ℓ} | K_{δ} (y) | \end{aligned}

By the properties of the good kernel, we have

\forall ε > 0, \exists θ > 0

s.t. when

δ < θ, \int_{| y | > ℓ} | K_{δ} (y) | < ε .

Then we obtain

‖ f * K_{δ} - f ‖ < (M + 2 ‖ f ‖) ε .

Hence

lim_{δ \to 0} ‖ f * K_{δ} - f ‖ = 0.

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