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4. Suppose that <math> G: [0,1] \times [0,1] \longrightarrow \mathbb{R} </math> is continuous. For <math> f \in L^2([0,1]) </math>, and <math> x \in [0,1] </math>, let | 4. Suppose that <math> G: [0,1] \times [0,1] \longrightarrow \mathbb{R} </math> is continuous. For <math> f \in L^2([0,1]) </math>, and <math> x \in [0,1] </math>, let | ||
Revision as of 13:42, 29 July 2009
Back to The_Pirate's_Booty
4. Suppose that $ G: [0,1] \times [0,1] \longrightarrow \mathbb{R} $ is continuous. For $ f \in L^2([0,1]) $, and $ x \in [0,1] $, let
$ (Tf)(x) := \int_0^1 G(x,y) f(y) dy $.
a) Show that $ Tf \in C([0,1]) $ and that $ T : L^2([0,1]) \longrightarrow C([0,1]) $ is a bounded operator.
Let $ \epsilon >0 , f \in L^2([0,1]) $. If $ ||f||_1 =0, Tf $ is clearly continuous. Assume $ ||f||_1 >0 $
$ G $ is uniformly continuous.
Thus, $ \exists \delta >0 $ so that $ | x-x'|<\delta $ implies that
$ \left| G(x,y)-G(x', y) \right| < \frac{\epsilon}{||f||_1} $
Then
$ \left| (Tf)(x')-(Tf)(x) \right| = \left| \int_0^1 (G(x',y)-G(x,y)) f(y) dy \right| \leq \frac{\epsilon}{||f||_1} \cdot ||f||_1=\epsilon $
To see that $ T $ is a bounded operator, we use Holder's inequality:
Let $ M >0 $ be so that $ |G| \leq M $
Then by Holder, for fixed $ x $, $ |(Tf)(x)| \leq ||G(x,y)||_2 \cdot ||f||_2\leq (M^2)^{\frac{1}{2}} ||f||_2=M||f||_2 $
So $ ||Tf||_\infty \leq M\cdot ||f||_2 $, and $ T $ is bounded.
b) Show that $ T $ takes bounded subsets of $ L^2([0,1]) $ to precompact sets in $ C([0,1]) $
Let $ \mathcal{F} \subset L^2([0,1]) $ be bounded by $ N>0 $.
Thus $ f \in \mathcal{F} \Rightarrow ||f||_2 \leq N $
Then from a), $ |(Tf)(x)| \leq MN $ where $ |G| \leq M $
Thus $ T(\mathcal{F}) $ is pointwise bounded by MN.
Given $ \epsilon >0 $,
let $ \delta >0 $ be so that $ |x'-x| < \delta \Rightarrow |G(x',y)-G(x,y)| < \frac{\epsilon}{M} $
Then for $ f \in \mathcal{F} $
$ \left| (Tf)(x')-(Tf)(x) \right| = \left| \int_0^1 (G(x',y) - G(x,y)) f(y) dy \right| \leq \frac{\epsilon}{M}\cdot ||f||_2 \leq \epsilon $
So $ T(\mathcal{F}) $ is equicontinuous.
Then for any sequence in $ T(\mathcal{F}) $, there is a subsequence convergent in $ L^\infty $ by Arzela-Ascoli.
Hence $ \overline{T(\mathcal{F})} $ is compact.
-Jacob Boswell
