Problem #7.4, MA598R, Summer 2009, Weigel

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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


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