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Let the aplication $T_\lambda : C^0([0,1], \mathbb{R}) \to C^0[(0,1), \mathbb{R})$ defined by

$$ T_\lambda \phi(x) = \lambda \int_0^1 \frac{x^2 + y^2}{1 + |\phi(y)|} \, dy $$

Show that $T_\lambda$ is contractive for $|\lambda| < 3/4$

My attemp:

I have only been able to show that $T_\lambda$ is contractive for $|\lambda | < 1/2$

$$ d ( T_\lambda(\phi), T_\lambda(\psi)) = || T_\lambda(\phi) - T_\lambda(\psi)||_\infty = \textrm{sup}_{x \in[0,1]} \{| \lambda \int_0^1\frac{x^2 + y^2}{1 + |\phi(y)|}dy - \lambda \int_0^1\frac{x^2 + y^2}{1 + |\psi(y)|}dy|\} $$

$$ = | \lambda \,\, | \textrm{sup}_{x \in [0,1]} \{ |\int_0^1\frac{(x^2 + y^2)(|\psi(y)| - |\phi(y)|)}{(1 + |\phi(y)|)(1 + |\psi(y)|)} dy |\} $$

Now let $f_x(y) =\frac{(x^2 + y^2)(|\psi(y)| - |\phi(y)|)}{(1 + |\phi(y)|)(1 + |\psi(y)|)} $ and let $y_0 \in [0,1]$ such that $f_x(y_0) = \textrm{max}_{y \in [0,1]} {f(y)}$ Then:

$$ | \lambda \,\, | \textrm{sup}_{x \in [0,1]} \{ |\int_0^1\frac{(x^2 + y^2)(|\psi(y)| - |\phi(y)|)}{(1 + |\phi(y)|)(1 + |\psi(y)|)} dy |\} \leq$$ $$ \leq | \lambda \,\, | \textrm{sup}_{x \in [0,1]} \{ |\frac{(x^2 + y_0^2)(|\psi(y_0)| - |\phi(y_0)|)}{(1 + |\phi(y_0)|)(1 + |\psi(y_0)|)} \leq 2|\lambda| |\psi(y_0) - \phi(y_0)| \leq 2 |\lambda| d(\phi, \psi) $$

Then is inmediate that $T_\lambda$ is ocntractive for $ |\lambda| < 1/2 $

Is there any errors in my attemp? How can I improve it to get $|\lambda| < 3/4$?

José D.
  • 1,324

1 Answers1

3

We have

$$\begin{align} \int_0^1 \frac{(x^2+y^2)(\lvert \phi(y) \rvert - \lvert \psi(y)\rvert)}{(1+\lvert\phi(y)\rvert)(1+\lvert\psi(y)\rvert)}\,dy &\leqslant \int_0^1 (x^2+y^2) \lvert \phi(y)-\psi(y)\rvert\,dy\\ &\leqslant \lVert \phi-\psi\rVert_\infty \int_0^1 x^2+y^2\,dy\\ &= \lVert \phi-\psi\rVert_\infty \left(x^2 + \frac13\right)\\ &\leqslant \frac43 \lVert \phi-\psi\rVert_\infty. \end{align}$$

The conclusion follows immediately from the last line.

Daniel Fischer
  • 206,697