I tried as follows:-
$|Tf(x)-Tg(x)|=|\int _0^xf^2-g^2dt| \leq ||f-g||_{\infty} \int _0^x|f+g|dt$
But there are functions of the form $K + x \in C[-r,r]$, $K$ is a constant, such that $\int _0^x|f+g|dt > 1$.
Is this approach somewhat correct?
$Tf(x)= \int_0^x(f(t))^2dt$ is not a contraction on $C[-r,r]$ for any r
0
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analysis
ordinary-differential-equations
3 Answers
2
Unless I'm mistaken, this reasoning isn't quite right. Even if you found such a function, the inequalities $$\lvert Tf(x) - Tg(x) \rvert \le \| f - g\|_\infty \int^x_0 \lvert f + g \rvert dt $$ and $$\int^x_0 \lvert f + g \rvert dt >1$$ do not imply that $$\lvert Tf(x) - Tg(x) \rvert > \| f - g\|_\infty. $$ However, the reasoning is on the right track. It suffices to find $f \in C[-r,r]$ such that $$\|Tf\|_\infty = C \|f\|_\infty$$ for some $C > 1$. This can be easily done with a constant function $f$.
1
Your approach is correct. To finish the question, it suffices to provide an example of two such functions. Keep in mind that to answer the question properly, you should construct functions $f$ and $g$ that fail to satisfy the contraction inequality.
We can do the same with a simpler construction, though: take $f$ and $g$ to be the constant functions $f(x) = 2/r$ and $g(x) = 0$.
1
Consider the function $f_r$ defined by $f_r(x)=K_rx$. You can check that $$ \|Tf_r\|_\infty=\max_{x\in [0,r]}|Tf_r(x)|=K_r^2\max_{x\in[0,r]}\int_0^xt^2\,dt=K_r^2\max_{x\in[0,r]}\dfrac13x^3=\dfrac13K_r^2r^3=\dfrac13K_rr^2\|f_r\|_\infty, $$ and if you choose $K_r>3r^{-2}$, then $$ \|Tf_r\|_\infty>\|f_r\|_\infty. $$