# Function of functions maps closed ball to itself

Let $$P: C[0,1]\to C[0,1]; P(f)(x)= 1+kxf(x)\int_0^1 \frac{f(s)ds}{x+s},$$

where $k$ is a constant, with $|k|<\frac{1}{4\ln 2}$. Let $f_1(x)\equiv 1$ on $[0,1]$. I want to prove that the set $\overline{B_r}(f_1):=\{f(x)\in C[0,1]:\|f(x)-1\|_\infty\le r\}$ is mapped to itself under $P$.

Here's how I'm proceeding.

Let $f(x)\in \overline{B_r}(f_1)$ and consider: $$\left|P(f(x))-1\right|=\left|kx f(x)\int_0^1 \frac{f(s)}{x+s}ds \right|\le |k|\|f\|_\infty\left|\int_0^1 \frac{ds}{x+s}\right|$$

$$\le|k|\|f\|_\infty\left\|\ln\left(\frac{x+1}{x}\right)\right\|_\infty<\frac14\|f\|_\infty$$

I don't see how to get $\|f-1\|_\infty$. Would appreciate a hint. Or is there perhaps a better method than inequalities?

In case of $r = 1$ for $f \in \overline{B_1(1)}$ you can write $$|P(f)(x) - 1| = |kxf(x)\int_0^1\frac{f(s)}{x+s}\, ds| \le |k|||f||_{\infty}^2\int_0^1\frac{x}{x+s}\, ds \le |k|||f||_{\infty}^2 \ln2\\ \le \frac{1}{4}||f||_\infty^2 \le \frac{1}{4}(||f-1||_\infty + ||1||_\infty)^2 \le 1.$$ and therefore $P(f) \in \overline{B_1(1)}$. For $r \ne 1$ the statement is false: take $f = 1+r$ and write $$|P(f)(1)-1| = |kf(1)\int_0^1\frac{f(s)}{1+s}\, ds| = |k(1+r)^2\ln2|.$$ If $P(f) \in \overline{B_1(1)}$, the last expression must be no greater than $r$. It is equivalent to $$|k| \le \frac{r}{(1+r)^2\ln2}$$ which is stronger than the given estimate for $r \ne 1$.

• This works only if $r \ge 1$ in @sequence's question. – Theoretical Economist Nov 6 '17 at 1:34
• I forgot about $r$. I'll edit my answer in the morning. But for $r>1$ the statement is false: take $f = 1+r$, all the inequalities become equalities (for $x=1$) and $||P(f)-1|| = (1+r)^2/4 > r$. – Petr Naryshkin Nov 6 '17 at 1:59
• @PetrNaryshkin How do we know that $\|f-1\|_\infty\le 1$? Or is this what we actually assume based on this result? – sequence Nov 6 '17 at 3:54
• @sequence, yes, we assume that $f \in B_1(1)$. – Petr Naryshkin Nov 6 '17 at 10:29