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Let $f(x)$ be a strictly positive continuous function in $[0,1]$ such that $f^2(t) \leq 1 + 2\int_0^t f(s)\, \mathrm{d}s$. Prove that $f(t)\leq t+1$.


Obviously, if $f(x)$ can be differentiated, it is true. But I can't deal with the case that it is not differentiated.

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I improved some formatting, please check if your question is alright. –  dtldarek Apr 21 '12 at 15:05

2 Answers 2

up vote 13 down vote accepted

Let

$$g(t) = \sqrt{1 + 2\int_{0}^{t} f(x) \text{d}x }$$

$g(t)$ is differentiable and $g(t) \ge 1$.

Squaring and differentiating we get

$g(t)g'(t) = f(t)$

Since we are given that $f(t) \le g(t)$ (after taking square roots)

we get that $g'(t) \le 1$.

Thus

$$f(t) \le g(t) = g(0) + \int_{0}^{t} g'(x)\ \text{d}x \le g(0) + \int_{0}^{t} 1\ \text{d}x = 1 + t$$

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Ok, this might be an overkill and there is nice solution by Aryabhata, but I wanted to present another approach using Banach fixed-point theorem. I am not 100% sure I didn't do any mistakes, so if somebody could comment on this I would really appreciate it.

To prove the main question let $\Gamma\phi(t) = \sqrt{1+2\int_0^t\phi(s)\,\mathrm{d}s}$. Then $f_0 = f$, $f_{n+1} = \Gamma f_n$ is non-decreasing sequence of functions for which $f_\infty = \lim_{n\to\infty}f_n$ is a fixed point of $\Gamma$, thus $f_\infty(t) = 1 + t$. To prove this, let's bound:

\begin{align*} (\Gamma\phi_1 - \Gamma\phi_2)(t) &\leq^{(*)} \int_0^{t}(\phi_1-\phi_2)(s)\,\mathrm{d}s \\\ &\leq t\|\phi_1 - \phi_2\|_\infty \end{align*}

so

\begin{align*} \|\Gamma^2\phi_1 - \Gamma^2\phi_2\|_\infty &= (\Gamma^2\phi_1 - \Gamma^2\phi_2)(t_\max) \\\ &\leq \int_0^{t_\max} (\Gamma\phi_1 - \Gamma\phi_2)(s)\,\mathrm{d}s \\\ &\leq \int_0^{t_\max} s\|\phi_1-\phi_2\|_\infty\,\mathrm{d}s \\\ &\leq \frac{1}{2}\|\phi_1 - \phi_2\|_\infty \end{align*}

Therefore $\Gamma^2$ is a contraction mapping on $C[0,1]$ with the standard $\|\cdot\|_\infty$ norm so $\Gamma^2$ has a unique fixed-point which happens to be $f_\infty$ (both even and odd indexes converge to the same fixed point). So the limit exists and we have that $f(t) \leq f_\infty(t) = 1 + t$ what completes the proof.


Small technical lemma: $$\sqrt{1+x} - \sqrt{1+y} \leq \frac{x-y}{2}\text{ for all }x \geq y \geq 0 \tag{$*$}$$ Proof: $\sqrt{1+x}-\frac{x}{2}$ is non-increasing for $x\geq 0$.

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Cool!I think of the fixed point theorem when I solve the problem. But I can just deal with the case of series. –  Joe Apr 22 '12 at 1:36
    
@Aryabhata Would you be so kind as to comment on my solution? –  dtldarek Apr 22 '12 at 10:42
    
Hi, Sorry I was never notified of your comment, and I just came across this. My functional analysis is a bit rusty, but it seems right at the first glance. –  Aryabhata Nov 5 '12 at 17:28

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