show the following $f:\mathbb{R}\rightarrow [0,1]$ be a continuous differentiable function. My approach: Assume via contradiction. Then $f$ is injective on $[0,\infty)$ and since $f(0)=1$ so $f$ is strictly decreasing on $[0,\infty)$. Thus $f$ has a limit as $x$ tends to infinity.
From this, how do I proceed? Please give some direction.
 A: Prove by contradiction. Suppose the contrary that $f'(x)\neq0$ for
all $x\in(0,\infty)$. Since $f'$ is continuous, we either have $f'(x)>0$
for all $x\in(0,\infty)$ or $f'(x)<0$ for all $x\in(0,\infty)$.
If $f'(x)>0$ for all $x\in(0,\infty)$, then $f$ is strictly increasing
on $[0,\infty)$. Hence, for any $x>0$, $f(x)>f(0)=1$, which is
a contradiction because $f(x)\in[0,1]$. Therefore $f'(x)<0$ for
all $x\in(0,\infty)$.
Let $M=\sup\{f'(x)\mid x\in[1,\infty)\}\leq0$. We assert that $M=0$.
Suppose the contrary that $M<0$. By mean-value theorem, for each $x>1$, there
exists $\xi_{x}\in(1,x)$ such that $f(x)-f(1)=f'(\xi_{x})(x-1)$.
That is, $f(x)=f(1)+f'(\xi_{x})(x-1)\leq1+M(x-1)$. Choose a sufficiently
large $x$, then $f(x)\leq1+M(x-1)<0$, which is a contradiction.
Choose a sequence $(x_{n})$ in $[1,\infty)$ such that $f'(x_{n})\rightarrow M=0$.
Since $f$ is strictly decreasing on $[0,\infty)$, we have that $0\leq f(x_{n})\leq f(1)<f(0)=1$.
By assumption,
\begin{eqnarray*}
1 & \leq & f^{2}(x_{n})+[f'(x_{n})]^{2}\\
 & \leq & f^{2}(1)+[f'(x_{n})]^{2}.
\end{eqnarray*}
Let $n\rightarrow\infty$, then we have $1\leq f^{2}(1)<1$, which
is a contradiction.
A: As you pointed out, $f'\neq 0$ implies $f$ is strictly monotonic(because $f'$ cannot change sign by continuity), and since $f\le 1$ is given, $f$ is a strictly decreasing function.
Then, $f(1)<1$.
$\forall x>1, 0\le f(x)<f(1)\Rightarrow |f'(x)|>\epsilon\Rightarrow f'(x)<-\epsilon$, where $\epsilon=\sqrt{1-f(1)^2}$.
($\because |f'(x)|^2+|f(1)|^2> |f'(x)|^2+|f(x)|^2\ge 1$ for $x>1$)
This implies $f(x)<-\epsilon(x-1)+f(1)$ for $x>1$.
Picking big $x$ you easily obtain a contradiction.
A: Indeed, we have that $f' < 0$ on $\langle 0,+\infty\rangle$ so
\begin{align}
f(x)^2+ f'(x)^2 \ge 1 &\implies f'(x)^2 \ge 1-f(x)^2 \\
&\implies f'(x) \le -\sqrt{1-f(x)^2} \\
&\implies \frac{f'(x)}{\sqrt{1-f(x)^2}} \le -1\\
&\implies (\arcsin f)'(x) \le -1\\
\end{align}
so for any $x_0 > 0$ we have
$$\arcsin f(x_0) = \int_0^{x_0} (\arcsin f)'(x)\,dx \le \int_0^{x_0} -1\,dx = -x_0 < 0$$
which is a contradiction since $0 \le f \le 1$ and so $\arcsin f(x_0) \in \left[0,\frac\pi2\right]$.
