If $0\le \phi\le 1$ then prove that $\int_0^1\phi(s)ds<1. $

Question

Assume that $\phi:[0,1]\to[0,1]$ is a continuous map which is not identically equal to 1. Then prove that:

$$\int_0^1 \phi(s)ds<1.$$ My attempt. $0\le \phi\le 1$ I directly get

$$\int_0^1 \phi(s)ds \le 1.$$ But the inequality is large rather than strict inequality. How can I prove it with the strict ineqaulaity?

Answer 1

Take a point $x_0$ where $f(x_0) \neq 1$. Since $f$ is continuous, there is a small enough neighborhood of $x_0$, $[x_0 - \frac{\delta}{2}, x_0 +\frac{\delta}{2}]$ where $f \neq 1$ on that neighborhood. This $f$ has a max, $C <1$, on that interval since it's compact. Thus $$\int_{0}^{1} \phi(s) ds \leq (1-\delta) + C\delta < 1$$

Answer 2

$$1-\int_0^1 \phi(s) ds = \int_0^1 (1-\phi(s))ds$$

Since $1-\phi \geq 0$, $\int_0^1 (1-\phi(s))ds\geq 0$ and if the integral is $0$, then $1-\phi = 0$ (which contradicts your assumption on $\phi$).

The last fact follows from a standard lemma for continuous functions over a closed interval.

Answer 3

Assume the equality hold, then \begin{align*} \int_{0}^{1}(1-\phi(x))=0. \end{align*} As $1-\phi\geq 0$ is also continuous, then $1-\phi(x)=0$ for all $x\in[0,1]$. This can be argued by some heavier machinery that at least it must be true that $1-\phi=0$ a.e., but continuity can remove the a.e. condition.

Answer 4

If $\varphi$ is not identically equal to $1$ then here is a $x_0\in [0,1]$ and $\epsilon>0$ such that $1-\varphi(x_0)>\epsilon$. By the continuity of $\varphi$ there exists an interval $[x_0-\delta,x_0+\delta]$ such that it is worth the inequality $$ 1-\varphi(x)>0,\quad \forall x\in [x_0-\delta,x_0+\delta]. $$ Supposing absurd that we have equality $ \int_{0}^{1}\varphi(x)\mathrm{\, d \,} x=1 . $ Note that $\int_{0}^{1} 1 \mathrm{\,d\,}x=1$ implies $$ \int_{0}^{1}\underbrace{(1-\varphi(x))}_{\geq 0} \rm d x =0 $$ But $$ \int_{0}^{1}\underbrace{(1-\varphi(x))}_{\geq 0} \rm d x \geq \int_{x_0-\delta}^{x_0+\delta}\underbrace{(1-\varphi(x))}_{\geq \epsilon} \rm d x \geq \int_{x_0-\delta}^{x_0+\delta}{ \epsilon}\;\; \rm d x = \epsilon \cdot 2\cdot \delta >0 $$ So we can only conclude that $$ \int_{0}^{1}\varphi(x) \,{\rm d } x<1. $$