I have this problem :
$f$ is defined and positive in $[0,1]$ for $1>a>0 \implies \inf f((a,1])>0$.
if $f$ continuous on the right at $x=0$ proof that $\inf f([0,1])>0$
My proof
Since $f$ is continous on the right at $x=0$ we know that : $$\lim_{x \to 0+} f(x)=f(0)$$
I thought to assume that $\inf f([0,1])\leq 0$, and using the limit inorder to get a contradiction but haven't managed to get there, Any ideas how to "move on" from this point?
Any help will be appreciated.
As it stands the result is not correct. Take, for example, $f(x)=x,\ x\in[0,1]$. Then $f$ is positive, continuous everywhere, $\inf f((a,1])=a>0$ for every $a\in(0,1]$ and $\inf f([0,1])=0$.
One way to correct it is to assume that $f(0)>0$. I will now prove the statement with this assumption. Since $f\ge 0$ we must have that $\inf f([0,1])\ge 0$. Suppose that $\inf f([0,1])=0$. Then, for every $a\in(0,1]$, it holds that $\inf f([0,a])=0$ since $0=\inf f([0,1])=\min\{\inf f([0,a]),\inf f((a,1])\}$ and $\inf f((a,1])>0$. Let $\epsilon>0$ be arbitrary, but fixed. Then, by the definition of the infimum, for every $n\ge 1$ one can find $x_n\in[0,\frac{1}{n}]$ such that $f(x_n)<\epsilon$ ($\inf f([0,\frac{1}{n}])=0$). We pass to the limit ($n\to\infty$) and use the continuity of $f$ in $0$ to deduce that $f(0)\le\epsilon$. $\epsilon$ was arbitrary, so this forces $f(0)=0$, which is a contradiction. It follows that $\inf f([0,1])>0$, as required.