# Thread: how can u prove this

1. ## how can u prove this

$\int_{0}^{1}{|f|}=0\Rightarrow f=0$
given f is continuous on [0,1]

2. Originally Posted by szpengchao
$\int_{0}^{1}{|f|}=0\Rightarrow f=0$
given f is continuous on [0,1]
this is not true in general. more conditions are needed for this to be true. i believe flyingsquirrel answered a similar question for you earlier. baring that in mind, i don't think it would be hard for you to come up with a counter-example to this

3. ## .

that question is about integrable one.

this is a continuous function f!!

4. Originally Posted by szpengchao
that question is about integrable one.

this is a continuous function f!!
you said continuous on [0,1], that is not enough, as the function $f: \mathbb{R} \mapsto \mathbb{R}$ defined by

$f(x) = \begin{cases} 0 &\mathrm{if}\,\,0\leq x \le1 \\ 1& \mathrm{otherwise}\end{cases}$

is continuous on [0,1] but is not identically zero, yet $\int_0^1 |f(x)|~dx = 0$

5. ## ..

should I make it more clear ?

$\int_{0}^{1}{|f|}=0\Rightarrow f(x)=0 \forall x\in [0,1]$

6. Originally Posted by szpengchao
should I make it more clear ?

$\int_{0}^{1}{|f|}=0\Rightarrow f(x)=0 \forall x\in [0,1]$
ok, now that makes sense.

try a proof by contradiction. assume the integral is zero but f is not. this would mean f is not continuous, a contradiction

7. ## 。。

$

\int_{0}^{1}{|f|}=0\Rightarrow \exists q\in[0,1], f(q)=0.$

assume there exists p in [0,1] , $f(p)\neq0,$ Then we can find a $\epsilon=|f(p)|>0 , |p-q|<\delta$ there discontinuous.
8. If $f$ is continuous on $\left[ {0,1} \right]$ so is $\left| f \right|$.
Suppose that $\left( {\exists c \in \left[ {0,1} \right]} \right)\left[ {\left| {f(c)} \right| > 0} \right]$ then $\left( {\exists \left[ {u,v} \right] \subseteq \left[ {0,1} \right]} \right)\left[ {\left( {\forall z \in \left[ {u,v} \right]} \right)\left[ {\left| {f(z)} \right| > 0} \right]}\right]$.
What can we say about $\int\limits_u^v {\left| {f(z)} \right|dz}$.