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Let $\mathbb{F}$ be either $\mathbb{R}$ or $\mathbb{C}$ and consider the vector space $C[0,1]$, the collection of continuous functions $f\colon[0,1]\to\mathbb{F}$. I want to show that $\|\cdot\|_2$ is a norm on $C[0,1]$. I have proven all the properties of a norm, but I still need to show that in that $\|f\|=0\implies f=0$.

EDIT: The norm $\|\cdot\|_2$ is assumed to be given by a Riemann integral.

If $\|f\|_2^2=0$, then $\int_0^1|f(x)|^2~\mathrm{d}x=0.$ I was wondering if the fact that $|f(x)|$ is nonnegative is sufficient to conclude that $|f(x)|=0$ (which ensures that $f(x)=0$ for all $x\in[0,1]$). Or do you need the fact that each $f$ is continuous?

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    @kevin: not only the non negativity but non negativity in the class of continuous function is needed to make the conclusion.2012-10-05

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We basically want to prove that if $f \in C[0,1]$ with $f \geq 0$ (the case for $f<0$ is similar) and $\int_0^1 |f(x)| dx=0$, then $f\equiv0$.

We aim at a contradiction: Suppose $f(x_0)\neq0$ at some $x_0 \in [0,1]$. Since $f$ is continuous, given $\epsilon= \frac{|f(x_0)|}2$, we can find $\delta>0$ such that whenever $ |x-x_0|<\delta$, $|f(x)-f(x)_0|< \epsilon=\frac{|f(x_0)|}2 $. This implies $f(x)>\frac{f(x_0)}2$ so $ \int_{x_0-\delta}^{x_0+\delta} f(x) dx>\frac{f(x_0)}{2}2\delta>0$. But this means $\int_0^1 |f(x)| dx>0$ which is a contradiction, so $f\equiv0$.

(I skipped some few technical details, but this is the essence of the proof.)

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To finish, we know that $\lvert f(x)\rvert^2=0$ almost everywhere on $[0,1]$, and since $f$ continuous implies $\lvert f\rvert^2$ continuous, we see that $\lvert f\rvert^2\equiv 0$ on $[0,1].$ Then we can use definiteness of $\lvert\cdot\rvert$ to see that $f\equiv 0.$