# on proving that $\|\cdot\|_2$ is a norm on $C[0,1]$

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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You are right. The fact that $|f(x)|\geq 0$ implies that $|f(x)|=0$. For intance, if you suppose that there exist some set with positive measure such that $|f(x)|>0$, then you can see that the integral must be positive. – Tomás Oct 5 '12 at 20:41
@Tomás I forgot to mention that the integral I defined is a Riemann integral. Is there a way another way (without using measure theory) to prove what I want? – kevin Oct 5 '12 at 20:46
Certainly you need the continuity assumption. Otherwise a function which is nonzero at a single point would have zero norm without being zero. – Chris Eagle Oct 5 '12 at 20:47
@Chris Eagle But these functions are not continuous right? And therefore not in C[0,1]. – kevin Oct 5 '12 at 20:48
Yes, that was my point. – Chris Eagle Oct 5 '12 at 20:50

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 $f\equiv 0.$

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