# Separates Points Property of Norm defined by Integral

Let $\mathcal{C}[0,1]$ be the space of continuous functions on the interval $[0,1]$ and define $$\| f \|_1 = \int_{0}^{1} |f(x)| \ dx \quad \text{and} \quad \| f \|_0 = \int_{0}^{1} x |f(x)| \ dx.$$ Assume that $\| \cdot \|_1$ is a norm on $\mathcal{C}[0,1]$. I was having a difficult time showing the property that $\| f \|_0 = 0 \implies f \equiv 0$. I can't help but feel there must be a slick way to do it since I didn't use the first norm $\| \cdot \|_1$ . Here is my proof (?) of the "separates points" property of a norm for $\| \cdot \|_0$ :

Suppose $f \not\equiv 0$. Using integration by parts, we set $u = x$ and $dv = |f(x)| \ dx$. Then, $\int_{0}^{1} x |f(x)| \ dx =\left[ x \cdot \int_{0}^{x} |f(t)| \ dt \right]_{0}^{1} - \int_{0}^{1} \int_{0}^{x} |f(t)| \ dt \ dx = \int_{0}^{1} |f(t)| \ dt - \int_{0}^{1} \int_{0}^{x} |f(t)| \ dt \ dx.$

By the fundamental theorem of calculus, $$v(x) = \int_{0}^{x} |f(t)| \ dt, \quad \text{with} \ x \in [0,1] .$$

Observe that for all $x \in [0,1]$, $$v(x) = \int_{0}^{x} |f(t)| \ dt \leq \int_{0}^{1} | f(t) | \ dt = M,$$ which follows from the fact that our function is positive. Now, since we're assuming $f(t)$ is not $0$, $M > 0$. And as $v(0) = 0$, the continuous function $v(x)$ is strictly bounded above" by $M$. Hence,

$$\int_{0}^{1} v(x) \ dx < \int_{0}^{1} M \ dx \iff \int_{0}^{1} \int_{0}^{x} |f(t)| \ dt \ dx < \int_{0}^{1} \int_{0}^{1} | f(t) | \ dt \ dx,$$ where the inequality here is strict. It follows that

$\|f \|_0 = \int_{0}^{1} |f(t)| \ dt - \int_{0}^{1} \int_{0}^{x} |f(t)| \ dt \ dx > \int_{0}^{1} |f(t)| \ dt - \int_{0}^{1} \int_{0}^{1} |f(t)| \ dt \ dx = 0.$

We conclude $f \not\equiv 0 \implies \|f \|_0 > 0$.

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Suppose that there is a $x_0$ such that $f(x_0)\ne 0$. By continuity there exists $\epsilon>0$ so that $x|f(x)|>0$ for all $x\in (-\epsilon+ x_0,x_0+\epsilon)$. Using the compactness of the closed interval $K=[-\frac \epsilon 2+x_0,x_0+\frac \epsilon 2]$ there is a minimum $m>0$ for $x|f(x)|$ in $K$. It follows that $m\cdot \epsilon\leq \int_K x|f(x)|dx\leq \int_0^1 x|f(x)|dx.$

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You are correct, this is a slick way using the first norm:

Since $\| \cdot \|_1$ is a norm, for a continuous function $g$ we have $\int^1_0 |g(x)| dx =0$ if and only if $g = 0$ identically.

If $f$ is continuous then so is $xf(x)$ so then assume $0 = \int^1_0 x |f(x) | dx = \int^1_0 |x f(x) | dx.$ Then by the above result, $xf(x)=0$ identically. Thus, we already have $f(x)=0$ on $(0,1]$. Since $f$ is continuous, we must also have $f(0)=0$ so $f=0$ identically, as required.

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