Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $.

share|improve this question

2 Answers 2

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. $

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.