Prove that the function $\|\cdot\|_1:C[a,b]\to\mathbb{R}$ defined by $\|f\|_1 = \int_a^b|f|d\lambda$ satisfies $\|f\|_1=0$ if and only if $f=0$.

Question

Problem

Let $[a,b]$ be a closed and bounded interval, and let $C[a,b]$ be the vector space of all continuous real-valued functions on $[a,b]$. Define the function $\|\cdot\|_1:C[a,b]\to\mathbb{R}$ by letting $$ \|f\|_1=\int_a^b|f|d\lambda. $$ Prove that if a function $f\in C[a,b]$ satisfies $\|f\|_1=0$, then $f=0$.

(Note that $\lambda$ is the Lebesgue measure.)

My Attempt

Let $f\in C[a,b]$. If $f=0$, then $\|f\|_1 = \int_1^b0d\lambda = 0$. Now suppose that $$\|f\|_1=\int_a^b|f|d\lambda= \sup\left\{\int hd\lambda:h\in\mathscr{S}_+\ \text{and}\ h\leq|f|\chi_{[a,b]}\right\} = 0.$$ Assume to the contrary that $f(x)\neq0$ for some $x\in[a,b]$. Then $|f(x)|>0$. Since $f$ is continuous, $|f|$ must be continuous and there must exists a closed and bounded interval $[c,d]\subseteq[a,b]$ such that $|f(z)|>0$ for all $z\in[c,d]$. Define a function $h$ by letting $h(z)=0$ for all $z\notin[c,d]$ and $h(z)=\min_{z\in[c,d]}|f(z)|$ (note that the minimum exists because $|f|$ is a continuous function on a compact set $[c,d]$). Then $h\in\mathscr{S}_+$ and $h\leq|f|\chi_{[a,b]}$, but $\int hd\lambda>0$, which is a contradiction. Thus, $f=0$.

Some definitions:

Definition$\quad$ Let $(X,\mathscr{A})$ be a measurable space. $\mathscr{S}_+$ is the set of simple nonnegative real-valued $\mathscr{A}$-measurable functions on $X$.

Definition$\quad$ $\chi_A$ is the characteristic function of the set $A$.

My Question

I am not sure if my attempt is correct. Could someone please help me check? Especially, I claimed that "there must exists a closed and bounded interval $[c,d]\subseteq[a,b]$ such that $|f(z)|>0$ for all $z\in[c,d]$" without proof. (In fact, I am not sure how to prove this rigorously, but I seriously believe this is correct.)

Thank you very much!

Answer 1

Just a hint to get you in the right direction, but you can make the argument a bit more rigorous and fill in the fact about the interval $[c,d]$ by appealing to the $(\epsilon,\delta)$ definition of continuity.

For example, if $f(z) = \alpha > 0$ then since $f$ is a continuous function we can set $\epsilon = \alpha/2 > 0$ and there must be a $\delta > 0$ such that for every $z' \in [z-\delta,z+\delta] \cap [a,b]$ it holds $$|f(z) - f(z')| < \epsilon = \alpha/2$$ In particular this implies for every $z$ in the interval $$f(z') > f(z) - \alpha/2 = \alpha - \alpha/2 = \alpha/2 > 0$$ This gives you both an interval $[z-\delta, z+\delta] \cap [a,b]$ and a strictly positive lower bound $\alpha/2$ of $f$ on that interval.

There are still a few details that I'll leave to you to fill in to handle the absolute value.

Answer 2

It seems that we may argue in this way: Let $f\in C[a,b]$ be such that $\int_{a}^{b}|f|d\lambda=0$, then $f=0$ a.e. (For, let $A=\{x\in[a,b]\mid|f(x)|>0\}$, and $A_{n}=\{x\in[a,b]\mid|f(x)|>\frac{1}{n}\}$. Note that $A=\cup_{n}A_{n}.$ If $\lambda(A_{n})>0$ for some $n$, then $\int_{a}^{b}|f|d\lambda\geq\frac{1}{n}\lambda(A_{n})>0$, which is a contradiction. Therefore, $\lambda(A_{n})=0$ for all $n$ and hence $\lambda(A)=0$.)

Let $A=\{x\in[a,b]\mid|f(x)|>0\}$. Since $\lambda(A)=0,$ $A$ does not contain any interior point (with respect to the relative topology of $[a,b]$). In another word, for each $x\in A$, there exists a sequence $(x_{n})\in[a,b]\setminus A$ such that $x_{n}\rightarrow x$. We argue that $A=\emptyset$. Prove by contradiction. Suppose that there exists $x\in A$. Choose a sequence $(x_{n})\in[a,b]\setminus A$ such that $x_{n}\rightarrow x$. Note that $f(x_{n})=0$, so $f(x)=\lim_{n\rightarrow\infty}f(x_{n})=0$, a contradiction.