Show that if $w$ is continuous on $[0,1]$ and $\int_0^1wv\, dx=0$ then $w=0$ everywhere in $[0,1]$

Question

I was hoping someone could check my proof of the following. The problem comes from section 1.1 of Finite Element Method by Claes Johnson.

Problem statement:

Let $$V:=\{v: v \text{ is a continuous function on } [0,1],~v' \text{is piecewise continuous and bounded on } [0,1] \text{, and } v(0)=v(1)=0\}.$$

Show that if $w$ is continuous on $[0,1]$ and $$ \int_0^1 wv\,dx=0~~~~\forall~v\in V$$ then $w(x)=0$ for $x\in[0,1]$.

Proof Attempt:

$\textit{Lemma:}$

For $w: [0,1] \mapsto \mathbb{R}$ which is continuous, there exists a collection of open intervals $P$ that contains all the points where $w(x)>0$. Similarly, there exists $M$ which captures the negative portions of $w$. $$P:=\{(a,b)\subset [0,1]~|~f(x)>0~\forall x\in(a,b)\}$$ $$M:=\{(a,b)\subset [0,1]~|~f(x)<0~\forall x\in(a,b)\}$$

$\textit{Lemma Proof:}$

Let $x\in[0,1]$ and suppose $f(x)>0$. Let $\epsilon=|f(x)|$. Then by continuity there exists $\delta>0$ such that if $y\in B_{\delta}(x)$ then $f(y)\in B_\epsilon(f(x))$. Thus for every $x\in[0,1]$, there is an associated open interval $B_\delta(x)$ such that $f(B_\delta(x)) \subset B_\epsilon(f(x)) >0$. Let $P$ be the collection of all such open intervals, and define $M$ in a similar way.

This completes the proof of the lemma.

$\textit{Proof:}$

Let $P,M$ be defined as above for $w$, and let $v\in V$. Then \begin{align} \int_0^1wv\,dx &=\int_Pwv\,dx+\int_Mwv\,dx \\ &=\int_Pwv\,dx-\int_M|wv|\,dx \end{align} By assumption, the same is true for $-v\in V$. \begin{align} \int_0^1wv\,dx &=-\int_Pwv\,dx-\int_M|wv|\,dx \\ \end{align} Subtracting the two equations above gives $$2\int_Pwv\,dx=0$$ which implies that $$\int_Mwv\,dx=0$$

Answer 1

You still have not shown $P=M=\emptyset$. Also (but this is a minor issue since such functions are sufficient for the proof) your proof requires that $v\ge 0$ or $v\le 0$ to work (otherwise $wv$ changes sign in $P$ or $M$).

The standard proof is by contradiction: choose $x_0$ with $w(x_0)\neq 0$, find an interval $(x_0-\delta,x_0+\delta)$ where $|w(x)-w(x_0)|<\frac{|w(x_0|}{2}$ and then construct a function $v$ that is positive only in $(x_0-\delta,x_0+\delta)$ such that $\int wv\, dx \neq 0$.

Answer 2

You can also apply Stone-Weirstrass here.

Consider $0<c<1$, and $V_c$ the restrictions of the elements of $V$ to $[c,1-c]$ $V_c$ contains the constants and separate point, so it is dense. Let $w_c$ be the restriction of $w$ to $[c,1-c]$, there exists $w_n\in V_c$ with $lim_nw_n=w_c$, this implies that $lim_n\int_c^{1-c}w_nw_c=\int_c^{1-c}w_c^2=0$. We deduce that $w_c=0$ and henceforth $w=0$.