1

Suppose f is continuous on $I=\{x|a \leq x \leq b\}$ and $\int_a^b \! f(x)g(x) \, \mathrm{d}x = 0$ for all functions $g$ which are continuous on $I$. Prove that $f(x) = 0$.

In this case, the integral is the Darboux integral with the definition $\int_a^b \! f(x) \, \mathrm{d}x = \inf\{S^{+}(f,\bigtriangleup)\} = \sup\{S_{\_}(f,\bigtriangleup)\}$, where $S^{+}(f,\bigtriangleup) = \sum\limits_{i=1}^n M_{i}(x_{i}-x_{i-1})$ and $S_{\_}(f,\bigtriangleup) = \sum\limits_{i=1}^n m_{i}(x_{i}-x_{i-1})$, and $M_{i}$ and $m_{i}$ are the l.u.b and g.l.b of f on $I_{i}$ respectively.

What I want to say is that you cannot force $S^{+}(f,\bigtriangleup)$ to $0$ by forcing $(x_{i}-x_{i-1})$ to $0$, since the decrease in subinterval length is offset by the increase in the number of subintervals. Instead, you must force $f(x)g(x)$ to $0$, and since $g(x)$ can be any (non-zero) function continuous on $I$, $f(x)$ must be $0$.

Is that right? If so, how can I formally state the first sentence of that?

Martin Sleziak
  • 56,060
ztforster
  • 259
  • The last sentence is the meat of the argument, and you just asserted it without proof. You need to show that the family of continuous functions is large enough that it can detect when a function vanishes. For example, there are other families of functions that cannot detect this (e.g. if you could only integrate w.r.t. functions supported on a subinterval, then $f$ might not vanish outside the subinterval). – pre-kidney Feb 19 '15 at 06:25
  • 1
    See also: http://math.stackexchange.com/questions/1128621/how-can-i-show-that-f-must-be-zero-if-int-fg-is-always-zero and http://math.stackexchange.com/questions/331286/why-does-int-ab-fg-dx-0-imply-that-f-0 – Martin Sleziak Feb 19 '15 at 10:51

3 Answers3

2

We show that $f^2=0$ and hence that $f=0$. Assume not. Then there is some subinterval $[\alpha,\beta]\subset I$,$\alpha<\beta$ such that $f(x)^2\geq c$ for some $c>0$ and all $x\in J$. If we take $g=f$ we then have \begin{align*} 0=\int_0^1f(x)^2dx=\int_{I\backslash J}f(x)^2dx+\int_\alpha^\beta f(x)^2dx\geq(\beta-\alpha)c>0, \end{align*} a contradiction.

sranthrop
  • 8,696
1

Suppose for contradiction that $f$ was not identically 0. Then without loss of generality, there is some $x$ for which $f(x)>0$ (otherwise replace $f$ with $-f$). Since $f$ is continuous, this implies that there is an open interval $(x_-,x_+)$ on which $f>0$. Now take $g$ to be a triangular bump function peaked at the midpoint of the interval, i.e. $$ g(x)=\begin{cases} 0,& x\not\in (x_-,x_+)\\ \frac{x_-+x_+}{2}-x,& x_-\leq x\leq \frac{x_-+x_+}{2}\\ x-\frac{x_-+x_+}{2},& \frac{x_-+x_+}{2}\leq x\leq x_+ \\ \end{cases} $$ Choosing a partition of $I$ involving $x_-$ and $x_+$, we see that the lower bound $S_-(fg)$ is strictly positive. Thus $\int_a^b f(x)g(x)\ dx>0$, which is a contradiction.

pre-kidney
  • 30,884
1

Since $g(x)$ can be any continuous function, try $g = f$. Then the result is that $f^2(x)$ is a non-negative function whose integral over $[a,b]$ is zero...therefore $f(x)$ is identically zero (or does this fact require proof?)

grand_chat
  • 40,909