# Proving a function is identically zero

#### Pinkk

If $$\displaystyle f$$ is a continuous function on $$\displaystyle [a,b]$$ such that $$\displaystyle \int_{a}^{b}fg = 0$$ for all continuous functions $$\displaystyle g$$. Then $$\displaystyle f=0$$ on $$\displaystyle [a,b]$$.

So yeah, I know I have to invoke that $$\displaystyle U(fg)=L(fg)=0$$ but after that I'm stumped. Thanks.

#### mabruka

Here is an idea.

We have $$\displaystyle \int_a^b f^2=0$$ where $$\displaystyle f^2\geq 0$$, so $$\displaystyle f^2=0$$ on $$\displaystyle [a,b]$$.

Hence f=0 on [a,b]

What do you think?

• Pinkk

#### Pinkk

I think that makes sense, yeah.

#### Pinkk

And I just want to make sure that proving there cannot be some point $$\displaystyle x_{0}\in [a,b]$$ where $$\displaystyle f^{2} > 0$$. So suppose there is (at least) one point $$\displaystyle x_{0}$$ such that $$\displaystyle f^{2}(x_{0}) > 0$$. Then since $$\displaystyle f^{2}$$ is continuous on $$\displaystyle [a,b]$$, $$\displaystyle f^{2} > 0$$ on some interval $$\displaystyle (x_{0} - \delta, x_{0} + \delta), \delta > 0$$, so $$\displaystyle f^{2} > 0$$ on the closed interval $$\displaystyle [x_{0} -\frac{\delta}{2}, x_{0} + \frac{\delta}{2}]$$. Clearly $$\displaystyle \inf \{f^{2}(x): x\in [x_{0} - \frac{\delta}{2}, x_{0} + \frac{\delta}{2}]\} > 0$$ since the function must achieve a minimum at a point on the interval by its continuity on a bounded interval, so for any partition $$\displaystyle P$$, $$\displaystyle L(f^{2}, P) > 0$$, so $$\displaystyle \int_{x_{0} - \frac{\delta}{2}}^{x_{0} + \frac{\delta}{2}} f^{2} > 0$$ and since the function is zero everywhere else, then $$\displaystyle \int_{a}^{b}f^{2} > 0$$, which is a contradiction.

I think this seems a little overboard but I just want to make sure if this is correct or if there is an easier proof that shows $$\displaystyle f^{2}$$ cannot be positive at any point. Thanks.

#### Drexel28

MHF Hall of Honor
And I just want to make sure that proving there cannot be some point $$\displaystyle x_{0}\in [a,b]$$ where $$\displaystyle f^{2} > 0$$. So suppose there is (at least) one point $$\displaystyle x_{0}$$ such that $$\displaystyle f^{2}(x_{0}) > 0$$. Then since $$\displaystyle f^{2}$$ is continuous on $$\displaystyle [a,b]$$, $$\displaystyle f^{2} > 0$$ on some interval $$\displaystyle (x_{0} - \delta, x_{0} + \delta), \delta > 0$$, so $$\displaystyle f^{2} > 0$$ on the closed interval $$\displaystyle [x_{0} -\frac{\delta}{2}, x_{0} + \frac{\delta}{2}]$$. Clearly $$\displaystyle \inf \{f^{2}(x): x\in [x_{0} - \frac{\delta}{2}, x_{0} + \frac{\delta}{2}]\} > 0$$ since the function must achieve a minimum at a point on the interval, so for any partition $$\displaystyle P$$, $$\displaystyle L(f^{2}, P) > 0$$ so then $$\displaystyle \int_{x_{0} - \frac{\delta}{2}}^{x_{0} + \frac{\delta}{2}} f^{2} > 0$$ and since the function is zero everywhere else, then $$\displaystyle \int_{a}^{b}f^{2} > 0$$, which is a contradiction.

I think this seems a little overboard but I just want to make sure if this is correct or if there is an easier proof that shows $$\displaystyle f^{2}$$ cannot be positive at any point. Thanks.
That's similar to what I do. You can weaken this to something like if the above is true for any polynomial $$\displaystyle g(x)$$

• Pinkk