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**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, 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.