Proving a function is identically zero

Mar 2009
419
64
Uptown Manhattan, NY, USA
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.
 
Jan 2010
150
29
Mexico City
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?
 
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Mar 2009
419
64
Uptown Manhattan, NY, USA
I think that makes sense, yeah.
 
Mar 2009
419
64
Uptown Manhattan, NY, USA
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

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Nov 2009
4,563
1,566
Berkeley, California
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 \(\displaystyle g(x)\)
 
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