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i need help with this exercise.

Suppose $f$ is continuous in $[a,b]$ and $\int_{a}^{b}fg=0$ for all functions continuous $g$ about $[a,b]$ prove $f=0$

I try make this:

Proof: Proof: Let $g=f$ Then $\int_{a}^{b}fg=\int_{a}^{b}f^{2}=0$ and $f=0$

Can someone help me?

rcoder
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  • So you've proven that $\int_a^b f^2 = 0$. Why does that imply $f^2 = 0$? – Arthur Nov 12 '16 at 00:14
  • Since $f^2$ is positive $\int_{a}^b f^2 = 0$ implies $f^2 = 0$ and then, $f=0.$ – L.F. Cavenaghi Nov 12 '16 at 00:14
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    @frusciante14 $f^2$ is non-negative :D – user251257 Nov 12 '16 at 00:17
  • @frusciante14 I am aware of that. What I'm really saying is that one needs to expand on that point a bit. I can think of non-negative functions which aren't $0$ everywhere, but integrate to $0$, for instance$$h(x) = \cases{1 & if $x = (a+b)/2$\0& otherwise}$$Why is continuity of $f$ enough to ensure that $f$ truly is $0$? – Arthur Nov 12 '16 at 00:19
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    By contraposition, http://math.stackexchange.com/questions/82839/prove-the-integral-of-f-is-positive-if-f-%e2%89%a5-0-f-continuous-at-x-0-and-f is the same as your question. – sTertooy Nov 12 '16 at 00:21
  • @Arthur, because $f^2 \neq 0$ for a point implies that it is not zero for a neighborhood. So $f^2$ must be zero everywhere. – L.F. Cavenaghi Nov 12 '16 at 00:36
  • @frusciante14 Exactly. Now, the last question is: Why are you answering this, and not the OP? I wasn't actually wondering about these things, I was trying to get the OP to think a bit about what continuity had to do with anything. – Arthur Nov 12 '16 at 00:39
  • I am sorry, I made the mistake of think that it was clear. My bad. What do you mean by OP? – L.F. Cavenaghi Nov 12 '16 at 00:43
  • See this. –  Nov 12 '16 at 03:09

4 Answers4

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Here is an answer based on the fundamental theorem of calculus. (Note: A answer based on the fact that if a continuous, non negative function satisfies $\int_a^b \phi = 0$, then $\phi = 0$ on $[a,b]$ is much more straightforward. This answer is for some comic relief :-).)

The idea is to approximate a step function by a collection of continuous functions so that we can compute $\int_0^t f(x)dx = 0$, and by differentiating, we get $f(x) = 0$.

Pick $t \in (a,b)$ and let $g_{t,n}(x) = \begin{cases} 1, & x < t \\ (1-n (x-t)), & t \le x < t+ {1 \over n} \\ 0, & \text{otherwise}\end{cases}$.

Since $f$ is continuous, we have $|f(x)| \le B$ for $x \in [a,b]$.

Note that $| \int_a^t f(x)dx - \int_a^b f(x)g_{t,n}(x) dx | \le {1 \over n}B$ and so $\int_a^b f(x)g_{t,n}(x) dx \to \int_a^t f(x)dx$.

Since $\int_a^b (f(x)g_{t,n}(x)) dx = 0$, we see that $F(t)=\int_a^t f(x)dx = 0$, and from the fundamental theorem of calculus we see that $F'(x)=f(x) = 0$ for $x \in (a,b)$ and hence $f(x) = 0$ for $x \in [a,b]$.

copper.hat
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I assume that $f$ and $g$ are real-valued functions. Then the integral

$$ \int_a^b f(x)g(x)\,dx $$

defines an inner product $\langle f,g \rangle$ on the vector space of continuous real-valued functions on the interval $[a,b]$. Since

$$ \int_a^b f^2(x)\,dx = 0 $$

it follows by properties of inner product spaces that $f(x) = 0$.

K. Miller
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Assume $f$ is nonzero. Consider $g=f$. $f^2$ is non-negative and thus the integral is positive, a contradiction.

Jacob Wakem
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To finish your proof (assuming $a< b$): If $x_0\in [a,b]$ with $f(x_0)\ne 0,$ the continuity of $f^2$ implies there exists $c,d$ with $a\leq c<d\leq b,$ and with $x_0\in [c,d]$, such that $y\in (c,d)\implies |f(y)-f(x_0)|< \frac {1}{2}|f(x_0)|.$ This implies $f(y)^2>\frac {1}{4}f(x_0)^2$ for all $y\in (c,d).$ Then $$\int_a^bf(y)^2\;dy\geq \int_c^d f(y)^2 \; dy>\frac {1}{4} (d-c)f(x_0)^2\ne 0.$$

DanielWainfleet
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