I'm trying to prove that if $f$ and $g$ (real numbers only) are each integrable on $[a,b]$, and $\int_a ^b g^2 = 0$, then $\int_a ^b fg = 0$.
Edit: Here, I'm using a definition of integrability taken from Chapter 13 of Spivak's Calculus:
A function $f$ which is bounded on $[a,b]$ is integrable on $[a,b]$ if
$$\sup\{L(f,P): P\text{ a partition of } [a,b]\} = \inf\{U(f,P) : P\text{ a partition of } [a,b]\}.$$In this case, this common number is called the integral of $f$ on [a,b] and is denoted by
$$ \int_a ^b f.$$
Edit: I realize the Cauchy-Schwarz inequality for integrals immediately handles this problem. I was interested in trying to show a different route. In Calculus by Spivak, 3rd edition Chapter 13 problem 39(b) the second proof of Cauchy-Schwarz (not the quadratic one) uses $2xy \leq x^2 + y^2$ with
$$x = \frac{f(x)}{\sqrt{\int_a ^b f^2}}, y = \frac{g(x)}{\sqrt{\int_a ^b g^2}},$$
but fails to deal with the case where either denominator might be $0$. Here's a case where showing $\int_a ^b g^2 = 0 \implies \int_a ^b fg = 0$ is handy.
Note: I thought of a shorter proof, which I'll add to the end of this post. While this new proof is more direct, the original proof contains a few useful offshoots, namely, the lemma, and the proof that $\int g^2 = 0 \implies \int |g| = 0$.
Original Proof
I will make use of the following theorems:
Theorem 1: If $f$ and $g$ are each integrable on $[a,b]$ then their product $fg$ is integrable on $[a,b]$.
Theorem 2: If $f$ is integrable on $[a,b]$ and $f > 0$ on $[a,b]$, then $\int_a ^b f > 0$. (Note this theorem actually not needed. See bottom edit.)
Theorem 3: If $f$ is integrable on $[a,b]$ and $f \geq 0$ on $[a,b]$, then $\int_a ^b f \geq 0$.
Theorem 4: If $f$ is integrable on $[a,b]$ then $|f|$ is integrable on $[a,b]$.
Theorem 5: If $f$ is integrable on $[a,b]$, then
$$\left| \int_a ^b f \right| \leq \int_a ^b |f|.$$
We begin by proving the following Lemma:
Lemma: If $f$ and $g$ are each integrable on $[a,b]$, with $\int_a ^b g = 0$, and if $g\geq 0$ on $[a,b]$, then $\int_a ^b fg = 0$.
Lemma Proof:
From Theorem 1, we see that $\int_a ^b fg$ exists. Proof that the integral equals zero will be by given by contradiction. We begin by assuming $\int_a ^b fg > 0.$
Case 1: $\int_a ^b fg > 0$
If $\int_a ^b fg > 0$ then there must be some partition $P$ such that the lower sum $L(fg,P) > 0.$
This implies the existence of some subinterval $[t_{i-1}, t_i]$ on which the infimum of $fg$ is greater than $0$. (Otherwise, every lower sum would be $\leq 0$, and the integral could not be positive).
So, on this subinterval we have
\begin{align}
m_i &= \inf\{f(x)g(x)\text{: } t_{i-1} \leq x \leq t_i \}, \\
m_i &> 0,
\end{align}
and
$$f(x)g(x) \geq m_i > 0.$$
This tells that $f$ and $g$ are each nonzero on the subinterval. Furthermore, since $g$ is nonnegative by hypothesis, we see that both $f$ and $g$ must be positive.
But this implies (theorem 2) that the integral of $g$ over this subinterval:
$$\int_{t_{i-1}} ^{t_i} g > 0,$$
and this implies (because $\int g \geq 0$ everywhere else on $[a,b]$) $\int_a ^b g > 0$, which contradicts the hypothesis.
Case 2: $\int_a ^b fg < 0$
Here, we can apply the Case 1 arguments to the function $(-f)g$ to show
$$\int_a ^b fg < 0 \implies \int_a ^b (-f)g > 0 \implies \int_a ^b g > 0,$$
which again, is a contradiction.
Lemma conclusion:
Since we cannot have $\int_a ^b fg > 0$ or $< 0$, we must have $\int_a ^b fg = 0$.
$$\blacksquare$$
With the lemma in hand, we begin our proof. Many of the steps will be similar to those used in the lemma proof.
To restate the desired result, we wish to show:
If $f$ and $g$ are each integrable on $[a,b]$, and $\int_a ^b g^2 = 0$, then $\int_a ^b fg = 0$.
Since $g$ is integrable on $[a,b]$, theorem 4 tells us that $|g|$ is also integrable on $[a,b]$. Theorem 3 tells us that $\int_a ^b |g| \geq 0.$
Now, suppose $\int_a ^b |g| > 0$.
Similar to the lemma, this would mean there exists some partition $P$ such that the lower sum $L(|g|,P) > 0$. This would imply (like in the lemma) the existence of some subinterval $[t_{i-1}, t_i]$ on which the infimum of $|g|$ is greater than $0$.
So, on this subinterval we have
\begin{align}
m_i &= \inf\{|g(x)|\text{: } t_{i-1} \leq x \leq t_i \}, \\
m_i &> 0,
\end{align}
and
\begin{align}
|g(x)| &\geq m_i > 0, \\
g(x)^2 = |g(x)|^2 &\geq m_i ^2 > 0.\\
\end{align}
This, combined with Theorems 2 and 3 would imply that $\int_{t_{i-1}} ^{t_i} g^2 > 0$, and hence $\int_a ^b g^2 > 0$, contradicting the hypothesis.
Therefore, we must have $\int_a ^b |g| = 0$.
We can now apply the lemma to the functions |f| and |g|, where |g| plays the role of $g$ in the lemma.
Since $|f|$ and $|g|$ are each integrable on $[a,b]$, with $\int_a ^b |g| = 0$, and $|g|\geq 0$ on $[a,b]$, we have
$$\int_a ^b |f||g| = 0.$$
Finally, applying theorem 5, we see
$$\left|\int_a ^b fg \right| \leq \int_a ^b |f||g| = 0,$$
$$0 \leq \left|\int_a ^b fg \right| \leq 0,$$
$$\int_a ^b fg = 0.$$
$$\blacksquare$$
Look mostly ok? Any glaring errors?
Modified Original Proof without Theorem 2 I think we don't actually need Theorem 2.
Below are the relevant parts of the proof, revised to work without Theorem 2.
Lemma Case 1: $\int_a ^b fg > 0$
If $\int_a ^b fg > 0$ then there must be some partition $P$ such that the lower sum $L(fg,P) > 0.$
This implies the existence of some subinterval $[t_{i-1}, t_i]$ on which the infimum of $fg$ is greater than $0$. (Otherwise, every lower sum would be $\leq 0$, and the integral could not be positive).
So, on this subinterval we have
\begin{align}
m_i &= \inf\{f(x)g(x)\text{: } t_{i-1} \leq x \leq t_i \}, \\
m_i &> 0,
\end{align}
and
$$f(x)g(x) \geq m_i > 0.$$
This tells that $f$ and $g$ are each nonzero on the subinterval. Furthermore, since $g$ is nonnegative by hypothesis, we see that both $f$ and $g$ must be positive.
Since $f$ is bound on the subinterval there is some $M_{f,i}$ with
\begin{align}
M_{f,i} &= \sup\{f(x)\text{: } t_{i-1} \leq x \leq t_i \}, \\
M_{f,i} &> 0,
\end{align}
so
\begin{align}
M_{f,i}\cdot g(x) &\geq f(x)g(x) \geq m_i > 0, \\
g(x) &\geq \frac{m_i}{M_{f,i}} > 0.
\end{align}
Thus, $\frac{m_i}{M_{f,i}}$ is a lower bound for $g(x)$ on the subinterval. Since this lower bound is $> 0$, we must have $\inf\{g\} \cdot (t_i – t_{i-1}) > 0 \implies L(g,P) > 0 \implies \int_a ^b g > 0$, a contradiction.
We can apply a similar argument to dispense with the other later use of Theorem 2 in the main theorem proof. There, we show that $g(x)^2 \geq m_i ^2$ on the subinterval implies that $\inf\{g^2\} \geq m_i^2 > 0$, which in turn implies $\int_a^b g^2 > 0$.
More Direct Approach
Suppose $f$ and $g$ are each integrable on $[a,b]$.
Prove that if $\int_a ^b g^2 = 0$, then $\int_a ^b fg = 0$
Proof
From Theorem 1, we see that $\int_a ^b fg$ exists. Proof that the integral equals zero will be by given by contradiction. We begin by assuming $\int_a ^b fg > 0.$
Case 1: $\int_a ^b fg > 0$
If $\int_a ^b fg > 0$ then there must be some partition $P$ such that the lower sum $L(fg,P) > 0.$
This implies the existence of some subinterval $[t_{i-1}, t_i]$ on which the infimum of $fg$ is greater than $0$. (Otherwise, every lower sum would be $\leq 0$, and the integral could not be positive).
So, on this subinterval we have
\begin{align}
m_i &= \inf\{f(x)g(x)\text{: } t_{i-1} \leq x \leq t_i \}, \\
m_i &> 0,
\end{align}
and
$$f(x)g(x) \geq m_i > 0.$$
This tells that $f$ and $g$ are each nonzero on the subinterval.
But if this were the case, then $g^2$ would be $> 0$ on this same subinterval, which would mean $\int_a ^ b g^2 > 0$, a contradiction.
Case 2: $\int_a ^b fg < 0$
Here, we can apply the Case 1 arguments to $(-f)g$ to show
$$\int_a ^b fg < 0 \implies \int_a ^b (-f)g > 0 \implies (-f)g \neq 0 \text{ on } [t_{i-1}, t_i] \implies \int_a ^b g^2 > 0,$$
which again, is a contradiction. (Alternatively, we can make arguments identical to those in Case 1, but with upper sums/supremums swapped in for lower sums/infimums.)
Proof conclusion:
Since we cannot have $\int_a ^b fg > 0$ or $< 0$, we must have $\int_a ^b fg = 0$.
$$\blacksquare$$
Best Answer
This seems like a lot of work for something that is trivial to prove: right away you get that $g=0$ a.e., and so $fg=0$ a.e.
Even more direct, you can use Cauchy-Schwarz to immediately get (since $f$ is bounded so $f^2$ is bounded) $$ \bigg|\int_a^b fg\bigg|\leq\bigg(\int_a^b g^2\bigg)^{1/2}\bigg(\int_a^b f^2\bigg)^{1/2}=0. $$