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While I was solving exercise problems from "Calculus on Manifolds by Michael Spivak", I got stuck with this one:

Given that $f$ and $g$ are integrable on $[a,b]$, prove that $$\left|\int_a^b f.g\right| \leq \sqrt{\int_a^b f^2}.\sqrt{\int_a^b g^2}$$

Here's what I did:

Case 1: $\forall \lambda \in \mathbb{R}$, consider

\begin{align*} \int_a^b (f-\lambda g)^2 &> 0 \\ \implies \lambda^2\int_a^b g^2 - 2\lambda \int_a^b f.g + \int_a^b f^2 &> 0 \\ \end{align*}

which is a quadratic in $\lambda$ with no real roots, therefore, the discriminant is negative

\begin{align*} \left(2\int_a^b f.g\right)^2 - 4\left(\int_a^b f^2\right).\left(\int_a^b g^2\right) &< 0 \\ \implies \left|\int_a^b f.g\right| \lt \sqrt{\int_a^b f^2}.\sqrt{\int_a^b g^2} \end{align*}

Case 2: $\exists\lambda \in \mathbb{R}$ such that $\int_a^b (f-\lambda g)^2 = 0$

At first, I thought, it should definitely be $f - \lambda g = 0$ which gives equality case of the given inequality.

But then I realized that it is not always true, when I came across the following example: Let $f: [0,1] \rightarrow \mathbb{R}$

$$f(x)= \begin{cases}1 &\text{ if } x=0\\ \frac{1}{\sqrt{n}} &\text{ if } x=\frac{m}{n} \in \Bbb{Q} \text{ with } \gcd(n,m)=1 \text{ and } n>0 \\ 0 &\text{ otherwise } \\ \end{cases}$$

Well, I evaluated that $\int_0^1 f^2 = 0$ in this case (if I am not going wrong here?!).

So, what conditions are implied on $f$ when $\int_a^b f^2 = 0$ given that $f$ is integrable?

For that exercise ($\int_a^b (f - \lambda g)^2 = 0$), Spivak gives the hint as to consider, what happens when $f$ and $g$ are continuous, but I couldn't see how that helps.

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    Borrowing a bit of facts from Lebesgue integration, an integrable function $f : [a, b] \to [0,\infty)$ satisfies $\int_{a}^{b} f = 0$ exactly when $f$ vanishes almost everywhere. – 2017-02-05
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    You don't actually need to consider two cases. Instead of proving strict inequality in the first case, just note that $\int |f-\lambda g|^2 \ge 0$, and say $\Delta \le 0$. Supposing $f$ is non-negative; if $f$ is merely integrable and $\int f = 0$, then $f=0$ almost everywhere for Lebesgue's measure. If $f$ is continuous, though, then $\int f = 0$ only if $f$ is identically zero. – 2017-02-05
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    @SangchulLee Are there techniques from Riemann integration to solve this one? Because I haven't been taught Lebesgue integration till yet. Although it would be a whole lot fun in studying Lebesgue theory now but I guess I might lack some prerequisites? (I am following Principles of Mathematical Analysis by Rudin and Calculus by Spivak for the time being) – 2017-02-05
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    @OpenBall So when exactly does equality holds in the given inequality? – 2017-02-05
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    Exactly if there is some $\lambda$ and some set $A$ of Lebesgue measure zero such that $f(x) = \lambda g(x)$ for all $x \notin A$. – 2017-02-05
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    @OpenBall I hope it doesn't sounds rude but can your above-said statement be translated in terms of Riemann integral (as I already pointed out my difficulty in one of the above comments)? – 2017-02-05
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    No, it can't. I am not making things unnecessarily complicated. – 2017-02-05
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    Oh ok! Nevermind then. Thanks for answering. – 2017-02-05

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