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Let $f(x)$ and $g(x)$ are two integrable function, $x \in [1,3]$ satisfying

$$\left( \int_1^3 f(x) g(x) dx \right)^2 = \left( \int_1^3 f^2(x) dx \right) \left( \int_1^3 g^2(x) dx \right)$$

Given that f(1)=2 and g(1)=4

My sir told me that

$$\frac{g(3)}{f(3)}= \frac{g(1)}{f(1)}=2$$

But I could not understand how he has said that ?

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    @PiotrBenedysiuk why ? – 2017-01-08
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    You will need to assume that $f(3)$ and $g(3)$ are both nonzero. – 2017-01-08
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    @edm why we have to assume like that – 2017-01-08
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    @user135482 or else you divide by zero. – 2017-01-08
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    @edm I want to know how they got the ratio – 2017-01-08

2 Answers 2

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Cauchy-Schwarz inequality implies that $f$ and $g$ are linearly dependent. That is, there exist two nonzero constants $\lambda$ and $\mu$ such that $\lambda f=\mu g$ almost everywhere. With no further information, little more can be said.

For example, if $f$ and $g$ are continuous we can delete the word "almost". If in addition, if both $f(1)$ and $f(3)$ are nonzero, then $$\frac{g(3)}{f(3)}=\frac{g(1)}{f(1)}=\frac\lambda\mu$$

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    Can we say like this $\frac{g(3)}{f(3)}=\frac{g(2)}{f(2)}=\frac\lambda\mu$. That is instead of 1 we can use any other value – 2017-01-06
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    The two functions are simply assumed to be integrable instead of continuous, so the binary function $\langle f,g\rangle:=\int_{1}^{3}f(x)g(x)dx$ is not an inner product on the space of integrable functions defined on $[1,3]$, and in the strict sense, we can't really use Cauchy-Schwarz inequality to show that $f$ and $g$ are linearly dependent. A counterexample is given by two functions $f$ and $g$ that are zero everywhere except $f(1)=2,f(3)=1,g(1)=4,g(3)=1$. This example also shows that the binary function is not an inner product. – 2017-01-08
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    As for the linear dependence almost everywhere, I think you need a proof other than proofs for Cauchy-Schwarz inequality. – 2017-01-08
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You cannot prove that the ratio is $2$. Consider two functions $f$ and $g$ that are zero everywhere except $f(1)=2,f(3)=1,g(1)=4,g(3)=1$. These functions are integrable and satisfy the equality as both sides are zero. But $$\frac{g(3)}{f(3)}=1.$$

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    From where do you get f(3) =1 and g(3)=1 – 2017-01-08
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    @user135482 They are constructed to be. – 2017-01-08
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    Can you explain it more – 2017-01-08
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    Does it mean my sir is wrong – 2017-01-08
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    @user135482 Do I explain how I come up with the functions or how they satisfy the equality? I don't know if you have wrote the question exactly as he said, but usually we would need to assume further that $f$ and $g$ are continuous and at least one of $f(3)$ and $g(3)$ is nonzero or else the example as I wrote would be a counterexample. – 2017-01-08
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    If possible could you explain both of them – 2017-01-08
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    Functions with only removable discontinuities and no other types of discontinuities are known to be integrable, and I can easily manipulate the ratio $\frac{g(x)}{f(x)}$ at finitely many $x$ without altering the value of integral. As for the integrals being zero, notice that the points where $f$ and $g$ are nonzero are too few and the total length of those points is too short to contribute any significant value to the integral. – 2017-01-08