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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 ?

user135482
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2 Answers2

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

ajotatxe
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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 – user135482 Jan 06 '17 at 16:47
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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. – edm Jan 08 '17 at 11:26
  • As for the linear dependence almost everywhere, I think you need a proof other than proofs for Cauchy-Schwarz inequality. – edm Jan 08 '17 at 11:27
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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 $$\frac{g(3)}{f(3)}=1.$$

edm
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  • From where do you get f(3) =1 and g(3)=1 – user135482 Jan 08 '17 at 11:27
  • @user135482 They are constructed to be. – edm Jan 08 '17 at 11:28
  • Can you explain it more – user135482 Jan 08 '17 at 11:29
  • Does it mean my sir is wrong – user135482 Jan 08 '17 at 11:29
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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.

     – edm Jan 08 '17 at 11:35
  • If possible could you explain both of them – user135482 Jan 08 '17 at 11:41
  • 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. – edm Jan 08 '17 at 11:49