# Prove that the Cauchy - Schwarz inequality is true for complex valued functions

In our course, we proved that this inequality works for definite integrals of real valued functions. How do we show that the inequality remains true for complex valued functions. Is it possible to build a proof without referring to complexity analysis? I mean, of course, complex numbers and their basic properties are well-known, but the theory of integration for complex valued functions has not beam covered yet.

General idea: if $$f$$ is complex valued then $$f=f_1+if_2$$ with $$f_1,f_2$$ real valued. Define $$\int_a^{b} f(x) \, dx$$ as $$\int_a^{b} f_1(x) \, dx +i\int_a^{b} f_2(x) \, dx$$. Using Riemann sums and the fact that $$|a+b| \leq |a|+|b|$$ for any two complex numbers $$a$$ and $$b$$ you can show that $$|\int_a^{b} f(x) \, dx| \leq \int_a^{b} |f(x)| \, dx$$. Once you have this you can apply C-S for real functions to get C-S for complex functions.
• If I understand correctly, shouldn't this leave you with a $\sqrt 2$, owing to having to rely on $\|a_i\|_{1} \le \sqrt 2 \|a_i\|_2$? – Calvin Khor Oct 11 '18 at 9:42
• I don't understand why you are bringing in $\sqrt 2$. What we have is $|\int fg|\leq \int |f|\, |g|$ and since $|f|,|g|$ are real valued we can apply C-S to $\int |f|\, |g|$ which gives the bound $\sqrt {\int |f|^{2} } {\sqrt {\int |g|^{2}} }$. – Kavi Rama Murthy Oct 11 '18 at 9:47
• to clarify, I think your path to the first inequality $|\int fg| \le \int |fg|$ brings a $\sqrt 2$ in it, i.e. this only proves $|\int fg| \le \sqrt 2 \int |fg|$? – Calvin Khor Oct 11 '18 at 9:48
• My guess is you are not using the fact that $|a+b| \leq |a|+|b|$ holds for all complex numbers $a$ and $b$. Do not make the mistake of writing $|f_1+if_2|$ as $\sqrt {f_1^{2}+f_2^{2}}$. That would complicate matters. – Kavi Rama Murthy Oct 11 '18 at 9:51
• No. I am avoiding this type of inequality by approximating integrals by Riemann sums and applying the inequality $|\sum a_k| \leq \sum |a_k|$. – Kavi Rama Murthy Oct 11 '18 at 9:55