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Let $PC_2[0,1]$ denote the vector space of piecewise continuous functions $f$ on $[0,1]$ such that $$\int_0^1|f(t)|^2dt < \infty.$$ Then I wish to show that $$\langle f, g\rangle =\int^1_0 f(t)\overline{g(t)} dt$$ defines an inner product on $PC_2[0, 1]$.

My question is how to show conjugate symmetry $\langle f,g\rangle=\langle g,f\rangle$? Is it just $$\int_0^1f(t)\overline{g(t)}dt=\int_0^1\overline{\overline{f(t)}g(t)}=\overline{\langle g,f\rangle}\;?$$

For positive definiteness, do I have to use epsilon delta definition of continuity to show it?

Pedro
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Homaniac
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1 Answers1

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Your proof that inner product is conjugate symmetric is correct. $ \langle f, f \rangle =\int_0^{1}|f(x)|^{2} \, d x \geq 0$ and equality holds iff $f =0$. [ If $f(x) \neq 0$ for some $x$ then there is an interval $(a,b) \subset (0,1)$ and a positive number $\delta$ such that $|f(y)| >\delta$ for all $y \in (a,b)$. But then $\int_0^{1}|f(x)|^{2} \, d x \geq \int_a^{b}|f(x)|^{2} \, d x \geq \delta^{2} (b-a)>0$, a contradiction].

Kavi Rama Murthy
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  • Thank You! I have a related qns at https://math.stackexchange.com/questions/3178840/functional-analysis-inner-product-and-haar-rademacher-walsh-functions if you are so kind to give some tips too :) – Homaniac Apr 08 '19 at 00:57
  • @Homaniac, Kavi: The argument in this post does not work for the function defined by $f(1/2)=1$, $f(x)=0$ if $x\neq 1/2$ which is usually considered piecewise continuous on $[0,1]$. So, what is the definition of piecewise continuous that you both are using? – Pedro Aug 24 '20 at 12:43
  • $f$ is piece-wise continuous on $[a,b]$ if there exist points $a=x_0<x_1<x_2<....<x_n=b$ such that $f$ is continuous on $[x_{i-1},x_i]$ for each $i$. @Pedro – Kavi Rama Murthy Aug 24 '20 at 23:14