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Consider the vector space $V$ where $V$ is the set of continuous functions $f\colon [0,2] \to \mathbb{C}$. Prove that the following defines an inner product: $$(f \mid g) = \int_{1}^{2} f(x)\overline{g(x)}(1 + x^2)\,dx$$

My issue is the bounds of the integral. If we were dealing with $$(f \mid g) = \int_{0}^{2} f(x)\overline{g(x)}(1 + x^2)\,dx$$ then I would know how to continue but I'm confused as to how to do it when the bounds don't match.

Axioms for inner products:

  1. $(f \mid f)$ $\geq$ 0 for all a $\in$ V
  2. $(f \mid f)$ = 0 $\implies$ f = 0$_V$
  3. Linearity in the first slot: $(f + \lambda h \mid g)$ = $(f \mid g) +\lambda(h \mid g)$
  4. $(f \mid g)$ = $\overline{(g \mid f)}$
juliana paternoster
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1 Answers1

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This isn't going to be an inner product space due to the bounds issue. Note that if you let $f$ be piecewise linear, $f(x)=1-x$ on $[0,1]$ and then $f(x)=0$ on $[1,2]$, you get $(f|f)=0$.

The closest this could get to be an inner product space is if you took the equivalence class of functions that agree on $[1,2]$ as your vectors. This would be isomorphic to the space of functions only defined on $[1,2$] Likely there was a typo.

Alan
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