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$H=\{f\in C[0,1]:f(0)=0, f \text{ is absolutely continuous, and } f' \in L^2[0,1]\}$

(a). Prove that $\langle f,g\rangle =\int_0^1 f'g'dx $ defines an inner product on $H$.

(b). Prove that the injection $i : H \to C[0,1]$ is continuous.

For the first part, I know to show that it defines an inner product, I show that for $f,g,h \in H$ and $\alpha, \beta \in R$, $$(i). \langle f,f\rangle \geq 0,\text{ and }\langle f,f\rangle=0 \iff f=0 $$ $$(ii). \langle \alpha f+\beta g ,h\rangle =\alpha\langle f,h\rangle +\beta\langle g,h\rangle $$ $$(iii). \langle f,g\rangle =\langle g,f\rangle $$ $(ii)$ and $(iii)$ seem to be easy to show. But regarding $(i)$ this is what I have $$\langle f,f\rangle =\int_0^1 (f')^2dx \geq 0$$ since $(f')^2 \geq 0$.

Assuming $\langle f,f\rangle=0$, I need to show that $f=0$. $$\langle f,f\rangle =\int_0^2 (f')^2=0 \Rightarrow (f')^2=0 \Rightarrow f'=0.$$ My concern is that $f'=0$ does not necessarily imply $f=0$.

Does the fact that $f(0)=0, f\in H$ do any help here?

Regarding problem (b), I don't really know how to go by that, can someone provide any help?

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  • $\begingroup$ You assume that any $f\in H$ satisfies $f(0)=0$ and is differentiable everywhere. So, if its derivative is $0$ everywhere, then certainly it remains $0$. throughout the interval. $\endgroup$ – Nick Peterson Jul 10 '17 at 14:59
  • $\begingroup$ For (b), you have proved that $H$ is a metric space, therefore, it has standard topology. To prove the continuity, verifying the definition is ok. $\endgroup$ – Aolong Li Jul 10 '17 at 15:03
  • $\begingroup$ Thank you all. I get the (a) part now. but @Nick I never proved that $H$ is a metric space. Unless every inner product induces a metric. Is that correct? $\endgroup$ – J. Kyei Jul 10 '17 at 15:24
  • $\begingroup$ Yes, every inner product produces a metric: it creates a norm defined by $\| a\|^2=\langle a,a\rangle$, which in turn gives you a metric defined by $d(a,b)=\|a-b\|$. $\endgroup$ – Nick Peterson Jul 10 '17 at 20:32
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For the first part you are correct in saying that '$f(0)=0$ for $f \in H$ helps'. The condition $<f,f> = 0$ implies that $f'=0$ (for all $x \in [0,1]$) which in turn implies $f$ is a constant function (by a simple application of the mean value theorem). Since this constant function must satisfy $f(0)=0$ to be a member of $H$, $f$ must be the $0$ function.

As for a hint for part $b)$: What does it mean for the map $i: H \to C([0,1])$ to be continuous? It means that given $\epsilon > 0$ there exists $\delta >0$ such that when $\|f-g\| = <f-g,f-g>^{1/2} < \delta$ then $$\|f-g\|_\infty = \sup_{x \in [0,1]} |f(x)-g(x)| < \epsilon.$$ Given some $\epsilon$ can you find an appropriate $\delta$? Hope this helps.

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  • $\begingroup$ it helps a lot, especially the first explanation. I will consider what you said about the (b) and post my idea back. thank you. $\endgroup$ – J. Kyei Jul 10 '17 at 15:20
  • $\begingroup$ Great, glad I could help! $\endgroup$ – Zestylemonzi Jul 10 '17 at 15:22

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