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