# $H_0=\{f\in C[0,1] \mid f(0)=f(1)=0, f' \text{ piecewise continuous}\}$ and supremum norm

For $f\in H_0$ if we define $\langle f,g\rangle_{H_0}=\int^1_0f'(x)g'(x)dx$, and assume $f'\in L^2[0,1]$, how do we show that $f$ is continuous and $\lVert f \rVert_{C[0,1]} \leq \lVert f \rVert_{H_0}$?

• I'm confused. By $C[0, 1]$, do you mean the space of continuous functions under the supremum norm? If so, then how do we guarantee the existence of the derivatives for the inner product? Nov 8, 2017 at 15:11
• Right derivative I believe, fixed. Yet even in this scenario this part is still confusing to me too, seems like f being continuous is like given. Nov 8, 2017 at 15:24
• It's more than that. Consider the sequence of functions $f_n(x) = \sin(2\pi n x)$. Then $\|f_n\|_{H_0} \rightarrow \infty$, but $\|f_n\|_\infty = 1$. The inequality can't hold. Nov 8, 2017 at 15:30
• The inequality holds for this case right? $H_0$ is the larger one. I can't prove it, tried those obvious ways, like Schwarz, Holder, bounding integrand by its maximum... Nov 8, 2017 at 15:41
• Oh yeah, sorry about that. I'll have another think. Nov 8, 2017 at 15:42

Finally, I got it! Sorry about the litany of mistakes. For $0 \le x \le 1$, by the Cauchy-Schwarz inequality and the Fundamental Theorem of Calculus, $$f(x)^2 = \left(\int_0^x f'(t) \mathrm{d}t\right)^2 \le \int_0^x f'(t)^2 \mathrm{d}t \int_0^x 1\mathrm{d}t = x\int_0^x f'(t)^2 \mathrm{d}t \le \int_0^1 f'(t)^2 \mathrm{d}t.$$ Taking the supremum over $x$ and square roots of both sides yields the result.