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 $\lVert f \rVert_{C[0,1]} \leq \lVert f \rVert_{H_0}$?

  • $\begingroup$ 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? $\endgroup$ Nov 8, 2017 at 15:11
  • $\begingroup$ 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. $\endgroup$
    – MonkeyKing
    Nov 8, 2017 at 15:24
  • $\begingroup$ 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. $\endgroup$ Nov 8, 2017 at 15:30
  • $\begingroup$ 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... $\endgroup$
    – MonkeyKing
    Nov 8, 2017 at 15:41
  • $\begingroup$ Oh yeah, sorry about that. I'll have another think. $\endgroup$ Nov 8, 2017 at 15:42

1 Answer 1


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.


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