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Let $\mathcal C^1([a,b],\mathbb R)$ be endowed with the $C^1$-norm $\|\cdot\|$ where $\|f\| = \| f \|_\infty + \| f' \|_\infty$ for $f \in \mathcal C^1([a,b],\mathbb R)$.

In class, we have proved that $\langle \mathcal C^1([a,b],\mathbb R), \|\cdot\| \rangle$ is a Banach space. I could not figure out if $\langle \mathcal C^1([a,b],\mathbb R), \|\cdot\| \rangle$ is a Hilbert space or not. In other words, I'm not sure if $\|\cdot\|$ is induced by some inner product.

Could you please shed me some light on this issue? Thank you for your help!

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    $\begingroup$ Check the parallelogram identity. $\endgroup$
    – amsmath
    Oct 30, 2019 at 21:12
  • $\begingroup$ Hi @amsmath, I've tried to use the parallelogram identity, but I don't know if there exist functions $f,g$ that violate the parallelogram identity. $\endgroup$
    – Akira
    Oct 30, 2019 at 21:15
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    $\begingroup$ Hint: there surely are. $\endgroup$
    – amsmath
    Oct 30, 2019 at 21:15
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    $\begingroup$ Check for example $f(x) = x$ and $g(x)=1$ on $[0,1]$. $\endgroup$
    – amsmath
    Oct 30, 2019 at 21:26
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    $\begingroup$ Also, you can check that the minimum principle fails in $C^1$, while it holds in every Hilbert space. That is, there exist non-empty closed convex sets in $C^1$ which have no vector of minimal norm (in contrast to Hilbert spaces), and/or have more than one such. $\endgroup$
    – paul garrett
    Oct 30, 2019 at 21:28

1 Answer 1

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On $[a,b]=[0,1]$ set $f(x) = x$ and $g(x) = 1$. Then $$ \|f-g\|^2+\|f+g\|^2 = (\|x-1\|_\infty + 1)^2 + (\|x+1\|_\infty + 1)^2 = 4+9=13. $$ But $$ 2\|f\|^2+2\|g\|^2 = 2(1+1)^2 + 2 = 10. $$

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    $\begingroup$ Thank you so much for your help!, I just did the calculation and cross-check it with your answer :)) $\endgroup$
    – Akira
    Oct 30, 2019 at 22:06

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