# Is $\mathcal C^1([a,b],\mathbb R)$ endowed with the $C^1$-norm a Hilbert space?

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!

• Check the parallelogram identity. Oct 30, 2019 at 21:12
• 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. Oct 30, 2019 at 21:15
• Hint: there surely are. Oct 30, 2019 at 21:15
• Check for example $f(x) = x$ and $g(x)=1$ on $[0,1]$. Oct 30, 2019 at 21:26
• 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. Oct 30, 2019 at 21:28

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 $$2\|f\|^2+2\|g\|^2 = 2(1+1)^2 + 2 = 10.$$