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Jun
21
comment Can any symmetric bounded bilinear mapping occur as the second derivative?
@I disagree: Thanks a lot. The assumption that $E$ is finite-dimensional does not seem to be necessary, as one can simply define $f(x)=\frac{1}{2}\nu(x,x)$ in the general case. Would you like to write that as an answer or should I?
Jun
19
comment Can any symmetric bounded bilinear mapping occur as the second derivative?
Feel free to change the tags, as I'm not sure.
Jun
19
asked Can any symmetric bounded bilinear mapping occur as the second derivative?
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reviewed Approve Prove by induction: $1(1!)+\cdots + n\cdot n!$ = (n+1)! - 1
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Jun
18
comment Connectedness of $\mathbb R$
The first $\cap$ should be $\cup$. You'll have to add your definition of connectedness.