# Can any symmetric bounded bilinear mapping occur as the second derivative?

Let $E$ and $F$ be two real Banach spaces.

By Schwarz' theorem (or a generalization of it), for $f\in C^2(E,F)$ and $x\in E$, $D^2f(x)$, viewed as a bounded bilinear mapping $E\times E\rightarrow F$, is symmetric.

Is that all we know about the values of the second derivative? In other words, for any symmetric bounded bilinear mapping $\nu:E\times E\rightarrow F$, is there a twice Fréchet differentiable mapping $f:E\rightarrow F$ such that $D^2f(x)=\nu$ for some $x\in E$?

In case it makes a difference, I would be also interested in the special case where $E$ is finite-dimensional.

• Feel free to change the tags, as I'm not sure. – Stefan Jun 19 '15 at 7:38
• If $E$ is finite-dimensional, you are asking about a function with prescribed Hessian matrix $H$ at some point, where the entries of $H$ are elements of $F$. The quadratic function $f(x) = \frac12 x^T Hx$, $x\in E$, does the job then. – user147263 Jun 21 '15 at 0:13
• @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? – Stefan Jun 21 '15 at 10:00