Determine linear map $T$ such that $T(f)(x)\geq 0$. 
Determine the linear mappings $T$ of $(C^2([0,1],\mathbb R)$ in $C^0([0,1],\mathbb R)$
  such that for every $f \in C^2([0,1],\mathbb R)$ and all $x \in(0.1)$ if $f$ has a local minimum at $x$, then $T(f)(x)\geq 0$.

My attempt :
If $f$ is $\mathcal {C}^2$ and admits a local minimum at $x$, we can probably show that $f$ is convex in the vicinity of $x$ as $x$ is within the segment $[0,1]$.
$$
T(f)(x) = f''(x)
$$
It is linear and it satisfies the hypotheses.
Nevertheless, I have not managed to prove this rigorously.
Any help will be very appreciated.
Thank you in advance.
 A: Your conjecture is true: For $f\in C^2([0,1])$ put
$$Tf(x):=f''(x)\ .$$
This $T$ is linear, and $Tf\in C^0([0,1])$.
Assume that an $f\in C^2([0,1])$ has a local minimum at a point $a\in(0,1)$, and let $\mu:=f''(a)$. Then $f'(a)=0$, and therefore by Taylor's theorem
$$f(x)-f(a)={\mu\over2}(x-a)^2+o\bigl((x-a)^2\bigr)=(x-a)^2\bigl(\mu+o(1)\bigr)\qquad(x\to a)\ .$$
This shows that when $\mu<0$ one has $f(x)-f(a)<0$ in a punctured neighborhood of $a$. Since $f$ is assumed to be locally minimal at $a$ it follows that necessarily
$$Tf(a)=\mu\geq0\ .$$
A: For any $a,b\in\mathcal C^0([0,1])$, let $T_{a,b}:\mathcal C^2([0,1])\to\mathcal C^0([0,1])$ be the linear map defined by $$T_{a,b}f(x)=a(x)\, f'(x)+b(x)\, f''(x)\, . $$
First, observe that if $b(x)\geq 0$ for all $x\in(0,1)$, then $T_{a,b}$ has the required property (for any $a$). Indeed, if $f\in\mathcal C^2([0,1])$ has a local minimum at some point $x_0\in (0,1)$, then $f'(x_0)=0$ and $f''(x_0)\geq 0$; so $T_{a,b}f(x_0)=b(x_0)f''(x_0)\geq 0$.
Conversely, let us show that any $T$ with the above property has the form $T=T_{a,b}$ with an everywhere nonnegative $b$.
Let us fix $x_0\in (0,1)$. 
First, observe that if a function $f\in\mathcal C^2([0,1])$ satisfies $f'(x_0)=0$ and $f'(x_0)>0$, then $f$ has a local minimum at $x_0$, and hence we must have $Tf(x_0)\geq 0$. Next, consider a function $f$ such that $f'(x_0)=0$ and $f''(x_0)\geq 0$. Then, for any $\varepsilon>0$, the function $f_\varepsilon$ defined by $f_\varepsilon (x)=f(x)+\varepsilon\, (x-x_0)^2$ satisfies $f'_\varepsilon (x_0)$ and $f''(x_0)=f''(x_0)+2\varepsilon >0$; so we must have $Tf_{\varepsilon}(x_0)\geq 0$. Now, denoting by $u_0$ the function $x\mapsto (x-x_0)^2$, we have $f_\varepsilon=f+\varepsilon u_0$, so that (by linearity) $Tf_\varepsilon=Tf+\varepsilon Tu_0$. It follows that $Tf(x_0)+\varepsilon Tu_0(x_0)\geq 0$, for any $\varepsilon >0$, and hence $Tf(x_0)\geq 0$.
So far, we have proved that for any $f$ such that $f'(x_0)=0$ and $f''(x_0)\geq 0$, we have $Tf(x_0)\geq 0$. From this, it follows that if a function $f$ satisfies $f'(x_0)=0$ and $f''(x_0)\leq 0$, then $Tf(x_0)\leq 0$: just apply the preceding result to $-f$. 
Hence, we have the following: if $f\in\mathcal C^2([0,1])$ satisfies $f'(x_0)=0$ and $f''(x_0)=0$, then $Tf(x_0)=0$. 
This can be rewritten as follows. Consider the linear functionals $\delta'_{x_0}$ and $\delta''_{x_0}$ defined (on $\mathcal C^2([0,1])$) by $\delta'_{x_0}(f)=f'(x_0)$ and $\delta''_{x_0}(f)=f''(x_0)$, and denote by $T_{x_0}$ the linear functional $f\mapsto Tf(x_0)$. Then, the result we have obtained reads:
$$\ker(\delta'_{x_0})\cap\ker(\delta''_{x_0})\subset\ker(T_{x_0})\, . $$
From this, one can deduce (this is a standard linear algebra result) that the linear functional $T_{x_0}$ is a linear combination of $\delta'_{x_0}$ and $\delta''_{x_0}$. In other words, one can find coefficients $a(x_0)$ and $b(x_0)$ such that
$$\forall f\in\mathcal C^2([0,1])\;:\; Tf(x_0)=a(x_0)\, f'(x_0)+b(x_0)\, f''(x_0)\, .$$
Now, let $x_0$ vary: this gives two functions $a,b$ on $(0,1)$ such that the linear map $T$ is given by
$$Tf(x)=a(x)\, f'(x)+b(x)\, f''(x)\, .$$
If we take $f_1(x)=x$, then $f'_1(x)\equiv 1$ and $f''(x)\equiv 0$, so that $Tf_1(x)=a(x)$ for all $x\in (0,1)$. This shows that the function $a$ is continuous on $[0,1]$ and, moreover, can be extended to a continuous function on $[0,1]$ (still denoted by $a$). If we take $f_2(x)=\frac{x^2}2$, then $f'_2(x)\equiv x$ and $f''(x)\equiv 1$, so that $Tf_2(x)\equiv a(x)\, x+b(x)$. Since $a$ is already known to be continuous, this shows that $b$ is continuous on $(0,1)$ and can be extended to a continuous function on $[0,1]$ (still denoted by $b$). Finally, if we fix $x_0\in (0,1)$ and consider the function $f(x)=\frac{(x-x_0)^2}2$, then $f$ has a local minimum at $x_0$ so we must have $Tf(x_0)\geq 0$; but since $f'(x_0)=0$ and $f''(x_0)=1$, we also have $Tf(x_0)=a(x_0)\times 0+b(x_0)\times 1=b(x_0)$, and hence $b(x_0)\geq 0$. Since $x_0$ is arbitrary in $(0,1)$, we can conclude that $b$ is everywhere nonegative on $(0,1)$, and hence on $[0,1]$ by continuity.
Altogether, the proof is complete.
