Consider the set

$\mathbb A=\{f\in C^1([-1,1]):f(-1)=-1,f(1)=1\}$.

Prove that there is no function in this function space that gives us the minimum of $S=\int_{-1}^1x^2f'(x)^2dx$. What is the infimum of $S$ for the functions of this space?
1. **Lemma Proof:**
   - Suppose \( f \in C[-1, 1] \) and \(\int_{-1}^{1} f(x) \eta'(x) \; dx = 0\) for all \(\eta \in C_{0}^{\infty} (-1, 1)\). We need to show that \( f \) is constant.
   - Let \(\varphi\) be a bump function, i.e., a smooth function whose support is compact and contains 0.
   - Define \(\varphi_{\epsilon}(x) = \frac{1}{\epsilon} \varphi \left( \frac{x}{\epsilon} \right)\). Then \(\int_{a}^{b} g(x) \varphi_{\epsilon} (x - c) \; dx \to g(c)\) for all \(g(x) \in C[a, b]\) and \(a < c < b\).
   - Now, let \(\eta_{\epsilon}(x) = \int_{-1}^{x} [\varphi_{\epsilon}(t - a) - \varphi_{\epsilon}(t - b)] \; dt\) for \(-1 < a < b < 1\). For sufficiently small \(\epsilon\), \(\eta_{\epsilon} \in C_{0}^{\infty} (-1, 1)\).
   - Therefore, \(0 = \int_{-1}^{1} f(x) \eta_{\epsilon}' (x) \; dx \to f(a) - f(b)\), which implies \(f(a) = f(b)\). Since \(a\) and \(b\) are arbitrary, \(f\) must be constant.

2. **Main Proof:**
   - Assume \(S\) achieves its minimum for \(f \in \mathbb{A}\). Then for any \(\eta \in C_{0}^{\infty}(-1, 1)\) and \(\epsilon \in \mathbb{R}\), we have \(S(f + \epsilon \eta) \geq S(f)\).
   - This implies \(\frac{d}{d\epsilon} S(f + \epsilon \eta) = 0\). A simple calculation shows that this implies \(\int_{-1}^{1} x^2 f'(x) \eta'(x) \; dx = 0\).
   - By the Lemma, the continuous function \(x^2 f'(x)\) must be constant. This is possible only if \(f'(x) = 0\), which contradicts the boundary conditions \(f(-1) = -1\) and \(f(1) = 1\). Thus, \(S\) has no minimum.

3. **Infimum Calculation:**
   - We claim that \(\inf_{f \in \mathbb{A}} S(f) = 0\). Clearly, \(S(f)\) is non-negative, so it suffices to show that the infimum cannot be positive.
   - For \(0 < \epsilon < 1\), let \(f_{\epsilon}\) be defined such that \(f_{\epsilon}(-1) = -1\) and 
     \[
     f'_{\epsilon}(x) = 
     \begin{cases} 
     \frac{2}{\epsilon} \left( 1 - \left| \frac{x}{\epsilon} \right| \right) & \text{if } |x| \leq \epsilon, \\
     0 & \text{otherwise}.
     \end{cases}
     \]
   - Then \(f_{\epsilon} \in \mathbb{A}\) and we have 
     \[
     S(f_{\epsilon}) = \int_{-\epsilon}^{\epsilon} x^2 f'_{\epsilon}(x)^2 \; dx = 8\epsilon \int_{0}^{1} t^2 (1 - t)^2 \; dt.
     \]
   - Evaluating the integral, we get 
     \[
     \int_{0}^{1} t^2 (1 - t)^2 \; dt = \int_{0}^{1} (t^2 - 2t^3 + t^4) \; dt = \left[ \frac{t^3}{3} - \frac{2t^4}{4} + \frac{t^5}{5} \right]_{0}^{1} = \frac{1}{3} - \frac{1}{2} + \frac{1}{5} = \frac{10 - 15 + 6}{30} = \frac{1}{30}.
     \]
   - Thus, 
     \[
     S(f_{\epsilon}) = 8\epsilon \cdot \frac{1}{30} = \frac{8\epsilon}{30} = \frac{4\epsilon}{15}.
     \]
   - By choosing sufficiently small \(\epsilon\), we can make \(S(f_{\epsilon})\) as small as we wish. This proves the claim.

The final answer is \( \boxed{ \inf_{f \in \mathbb{A}} S(f) = 0 } \).