I come accross this problem in my self-study for the calculus of variations and I think I solved the first point. However, the second point is not completely clear to me
Wirtinger's inequality Let $a \in [0,\pi]$. Then for any $u \in C^2 ([0,a])$ that satisfies the boundary conditions $u(0)=u(a)=0$, the following inequality holds;
$$\int_0^a u^2 (x) dx \leq \int_0^a (u'(x))^2 dx.$$
When $a > \pi$, there exists a function $u$, for which the inequality is false.
(1)- Show that the function $\overline{u} \equiv 0$ is stationary curve of the variational problem
$$\inf_{u \in X} I[u],\,\,\, I[u] = \int_0^a [(u')^2 - u^2] dx,\,\,\,\, X= \{ u \in C^2([0,a]); u(0)=u(a)=0 \}$$ and that satisfies Legendre's condition.
(2)- Using the homogeneity property $I[\lambda u] = \lambda^2 I[u],\,\, \forall u \in X, \forall \lambda \in \mathbb{R} $ of the functional, show that $\overline{u}$ is weak local minimizer of the variational problem if and only if it is a (global minimizer).
Soulution 1- The stationary curve it the solution when the integrand satisfies the Euler-Lagrange equation and the boundary conditions. We have $f(x,u,u')= {u'(x)}^2-u^2(x)$, and thus we require that
$$\frac{\partial f}{\partial u} - \frac{d}{dx} \frac{\partial f}{\partial u'}=0 \Rightarrow -2 u - 2u''=0 \Rightarrow u(x) = C_1 \cos(x) + C_2 \sin(x),$$ where $C_1, C_2$ are constants. From the boundary conditions we have
$$u(0)=0 \Rightarrow 0= C_1 \cos(0) + C_2 \sin(0) \Rightarrow C_1=0.$$ and
$$u(a)=0 \Rightarrow 0= C_2 \sin(a) \Rightarrow C_2=0, \text{ if } a \in (0,\pi), \text{or} \sin(a)=0 \text{ if } a=\pi.$$
Therefore, we have $u\equiv0$ is the stationary solution. The stationary solutions satisfies Legendre's condition for we have $f(x,u(x),u'(x)) = u'^2 - u^2$, and
$$\frac{\partial^ f(x,u(x),u'(x))}{\partial u'^2} =2 \Rightarrow \frac{\partial^ f(x,0,0)}{\partial u'^2} =2 > 0,\,\, x \in [0,a].$$
Then $u\equiv 0 = \inf_{u \in X} I[u].$
\textbf{(2):} ($\Rightarrow$) It is trivial that every global minimizer is already weak local minimizer.\($\Leftarrow$) Let $\overline{u}$ is a weak local minimizer. Then there is a neighbourhood
$$N_\delta(\overline{u})=\{ u\in X| \|u - \overline{u}\|_w \le \delta\},\quad \|u\|_w= \max_{x\in [0,a]}|u(x)|$$ such that $I[\overline{u}] \le I[u],\,\, \forall u \in N_\delta(\overline{u})$.
Now, for $u \in N_\delta(\overline{u})$ define $Y:= \operatorname{span} \{u\}$. Restricting the minimizer to $Y$ we obtain for every $v \in Y$ ($v= \lambda u; \, \lambda \in \mathbb{R}$) we have $I[\overline{u}]\le I[v] = \lambda^2 I[u]$. Assume that $\overline{u}$ is not global minimizer, then there is $\overline{v}=\lambda u \in Y$ such that
$$I[\overline{v}] < I[\overline{u}] \Rightarrow \lambda^2 I[u] < I[\overline{u}].$$
Since $\lambda$ is arbitrary, choose $\lambda \in (0,1)$ small enough such that $\|\overline{v} -\overline{u}\|_w \leq \delta$, which makes $\overline{v} \in N_\delta(\overline{u})$ which contradicts the fact that $\overline{u}$ is a weak local minimizer. Therefore, $\overline{u}$ is a global local minimizer.
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\textbf{(3):} Let $x \in [0,a]$, where $a$ is to be determined later. Since $u(a) = 0$ we can write, using the fundamental theorem of calculus: $$ u(x) = \int_0^x \frac{\mathrm{d}u(y)}{\mathrm{d}y}\mathrm{d}y,\ \ \ x\in[0,a], $$ hence $$ \left|u(x)\right| \le \int_0^x \left|\frac{\mathrm{d}u(y)}{\mathrm{d}y}\right|\mathrm{d}{y}. $$ Cauchy-Schwarz inequality ensures: \begin{align*} \int_0^x \left|\frac{\mathrm{d}u(y)}{\mathrm{d}y}\right|\mathrm{d}{y} &\le \left(\int_0^x\mathrm{d}y\right)^{1/2} \cdot \left(\int_0^x\left|\frac{\mathrm{d}u(y)}{\mathrm{d}y}\right|^2\mathrm{d}{y}\right)^{1/2}\\ &\le \left(\int_0^a\mathrm{d}y\right)^{1/2} \cdot \left(\int_0^a\left|\frac{\mathrm{d}u(y)}{\mathrm{d}y}\right|^2\mathrm{d}{y}\right)^{1/2}\\ &=(a)^{1/2}\left(\int_0^a \left|\frac{\mathrm{d}u(y)}{\mathrm{d}y}\right|^2\mathrm{d}{y}\right)^{1/2}. \end{align*} Thus we have
$$\left|u(x)\right| \leq \sqrt{a} \left(\int_0^a \left|\frac{\mathrm{d}u(y)}{\mathrm{d}y}\right|^2\mathrm{d}{y}\right)^{1/2} \Rightarrow \left|u(x)\right|^2 \leq a \int_0^a \left|\frac{\mathrm{d}u(y)}{\mathrm{d}y}\right|^2\mathrm{d}{y}.$$
Choose $a\in [0,1]$ we obtain
$$\left|u(x)\right|^2 \leq \int_0^a \left(u'(y)\right)^2\mathrm{d}{y}, \quad \forall x \in [0,a].$$
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\textbf{(4):} Consider the function $u(x) = x(x-a)$, now we find
\begin{align*} I[u] =\int_0^a (2x-a)^2 - (x^2-ax)^2 dx= -\frac{1}{30} a^3 (a^2-10).\\ \end{align*}
It is clear to see that for $a > \sqrt{10}$ we get $I[u]<0$.
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\textbf{(5):} Put $F(x,u(x),u'(x)) = u'(x)^2 -u(x)^2$. Let $P(x) = F_{u' u'}(\overline{u}) = 2$ and $Q(x) = F_{uu}(\overline{u}) -\frac{d}{dx}F_{uu'}(\overline{u})$=-2. Then the Jacobi equation
$$\frac{d}{dx} (P(x) u'(x)) = Q(x) u(x), \quad u \in C^2[0,a].$$ So we have
$$\frac{d}{dx} (2 u'(x))=-2 u(x) \Rightarrow u''(x)+ u(x)=0.$$
The solution for the equation is $u(x)=C_1 \cos(x) + C_2 \sin(x)$ which satisfies $u(0)=u(a)=0$ which results that $\sin(a)=0 \Rightarrow a=\pi$.
$$==================================================$$ \textbf{(6):} We have $u(0)=u(\pi)=0$ ($c_0 = \int_0^\pi u(x)dx=0$), then we expand $u$ in as a Fourier sine series; $$u(x)=\sum_{k=1}^\infty a_k \sin k x,$$
and hence $$ \int_0^\pi u(x)^2 dx= \frac{\pi}{2}\sum_{k=1}^\infty a_k^2 \qquad \text{while} \qquad \int_0^\pi (u'(x))^2 dx= \frac{\pi}{2}\sum_{k=1}^\infty k^2a_k^2, $$ and therefore $$ \int_0^\pi (u'(x))^2 dx\ge \int_0^\pi u(x)^2 dx.$$
Your solution is fine but you need to adjust your example because the case when $\pi < a < \sqrt{10}$ is not clear.
Consider the function $u(x) = \sin{\frac{\pi}{a}x}$, now we find
\begin{align*} I[u] &=\int_0^a \left( \frac{\pi^2}{a^2} \cos(\frac{\pi}{a}x)^2 - \sin(\frac{\pi}{a}x)^2\right) dx= \frac{\pi^2-a^2}{2a}. \end{align*}
Thus we have $I[u] <0$ when $a > \pi$.