Boundedness of Extremal This question was asked in CSIR NET December 2017.

$I[y]=\int_{0}^{1}\frac{1}{2}[(y')^{2}-4π^{2}(y)^{2}]dx$
Let $(P)m= \inf\{I[y]: y\in C^{1}[0,1], y(0)=0,y(1)=0\}$ Let $y_{0}$
satisfy the Euler-Lagrange Equation associated with $(P)$. Then which
of the following is /are true?

*

*

*

*$m= -\infty$, $I$ is not bounded



*


*$m\in R$ with $I[y_{0}]=m$



*


*$m\in R$ with $I[y_{0}]>m$



*


*$m\in R$ with $I[y_{0}]<m$

My Attempt
Here the Extremal function will be:

$y_{0}=c\sin(2πx)$

if we apply the boundary conditions. The answer should be the first option according to the answer key. But I am stuck around the boundedness of the extremal.  Any help would be appreciated.
Thank you.
 A: Poincaré - Wirtinger Inequality $\pi^2 \int_0^1 y^2 \leq \int_0^1 (y^\prime )^2$, which is valid for any $y \in C_0^1([0,1]):= C_0([0,1]) \cap C^1([0,1])$, entails:
$$I[y] = \frac{1}{2}\ \int_0^1 (y^\prime)^2 - 4\pi^2 y^2 \geq \frac{1}{2}\ \int_0^1 \pi^2 y^2 - 4\pi^2 y^2 = -\frac{3}{2}\pi^2 \int_0^1 y^2 $$
with equality iff $y(x) = y_C(x) := C \sin(\pi x)$; therefore:
$$I[y_C] = -\frac{3}{2}\pi^2 C^2 \int_0^1 \sin^2 (\pi x)\ \text{d}x \to -\infty \qquad \text{as } C \to +\infty$$
and $I$ is unbounded from below over $C_0^1([0,1])$.

If you are not familiar with inequalities like Poincaré - Wirtinger's, you can argue as follows.
Functions of the family $y_C(x) := C \sin (\pi x)$ belong to $C_0^1([0,1])$ for each $C \in \mathbb{R}$.
Evaluating $I[\cdot]$ on $y_C$ gives:
$$\begin{split} I[y_C] &= \frac{C^2}{2} \int_0^1 \Big[\pi^2 \cos^2(\pi x) - 4\pi^2 \sin^2(\pi x)\Big]\ \text{d} x\\ &= \frac{\pi^2 C^2}{2} \int_0^1 \Big[ 1 - 5 \sin^2 (\pi x)\Big]\ \text{d} x \\ &\stackrel{t=\pi x}{=} \frac{\pi C^2}{2} \underbrace{\int_0^\pi \Big[ 1 - 5 \sin^2 t\Big]\ \text{d} t}_{= - \frac{3\pi}{2}} \\ &= - \frac{3\pi^2}{4} C^2\end{split}$$
and the unboundedness of $I[\cdot]$ follows by letting $C \to +\infty$ as above.
A: *

*FWIW, one way to physically understand the result, consider the Fourier series
$$y(x)~=~\sum_{n\in\mathbb{N}}a_n \sin(\pi n x), \qquad a_n~\in~\mathbb{R}.\tag{A}$$
OP's action functional for the harmonic oscillator becomes
$$  I[y]~=~\frac{1}{2}\int_0^1 \! dx~(y^{\prime}(x)^2-4\pi^2 y(x)^2)~=~\ldots~=~\frac{\pi^2}{4}\sum_{n\in\mathbb{N}}(n^2-4)a^2_n.\tag{B}$$
We see that the only Fourier mode with negative contribution to the action (B) is the $n=1$ mode. By setting all the other modes to zero, and letting $|a_1|\to\infty$, we see that the action $I[y]$ is unbounded from below, i.e. option (1) is correct.


*Fun fact: We can read off the classical solutions/natural frequency of the harmonic oscillator from formula (B). It is the mode $n=2$ with arbitrary amplitude $a_2$. This agrees with OP's last formula. It is a stationary but not a minimum configuration for the action. See also e.g. this related Phys.SE post.
