I want to find a function $u \in C^2([0, 3])$ such that $$ \lVert u \rVert_{L^2([0, 3])} > 3\lVert u'' \rVert_{L^2([0, 3])} $$ and $u(0) = u'(0) = 0$ - if it exists. I know that those functions fulfil $$ \lVert u \rVert_{L^2([0, 3])} \leq \frac{9}{2} \lVert u'' \rVert_{L^2([0, 3])} $$ which means that I am just unsure if this bound is sharp. I also figured that observing monomials does not help.
Any help is appreciated.
Define $u:[0,3]\to[0,1]$ by $u(x)=\sin(\frac{x}{2})$. Note $u''(x)^2=\frac{1}{16}\sin^2(\frac{x}{2})$ and thus $u(0)=u''(0)=0$. Finally, $$\Vert u\Vert^2=\int_0^3\sin^2(x)\;dx=16\cdot\int_0^3\frac{1}{16}\sin^2(x)\;dx=16\cdot\Vert u''\Vert^2>0$$ $$\therefore\;\Vert u\Vert= 4\cdot\Vert u''\Vert>3\cdot\Vert u''\Vert$$ $$\therefore\;\Vert u\Vert= 4\cdot\Vert u''\Vert\leq\frac{9}{2}\cdot\Vert u''\Vert$$