Minimizing a distance to a set on Hilbert space

We are on E the vector space of continuous and 2 times differentiable functions on $$[0,1]$$ i.e., $$C^2[0,1].$$ I have the sets $$V = f$$ such that $$f(0)=f(1)=0$$ and $$W = f$$ such that $$f=f''.$$ And the function $$\langle f,g\rangle = \int_0^1[{f(t)g(t)+f'(t)g'(t)}]\,dt.$$ I have proven that they are complementary subspaces and that they are orthogonal by a scalar product defined as above.

With that I will be able to find the orthogonal projection of any function on $$W.$$

Now if I'm given another set $$G_{a,b} = f$$ such that $$f(0)=a$$ and $$f(1)=b$$ how can I determine $$\inf_{f\in G_{a,b}}\int_0^1\left[f^2(t)+f'^2(t)\right]dt?$$

Let $$\|\cdot\|$$ be the norm induced by the inner product $$\langle\cdot,\cdot\rangle.$$

I know it's the same as $$\left(\inf_{f\in G_{a,b}} {\|f\|}\right)^{\!2}$$
and that $$\inf_{f\in G_{a,b}}{\|f\|}$$ is $$d(0,G)$$ which is the projection of $$0$$ on $$G.$$

But $$G$$ is not even a subspace; I really don't know how to go about it. Any help would be appreciated, thanks!!!

• Can you please explain what the notation $\|\cdot\|_{<>}$ means? That is non-standard notation in functional analysis. Jan 14, 2020 at 22:25
• Sorry I just meant the norm associated to the scalar product <> Jan 14, 2020 at 22:26
• Oh, you probably don't need to write that in there. Or if you think there might be confusion, you can just insert language like, "Let $\|\cdot\|$ be the norm induced by the inner product $\langle\cdot,\cdot\rangle.$" Jan 14, 2020 at 22:28
• Thanks for the remark Jan 14, 2020 at 22:30

You can use the Calculus of Variations here. Let $$L(f,\dot{f};t)=f^2(t)+\dot{f}^2(t).$$ Then the Euler-Lagrange equation says that you must set $$\frac{\partial L}{\partial f}+\frac{d}{dt}\,\frac{\partial L}{\partial \dot{f}}=0.$$ This becomes \begin{align*} 2f+\frac{d}{dt}\left(2\dot{f}\right)&=0\\ f+\ddot{f}&=0\\ f(t)&=A\sin(t)+B\cos(t). \end{align*} Applying $$f(0)=a$$ shows us that $$B=a,$$ so we rewrite as $$f(t)=A\sin(t)+a\cos(t).$$ Applying $$f(1)=b$$ shows us that \begin{align*} b&=A\sin(1)+a\cos(1)\\ b-a\cos(1)&=A\sin(1)\\ b\csc(1)-a\cot(1)&=A. \end{align*} Hence, $$f(t)=\left[b\csc(1)-a\cot(1)\right]\sin(t)+a\cos(t).$$