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!!!

  • $\begingroup$ Can you please explain what the notation $\|\cdot\|_{<>}$ means? That is non-standard notation in functional analysis. $\endgroup$ Jan 14, 2020 at 22:25
  • $\begingroup$ Sorry I just meant the norm associated to the scalar product <> $\endgroup$ Jan 14, 2020 at 22:26
  • $\begingroup$ 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 $\langle\cdot,\cdot\rangle.$" $\endgroup$ Jan 14, 2020 at 22:28
  • $\begingroup$ Thanks for the remark $\endgroup$ Jan 14, 2020 at 22:30

1 Answer 1


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, $$f(t)=\left[b\csc(1)-a\cot(1)\right]\sin(t)+a\cos(t). $$


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