# Space of $f(0)=f(1)$: Is Hilbert space?

Let $S$ be space consisting of collection of square integrable continuous functions $f:[0,1]\rightarrow\mathbb{R}$ with the constraint $f(0)=f(1)$. So $S$ is an inner product space with the inner product $$\langle f,g \rangle =\int_0^1f(x)g(x)dx.$$

Now the question is whether $S$ is a Hilbert space or not??

Now, $S$ to be a Hilbert space it is necessary to show that $S$ is complete metric space with respect to its natural metric $$\|f-g\|=\sqrt{\langle f-g,f-g\rangle}.$$ To show completeness I need to show that every Cauchy sequence in that space is convergent.

Here I stuck. Specifically I have no idea how to show the sequence of $f$ has limit $f_0$ which has the property that $f_0(0)=f_0(1)$.

Is there any counter example to show that $S$ is not a Hilbert space? Thanks for any suggestion.

Added later: Thanks to @Aweygan for pointing out the role of the (previous) restriction of positivity to the function. Though primarily I was interested about positive valued function, with out loss of generality I am now considering the mean shift version of the functions which enable the new functions to take values in real line. I modified this point in the edited question.

• It is not even a vector space, as it is not closed under scalar multiplication. – Aweygan Jul 23 '16 at 6:21
• @Aweygan I can not see why it is not closed under scalar multiplication. Could you explain ? – nicomezi Jul 23 '16 at 6:40
• The OP requires functions $f:[0,1]\to\mathbb{R}^+$. For any such function $f\in S$ and any real number $a<0$, $af\notin S$ – Aweygan Jul 23 '16 at 6:44
• Ok, I have missed the $+$. By the way I believe there would also have been a problem with completeness, the constraint is not enough to prevent continuous functions converging to non-continuous ones. – nicomezi Jul 23 '16 at 6:45
• There is no square root. – nicomezi Jul 23 '16 at 6:54

## 1 Answer

Regarding convergence and completeness: For $n\in N$ let $f_n(x)=0$ for $x\leq 1/2-1/(n+2)$ and $f_n(x)=1$ for $x\geq 1/2.$ Let $f_n(x)$ be linear for $x\in [1/2-1/(n+2),1/2].$

Then $(f_n)_{n\in N}$ is a Cauchy sequence with respect to the norm $\|f-g\|=[\int_0^1|f(x)-g(x)|^2\;dx]^{1/2}.$

Let $h(x)=0$ for $x\leq 1/2$ and $h(x)=1$ for $x>1/2.$ The function $h$ is not continuous. And it has the property that for any continuous $g:[0,1]\to R$ there exists $r>0$ and here exist $a,b$ with either $0<a<b<1/2$ or $1/2<a<b<1$, such that $x\in [a,b]\implies |g(x)-h(x)|\geq r.$

But for all sufficiently large $n$ we have $x\in [a,b]\implies f_n(x)=h(x).$ So $\lim \sup \|g-f_n\|^2\geq \int_a^b |g(x)-h(x)|^2\;dx\geq (b-a)r^2>0.$ So $(f_n)_{n\in N}$ cannot converge in norm to any continuous $g.$

• My assumption was $f_n$ are continuous functions. In your example $f_n$ function is "not" continuous having discontinuity at two points (1/2-1/(n+2) and 1/2). – Janak Jul 23 '16 at 9:01
• @Janak In fact, the function $f_n$ is continuous. Images psoted in answers to this question might help you visualize the sequence $f_n$. – Martin Sleziak Jul 23 '16 at 10:20
• @MartinSleziak Thank you. I got the point. – Janak Jul 23 '16 at 10:45
• Could you please suggest me further: Is it possible to make this space($S$) a Hilbert space by imposing minimal conditions on $f$. Thanks a lot. – Janak Jul 23 '16 at 10:52
• A Hilbert-space norm must by def'n give a a complete metric. I don't think that any infinite-dimensional subspace of $S$ can be a complete metric space with this norm, but I haven't proved it. But I think it shouldn't be hard to do so. – DanielWainfleet Jul 24 '16 at 19:37