Is the space of all constant functions closed in $C^1[0,1]$ with respect to $L^2$-norm?

Question

Let $C$ be the space of all constant polynomials over $\Bbb R.$ Is $C$ closed in $C^1[0,1]$ with respect to $L^2$-norm?

Let $f \in C^1[0,1]$ be a limit point of $C.$ We need to show that $$\displaystyle\inf \left \{\int_{0}^{1} \left (f(x) - c \right )^2\ dx\ \bigg |\ c \in \Bbb R \right \} = 0 \implies f \in C.$$ Let us define a function $g : \Bbb R \longrightarrow \Bbb R$ by $$g(c) = \int_{0}^{1} \left (f(x) - c \right )^2\ dx,\ c \in \Bbb R.$$ Then $g$ is differentiable on $\Bbb R$ and by Leibnitz integral rule we have $$g'(c) = \int_{0}^{1} -2 \left (f(x) - c \right )\ dx.$$ Now if the infimum of $g$ is attained at $c_0 \in \Bbb R$ (the infimum has to be attained at some real number because if $c \to \pm \infty$ then $g(c)$ blows up to $+ \infty$) then $g'(c_0) = 0 \implies \displaystyle \int_{0}^{1} f(x)\ dx = c_0.$ Now if we compute $g(c_0)$ we find that $$g(c_0) = \int_{0}^{1} \left (f(x) \right )^2\ dx - \left ( \int_{0}^{1} f(x)\ dx \right )^2.$$ So by Cauchy-Schwarz inequality we have $$g(c_0) \geq 0\ \ \ \ (\because \left \lvert \left \langle f, 1 \right \rangle \right \rvert \leq \left \|f \right \|_2 \left \|1 \right \|_2 ).$$ So $g(c_0) = 0 \iff f$ is a scalar multiple of $1$ i.e. $f \in C.$ But this proves that $C$ is closed in $C^1[0,1].$

Is my above reasoning correct at all? Would anybody please verify it?

Thanks for your time.

I didn't read your solution, but I propose a much shorter one. $L^2$ convergence implies subsequential a.e. convergence, thus the $L^2$ limit of constant functions is the a.e. limit of constant functions, which must be a.e. constant, hence constant in $L^2$ (also in $C(0,1)$). — Physical Mathematics, Feb 04 '21 at 17:07
@Physical Mathematics very nice argument. Thanks for your valuable suggestion. One small point though. You are using Riesz-Fischer theorem to conclude a.e. subsequential convergence. Right? — Anacardium, Feb 04 '21 at 17:19
I do not really understand the end, how do you deduce from Cauchy-Schwarz that $g'(c_0) \ge 0$ ? — Falcon, Feb 04 '21 at 17:23
I have mentioned it in bracket @Falcon. That's because $|\langle f,1 \rangle | \leq |f|_2 |1|_2.$ — Anacardium, Feb 04 '21 at 17:24
Moreover your argument works only if there if $c \in \mathbb R$ that minimize $g$, which may not be the case. — Falcon, Feb 04 '21 at 17:25
the minimum might be $\pm \infty$ which is not in $\mathbb R$. — Falcon, Feb 04 '21 at 17:28
@Falcon we may restrict $g$ to a compact set which contains $\displaystyle \int_{0}^{1} f(x)\ dx$ as an interior point. That will do our job. — Anacardium, Feb 04 '21 at 17:32
@Falcon if $c \rightarrow \pm \infty$ then $g$ blows up to $+\infty.$ Have you noticed that? — Anacardium, Feb 04 '21 at 17:35
Ok for that point! But sorry, I am a bit slow, but I still don't understand how, from the fact that $$g(c_0) = \int f(x)^2 - \left(\int f(x)\right)^2$$ and Cauchy-Schwarz, you deduce that $g'(c_0) \ge 0$.. — Falcon, Feb 04 '21 at 17:38
@Anacardium Subsequential a.e. convergence is a standard result for $L^p$ spaces, the standard proof follows by noting that convergence in $L^p$ implies convergence in measure and then noting convergence in measure implies subsequential a.e. convergence. — Physical Mathematics, Feb 04 '21 at 17:41
@Falcon there was a typo. I corrected that. Please have a look at it now. — Anacardium, Feb 04 '21 at 18:10
@Falcon don't you understand why $\displaystyle \int_{0}^{1} \left (f(x) \right )^2\ dx \geq \left (\int_{0}^{1} f(x)\ dx \right )^2\ $? — Anacardium, Feb 04 '21 at 18:17
Now is the proof fine @Falcon? I have added the details also as to why the infimum of $g$ is attained at some real number. If any further modification is needed please let me know. I will edit it accordingly. — Anacardium, Feb 04 '21 at 18:25

score 2 · Accepted Answer · answered Feb 04 '21 at 18:23

2

Seems ok to me now. However, it could have been much shorter: $g(c) \ge 0, \inf_{c \in \mathbb R}g(c) = 0$ implies that there is $c_0 \in \mathbb R$ such that
$$g(c_0) = 0 = \int_0^1 (f(x) - c_0)^2 \quad \Rightarrow \quad f(x) = c_0 ~~\text{a.e.}$$

answered Feb 04 '21 at 18:23

Falcon

3,978

Yes you are right. Thanks for your valuable suggestion. – Anacardium Feb 04 '21 at 18:29
You're welcome! – Falcon Feb 04 '21 at 18:31

Is the space of all constant functions closed in $C^1[0,1]$ with respect to $L^2$-norm?

1 Answers1