Closedness and Minimal Norm Elements

Question

While studying I came upon this problem in which I don't really know how to solve.

Let $C[0,1]$ be the space of continuous functions on $[0,1]$ with the sup norm. Let

\begin{equation} M = \left\{f \in C[0,1] : \int_{1}^{\frac{1}{2}} f - \int_{\frac{1}{2}}^{1} f = 1\right\} \end{equation}

Show that $M$ is a closed convex set which contains no minimal norm element.

I have thought of the following solution. However, I am not sure if it is correct. Pretty sure I am missing something.

1. Closedness

Let $ M \text{ be defined as } M = \{ f \in C[0,1] : \int_{1}^{\frac{1}{2}} f - \int_{\frac{1}{2}}^{1} f = 1 \} $

To show that $M$ is closed it must contain all its limit points. Consider a sequence $f_n \in M$ converging to some $f$ in $C[0,1]$. We need to show that $f \in M$.

For any $f_n \in M$ we have $\int_{1}^{\frac{1}{2}} f_n - \int_{\frac{1}{2}}^{1} f_n = 1$.

Now, consider the limits as $n$ approaches infinity: $\lim_{{n \to \infty}} \left( \int_{1}^{\frac{1}{2}} f_n - \int_{\frac{1}{2}}^{1} f_n \right) = \int_{1}^{\frac{1}{2}} f - \int_{\frac{1}{2}}^{1} f.$

Since $f_n$ converges to $f$ , the limits on both sides are equal. Therefore, $ \int_{1}^{\frac{1}{2}} f - \int_{\frac{1}{2}}^{1} f = 1 $, and $f \in M$ . This implies that $M$ is closed.

2. No Minimal Norm Element:

To show that $M$ contains no minimal norm element. In other words, for any $ f \in M $, there exists $ g \in M $ such that $\|g\| < \| f \|$ .

Consider a function $ f \in M $ such that $ \int_{1}^{\frac{1}{2}} f - \int_{\frac{1}{2}}^{1} f = 1 $.

Now we define a function $g$ as $ g(x) = f(x) - \epsilon $, where $ \epsilon > 0 $. $g$ is also in $M$ because:

$ \int_{1}^{\frac{1}{2}} g - \int_{\frac{1}{2}}^{1} g = \int_{1}^{\frac{1}{2}} (f(x) - \epsilon) - \int_{\frac{1}{2}}^{1} (f(x) - \epsilon) = 1 - \epsilon(1-1) = 1. $

So, $g$ is also in $M$ , and $\| g \| < \| f \|$ because $g$ is obtained by subtracting a positive constant $ \epsilon $ from $f$.

Any feedback?

Thank you.

Answer 1

I'm going to assume the set you are looking at is

$$M := \left\{ f\in C[0,1] : \int_{0}^{1/2}f - \int_{1/2}^{1}f = 1\right\},$$

because the other set does indeed have elements of minimal norm.

Your work for showing that $M$ is closed looks good overall. However, it is worth emphasising that you can take the limit inside the integral because the sequence of functions $(f_{n})_{n\in\mathbb{N}}$ converges uniformly to $f$.

You appear to have left out showing that $M$ is convex. The process is similar to how you show that $M$ is closed.

For showing that $M$ does not have an element of minimal norm, what you have there does not work. You can't conclude from there that $\|g\| < \|f\|$. Try looking at the example $f:[0,1] \to \mathbb{R}$ defined by $f(x) := 1/2 - x$ to see why.

One approach you can try is to show that for any $\varepsilon >0$, you can always find a continuous function $f$ on $[0,1]$ in $M$ such that $\|f\| \leq 1 + \varepsilon$, and you can also show that there is no continuous function $f$ on $[0,1]$ with $\|f\| \leq 1$ such that $f\in M$. The desired result can be obtained from there by connecting those two results.

You can also look at this post which addresses the same question.