Let $X:=C([-1,1])$ and equip it with $\| \cdot \|_{1}$. Further, let $\delta_{n}: X \to \mathbb C$ be a linear functional, such that for $n \in \mathbb N$, $\delta_{n}(f)=\frac{n}{2}\int_{-\frac{1}{n}}^{\frac{1}{n}}dxf(x)$.

I want to show:

$1.$ $\| \delta_{n} \|_{*}=\frac{n}{2}$

$2.$ that $\delta_{\infty}$ where $\delta_{\infty}(f):=f(0)$ is not linear bounded functional on $X$.

$3.$ $\lim\limits_{n \to \infty}\delta_{n}(f)=f(0)$ for all $f\in X$.

My ideas:

$1.$ the easiest part is clear: for $f\in X$: $\vert\delta_{n}(f)\vert=\frac{n}{2}\vert\int_{-\frac{1}{n}}^{\frac{1}{n}}dxf(x)\vert\leq \frac{n}{2}\int_{-\frac{1}{n}}^{\frac{1}{n}}dx\vert f(x)\vert\leq \frac{n}{2}\| f\|_{1}\Rightarrow \| \delta_{n} \|_{*}\leq\frac{n}{2}$ for $n \in \mathbb N$.

for the converse $\geq $ I want to use, for a particular $n \in \mathbb N$, the function $f=\frac{n}{2}1_{[-\frac{1}{n}, \frac{1}{n}]}$. It is clear that $\delta_{n}(f)=\frac{n}{2}$ but my problem is that $f \notin X$. I am struggling to create a sequence of continuous functions $(f_{n})_{n}$ that converges pointwise to $f$. Any ideas?

$2.$ My idea is to find normalised functions $(f_{n})_{n}$ on $[-1,1]$, i.e. $\| f_{n} \|_{1}=1$ but that $f_{n}(0)=n$ for all $n \in \mathbb N$ but no idea how to formalize this into continuous functions.

for $3.$ I am think of using some form of Hölder but I am not sure how I would eventually reach $\lim\limits_{n \to \infty}\delta_{n}(f)=f(0)$

  • $\begingroup$ Your choice for $f$ in 1. is perfectly fine. $\endgroup$ – Jakobian Jul 7 at 19:00
  • $\begingroup$ But I would need to approximate it with continuous functions, how do I do that? $\endgroup$ – SABOY Jul 7 at 19:31
  • $\begingroup$ Just choose piecewise continuous ones $\endgroup$ – Jakobian Jul 7 at 19:39
  • $\begingroup$ Connecting points $(0, 0), (-1/n-\varepsilon, 0), (-1/n, 1), (1/n, 1), (1/n+\varepsilon, 1), (1, 1)$, a trapezoid $\endgroup$ – Jakobian Jul 7 at 19:43
  • $\begingroup$ @Jakobian can you give an example of one? $\endgroup$ – SABOY Jul 7 at 21:03
  1. You need a function $f \in X$ which is supported on $\left[-\frac1n, \frac1n\right]$. For example you can take $$f(x) = n\exp\left({\frac1{n^2x^2-1}}\right)\,1_{\left(-\frac1n, \frac1n\right)}(x)$$ Then we have $$\|f\|_1 = n\int_{-\frac1n}^\frac1n \exp\left({\frac1{n^2x^2-1}}\right)\,dx = \begin{bmatrix} t =nx \\dt = n\,dx \end{bmatrix} = \int_{-1}^1 \exp\left(\frac1{t^2-1}\right)\,dt$$ $$\delta_n(f) = \frac{n^2}2\int_{-\frac1n}^\frac1n \exp\left({\frac1{n^2x^2-1}}\right)\,dx = \begin{bmatrix} t =nx \\dt = n\,dx \end{bmatrix} = \frac{n}2\int_{-1}^1 \exp\left(\frac1{t^2-1}\right)\,dt$$ so $\|\delta_n\|_* \ge \frac{|\delta_n(f)|}{\|f\|_1} = \frac{n}2$.
  2. Consider $f_n(x) = (1-|x|)^n$. We have $f_n(0) = 1$ but $$\|f_n\|_1 = 2\int_0^1(1-x)^n\,dx = \frac2{n+1} \xrightarrow{n\to\infty} 0$$ so $\delta_\infty$ cannot be bounded.
  3. Let $\varepsilon > 0$. $f$ is continuous at $0$ so there exists $n_0 \in \mathbb{N}$ such that $$|x| < \frac1{n_0} \implies |f(x)-f(0)| < \varepsilon$$ For $n \ge n_0$ we have $$|\delta_n(f) - \delta_\infty(f)| = \left|\frac{n}2\int_{-\frac1n}^\frac1n f(x)\,dx-f(0)\right| \le \frac{n}2\int_{-\frac1n}^\frac1n |f(x)-f(0)|\,dx \le \varepsilon$$ so $\delta_n(f) \to \delta_\infty(f)$.

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