# Determining $\| \delta_{n} \|_{*}=\frac{n}{2}$

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 $$\lim\limits_{n \to \infty}\delta_{n}(f)=f(0)$$

• Your choice for $f$ in 1. is perfectly fine. – Jakobian Jul 7 at 19:00
• But I would need to approximate it with continuous functions, how do I do that? – SABOY Jul 7 at 19:31
• Just choose piecewise continuous ones – Jakobian Jul 7 at 19:39
• Connecting points $(0, 0), (-1/n-\varepsilon, 0), (-1/n, 1), (1/n, 1), (1/n+\varepsilon, 1), (1, 1)$, a trapezoid – Jakobian Jul 7 at 19:43
• @Jakobian can you give an example of one? – 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 $$\delta_n(f) \to \delta_\infty(f)$$.