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I was trying to solve the following problem. $$ \varphi(f) = \int_0^1 t f(t)dt, \quad f \in X = \{g: [0,1]\to \mathbb{R}, g \text{ continuous in }[0,1] \} $$ Consider X with the norm $ \| \cdot \|_1 $. Show that $ \varphi \in X^* $ and compute its norm.

I've done the following: $$ |\varphi(f)| \leq \int_0^1 t|f(t)|dt \leq \int_0^1|f(t)|dt=\|f\|_1 $$ $$ \implies \|\varphi\|\leq 1 \implies \varphi \in X^* $$ Then, I was trying to show that $ \| \varphi \| =1 $. It seemed to me that the sequence $ f_n = n \chi_{(1,1-1/n)} $ did the job, since: $$ \| f_n\| = 1 \quad \forall n \in \mathbb{N} $$ and $$ \int_0^1 tf_n(t)dt = n \int_{1-1/n}^1tdt = \frac n 2 t^2 \bigg|_{1-1/n}^1 = \frac n 2 \left[1-\left(1-\frac 1 n \right)^2\right] = \frac n 2 \left(- \frac{1}{n^2} + \frac 2 n \right) = 1 - \frac{1}{2n} $$ Hence, $ \forall n \in \mathbb{N} $, $$ \| \varphi \| = \sup_{f\neq0} \frac{|\varphi(f)|}{\|f\|_1} \ge \frac{|\varphi(f_n)|}{\|f_n\|_1}= 1-\frac{1}{2n} \implies \| \varphi \| \ge 1 $$ However, $f_n$ are not continuous, meaning that $f_n \not\in X$ and I cannot use them. How can I proceed? I was thinking about using the density of $C([0,1])$ in $L^1([0,1])$ but I do not know how. Can someone please help me? Thank you in advance.

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    $\begingroup$ You can change your $f_n$ to make them continuous without changing the integrals very much: Replace the jump at $1-1/n$ by a very steep line. $\endgroup$ – Jochen Sep 4 '18 at 12:24
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Let $f_n(t)=(n+1)t^n$.

Then $\|f_n\|_1=1$ and $\phi(f_n)=\frac{n+1}{n+2}$.

Hence $\|\phi\|\geq \frac{n+1}{n+2}$ for every $n$.

Therefore $\|\phi\|\geq 1$.

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  • $\begingroup$ Beat me to it by seconds! $\endgroup$ – MisterRiemann Sep 4 '18 at 12:28
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Hint

One way is to use functions looking like your $f_n$ but continuous.

For example: $$g_n(x)=\begin{cases}0 \text{ if } x \leq 1-\frac{1}{n}-\frac{1}{n^2}\\ \left(x- \left(1-\frac{1}{n}-\frac{1}{n^2}\right)\right) n^3 \text{ if } 1-\frac{1}{n}-\frac{1}{n^2} \leq x \leq 1-\frac{1}{n}\\ n \text{ if } x \geq 1-\frac{1}{n}\end{cases}$$

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