ways to calculate norm of the Bounded Linear Functional.

Question

$\Lambda$ be a linear functional on $C([0,1])$ defined by $$ \Lambda(f) = \int_0^1 xf(x)dx \;\;\; \text{ for } f \in C([0,1]). $$ and use $\|f\|_{sup} = \sup_{x \in [0,1]} |f(x)| $ for $f \in C([0,1])$.

I'm trying to find all $f \in C([0,1])$ with $\|\Lambda (f) \| = \|\Lambda\| \|f\|$.

I already solve a similar problem about the norm of bounded linear functional, but I still don't know how to calculate the norm in general case. Is there any way to calculate norm of bounded linear functional?

Answer 1

Since for every $f\in C[0,1]$ we have $|\Lambda f|\leq\frac{1}{2}\|f\|$, and there is equality if $f$ is the constant function $1$, the norm of $\Lambda$ is $\frac{1}{2}$. Let $f\in C[0,1]$ be such that $\|f\|=1$, and such that $\Lambda f=\frac{1}{2}$. Put $$A=\{x\in [0,1]| f(x)\geq 0 \}$$ Then $$\frac{1}{2}=\int_0^1 xf(x)\,dx=\int_A xf(x)\,dx + \int_{[0,1]\backslash A}xf(x)\,dx\leq \int_A xf(x)\,dx\leq \frac{1}{2}$$ Hence $f$ must be non-negative almost everywhere, and since it is continuous, it must be non-negative everywhere. Now, if there is a certain value of $0\leq x\leq 1$ for which $0\leq f(x)<1$, then by continuity there is some interval where $0\leq f(x)<1$, and so $$\int_0^1 xf(x)\,dx < \int_0^1x\,dx=\frac{1}{2}$$ Conclusion: the only norm-$1$ continuous function for which $\Lambda f=\frac{1}{2}$ is the constant function $1$. As a result, the functions satisfying $\|\Lambda f\|=\|\Lambda\|\|f\|$ are precisely the constant functions, and this is the answer to your question.

Answer 2

General case. Any linear bounded functional $A \colon C([0,1]) \mapsto \mathbb{R}$ can be represented in the form of Riemann–Stieltjes integral $$ A(f) = \int\limits_0^1 f(x) dg(x), $$ where $g$ is a function of bounded variation. Furthermore $\lVert A \rVert = \mathop{\mathrm{Var}}_0^1[g]$.

Your particular case. Let $g(t) = \dfrac{t^2}{2}$ and $\mu_g$ is Lebesgue–Stieltjes measure defined by the function $g$. We can represent $\Lambda$ in the form of Lebesgue–Stieltjes integral: $\Lambda(f) = \int\limits_0^1 x f(x)dx = \int\limits_0^1 f(x) d \mu_g$. Then $\lVert \Lambda \rVert = \mathop{\mathrm{Var}}_0^1[g] = \Big[g \in C^1([0,1])\Big] = \int\limits_{0}^{1} |g'(t)| dt = \dfrac{1}{2}$.

Answer 3

Clearly $f\equiv 0$ is a solution. So let's assume $f$ is not identically $0$.

The operator norm is given by $\|\Lambda\|=\sup_{\|g\|=1}\|\Lambda(g)\|$. So your equation becomes \begin{equation*} \frac{\|\Lambda(f)\|}{\|f\|}=\sup_{\|g\|=1}\|\Lambda(g)\|. \end{equation*}

Note that the LHS can be rewritten, due to linearity, as $\|\Lambda(\frac{f}{\|f\|})\|$. So the LHS is always less or equal than the RHS, with equality if and only if $\|\Lambda(g)\|=\|\Lambda\|$, where $g:=\frac{f}{\|f\|}$.

That means that if $f$ satisfies your equation, $f$ normalized is the maximizer in the computation of the norm of $\Lambda$. Now clearly, $\|\Lambda\|\leq \frac{1}{2}$ since $xf(x)\leq x\|f(x)\|\leq x$ whenever $\|f\|=1$ and taking $f\equiv 1$, we can see that $\|\Lambda(f)\|=\frac{1}{2}$, so $\|\Lambda\|=\frac{1}{2}$.

So now the question simply becomes what are $f\in C([0,1])$ with $\|f\|=1$ such that $\int_0^1 xf(x)=\frac{1}{2}$. Then all your solutions will be $\lambda f$ for arbitrary constants $\lambda$. Argue that this means $f\equiv\lambda$ using the continuity of $f$.