For the purpose of practice I took on the following problem about the Riesz representation theorem which asks me to find the unique element whose existence the representation theorem ensures.

Let $H = L^2(0,1)$ and for $u \in H$ let $F$ be the functional

$$ F(u) := \int\limits_{0}^{1/2} u(t)\, dt $$

Show that $F$ is continuous and linear on $H$, calculate its operator norm and find the unique element $f \in L^2(0,1)$ which represents $F$ by Riesz representation theorem

$$ \langle u, f \rangle = \int\limits_{0}^{1} f(t) u(t)\, dt \quad \forall u \in L^2(0,1)$$

The first part I think I solved. The linearity of the functional is quite clear due to the Lebesgue integral properties, i.e.

$$F(\alpha u_1 + u_2) = \int\limits_{0}^{1/2} \alpha u_1 + u_2 dt = \alpha F(u_1) + F(u_2) $$

Also, it's continuous since

$$F(u) = \int\limits_{0}^{1/2} u(t)\, dt \leq \int\limits_{0}^{1} \lvert u \rvert dt \leq \int\limits_{0}^{1} \lvert u(t)\rvert^2\, dt$$

and hence

$$F(u) \leq \Vert u \Vert_{L^{2}(0,1)}$$

What I struggle with is the operator norm and to actually find $f \in L^2(0,1)$. I read that the operator norm of a functional

$$F_g(u) = \int\limits_{\Omega} ug \, dt $$

is given by $\Vert F\Vert = \int\limits_{\Omega} \lvert g\rvert\, dt$. But what is it in my case and how do I find the unique element $f$?


The Riesz representation $f$ of $F$ should obey $$ F(u) = \int fu\,\mathrm dt. $$

Thus it can be seen that $f=\chi_{(0,\frac12)}$ has to be the representant of $F$.

The operator norm for functionals on $L^2(0,1)$ obeys $\| F\|^2 = \int |f|^2 \,\mathrm dt$. This is easy to calculate for the function $f$ above.

  • $\begingroup$ Thank you for your answer. With regard to the operator norm, is it then just the squared indicator function over (0, 1), i.e. the squared measure of the unit interval? $\endgroup$ – Taufi Jul 20 '17 at 12:29
  • $\begingroup$ note that $f$ is the indicator function over $(0,\frac12)$, not $(0,1)$. $\endgroup$ – supinf Jul 20 '17 at 13:02
  • $\begingroup$ Thanks! That helped! $\endgroup$ – Taufi Jul 20 '17 at 13:25

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