In this problem we have the function

$$L:L^2[0,1] \to \mathbb{C}$$ $$Lf = 2\int_0^1 x \left( \int_x^1f(y) \mathrm{d}y \right) \mathrm{d}x$$

for any function $f \in L^2[0,1]$.

Show that this is a Bounded Linear functional and obtain the norm of $L$.

The first part is not bad, for linearity let $a_1, a_2 \in \mathbb{C}$ and $f_1, f_2 \in L^2[0,1]$ and we simply use the linearity of the integral

$$L(a_1 f_1+a_2f_2) = 2 \int_0^1x \left( \int_x^1 \left(a_1 f_1 (y) + a_2 f_2(y) \right) \mathrm{d}y \right) \mathrm{d}x$$

$$= 2 \int_0^1 \left(a_1 x\int_x^1 f_1(y) \mathrm{d}y + a_2 x \int _x^1f_2(y) \mathrm{d}y \right) \mathrm{d}x$$

$$= 2a_1\int_0^1 x \left( \int_x^1 f_1(y) \mathrm{d}y \right) \mathrm{d}x + 2a_2\int_0^1 x \left( \int_x^1 f_2(y) \mathrm{d}y \right) \mathrm{d}x = a_1 Lf_1+a_2Lf_2$$

Now, for boundedness, let $f : ||f||=1$

$$\int_0^1 |f(x)| \mathrm{d}x = ||f|| = 1$$

Since we have $x \in [0,1]$, it follows that $[x,1] \subseteq [0,1]$ and $\chi_{[x,1]} \leq \chi _{[0,1]}$ so we have that $|f(x)|\chi_{[x,1]} \leq |f(x)| \chi_{[0,1]}$ and $\int_x^1 |f(x)| \mathrm{d}x \leq \int_0^1 |f(x)| \mathrm{d}x = ||f|| = 1$

Thus, we have the bound $$\sup_{||f||=1} |Lf| = \sup_{||f||=1} \left| 2\int_0^1 x \left( \int_x^1f(y) \mathrm{d}y \right) \mathrm{d}x \right| \leq \left| 2 \int_0^1 x \mathrm{d}x \right| = 1$$

Nevertheless, I can't explicitly calculate the norm $||L||$


1 Answer 1


You can interchange the integrals to write the function as $Lf=2\int_0^{1}\int_0^{y}xdx f(y)dy=\int_0^{1}y^{2}f(y)dy$. Applying Hoder's /C-S in equality we see that $|Lf| \leq \|f\|_2\sqrt {\int_0^{1}y^{4}}dy$ which gives $\|L\| \leq \frac 1{\sqrt 5}$. To see that equality holds take $f(x)=\sqrt 5 x^{2}$. Verify that $\|f\|_2=1$ and that $Lf=\frac 1{\sqrt 5}$.

  • $\begingroup$ Why does that specific function guarantee the equality for all $f$ with unit norm? $\endgroup$
    – The Bosco
    Commented Nov 14, 2019 at 4:51
  • $\begingroup$ The defintion of $\|L\|$ is $\sup |Lf|:\|f\|\leq 1$. If you find one $f$ with $\|f\|\leq 1$ and $|Lf|\geq \frac 1 {\sqrt 5}$ then $\|L\| \geq \frac 1 {\sqrt 5}$ becasue the supremum of a set is greatest than or equal to any element of that set. $\endgroup$ Commented Nov 14, 2019 at 5:06

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