I want to calculate the operator norm of the operator $A: L^2[0,1] \to L^2[0,1]$ which is defined by $$(Af)(x):=i\int\limits_0^x f(t)\,dt-\frac{i}{2} \int\limits_0^1 f(t)\, dt$$

I've already shown that this operator is compact and selfadjoint. I think maybe this helps me calculating the operator norm. Maybe through spectral theorem for compact self adjoint operators.

I also know that for integral operators of the form $$(Kf)(x)=\int\limits_0^1 k(x,t) f(t)\,dt$$ the inequality $\Vert K \Vert \leq \Vert k \Vert{}_{L^2}$ holds.

For $$(Af)(x)=i\int\limits_0^x f(t)\,dt-\frac{i}{2} \int\limits_0^1 f(t) \,dt = \int\limits_0^1 i\,\left(1_{[0,x]}(t)-\frac{1}{2}\right)f(t)\,dt$$ this gives me an upper bound:

$$\Vert A \Vert \leq \left\Vert i~1_{[0,x]}-\frac{i}{2} \right\Vert{}_{L^2}=\frac{1}{2}$$

Can someone help me?

  • $\begingroup$ I tried finding a function $f$ such that $\Vert Af \Vert = \frac{1}{2}$ this would give me $\Vert A \Vert =\frac{1}{2}$. But I couldn't find one. Can someone give me a hint? $\endgroup$ Jun 3, 2019 at 14:55
  • 1
    $\begingroup$ Hint: the Cauchy-Schwarz inequality $\lvert \langle u,v\rangle \rvert \le \| u \| \| v\|$ gives equality when $u = \alpha v$ for some constant $\alpha$. $\endgroup$
    – User8128
    Jun 3, 2019 at 15:14
  • $\begingroup$ @User8128 Thank you for your response. Can you tell me on which expression you apply Cauchy Schwarz? I guess the function $f$ which I mentioned in my previous comment is determined by the condition for equality for C.S.? $\endgroup$ Jun 4, 2019 at 10:14
  • $\begingroup$ I assume you calculated $\|A\|$ using CS: $$\| A \| = \sup_{\|f \|_{L^2} \le 1} \int^1_0 (i\mathbf{1}_{[0,x]} - \tfrac i 2) f(t) dt \le \sup_{\|f\|_{L^2}\le 1} \|i\mathbf{1}_{[0,x]} - \tfrac i 2\|_{L^2} \| f\|_{L^2} \le \|i\mathbf{1}_{[0,x]} - \tfrac i 2\|_{L^2} = \tfrac 1 2.$$ To make this into an equality, choose the $f$ so that $\|A(f)\|_{L^2} = \tfrac 1 2$; this amounts to choosing $f$ in the first inequality which produces an equality. The first inequality is just Cauchy-Schwarz. $\endgroup$
    – User8128
    Jun 4, 2019 at 18:29
  • $\begingroup$ @User8128 But all your expressions after the first equality are still dependent on $x$, aren't they? Also for the equality we have to choose the function $f = \alpha (i~1_{[0,x]}(t)-\frac{i}{2})$ so $f$ would also depend on both $x$ and $t$. Am I missing something? $\endgroup$ Jun 5, 2019 at 19:22

2 Answers 2


Let $k(t,x)=i\big(\mathbb{1}(0<t\leq x)-\frac12\big)$ and define the operator $A_k:f\mapsto\int^1_0k(t,x) f(t)\,dt$ in $L_2([0,1])$. As pointed out in the statement of the problem, $A_k$ is compact and self adjoint. Here is a short proof for completeness.

  • $A_k$ is compact in $L_2(0,1)$ because $\int_{[0,1]^2}|K(t,x)|^2\,dt\,dx =\frac14<\infty$.
  • To check self-adjointness of $A_K$, notice that $$A_Kf(x)=i\Big(\int^x_0f(t)\,dt-\frac12\int^1_0f(t)\,dt\Big)=\frac{i}{2}\Big(\int^x_0 f(t)\,dt-\int^1_x f(f)\,dt\Big)$$ while \begin{aligned} A^*_Kf(x)&=\int^1_0\overline{k(x,t)}f(t)\,dt=-i\Big(\int^1_xf(t)\,dt-\frac12\int^1_0f(t)\,dt\Big)\\ &=\frac{i}{2}\Big(\int^x_0f(t)\,dt-\int^1_xf(t)\,dt\Big)=A_kf(x) \end{aligned}

With all these, we have that the spectrum of $A_f$ consists of countable eigenvalues converging to $0$ (and possibly zero). The larges eigenvalue (in magnitude) is also the norm of $A_k$.

For each $n\in\mathbb{Z}$, the function $\phi_n(t)=e^{i(2 n + 1)t}$ is an eigenvector of $A_K$ corresponding to the eigenvalue $\frac{1}{(2n+1)\pi }$. At least this gives $\frac{1}{\pi}\leq\|A_K\|\leq 2$. One needs to check that $\{\frac{1}{(2 n+1)\pi}:n\in\mathbb{Z}\}$ are the only eigenvalues. Once this is verified, it turns out that $\|A_k\|=\frac{1}{\pi}$.

A side note: The functions $f_\alpha(t)=\sqrt{2\alpha+1}\,t^\alpha$ with $\alpha>-\frac12$, although they are not eigenfunctions, give an interesting bound: $\|A_K f_\alpha\|^2_2=\frac{2\alpha+1}{(\alpha+1)^2}\Big(\frac{1}{2\alpha+3}-\frac{1}{\alpha+2}+\frac14\Big)$. This attains a maximum at $\alpha=0.56807...$ and which gives a lower bound of $0.298225...$ for $\|A_k\|$. That is optimal as $\frac{1}{\pi}=0.3183099...$.


The eigenfunctions of the operator $A$ form an orthonormal system, therefore we can write: $$Af = \sum\limits_{k\in\mathbb{Z}} \lambda_k (f,e_k)e_k$$ Where $\lambda_k = \frac{1}{(2k+1)\pi}$ are the eigenvalues of $A$ with the corresponding eigenfunctions $e_k = e^{(2k+1)\pi i}$. Now we define $$c:=\max\limits_{k\in\mathbb{Z}}(\vert\lambda_k\vert)$$

$$\Vert Af\Vert^2 = \sum\limits_{k\in\mathbb{Z}} \vert \lambda_k (f,e_k) \vert^2\leq c^2\sum\limits_{k\in\mathbb{Z}} \vert(f,e_k) \vert^2=c^2 \Vert f \Vert^2$$

Hence, $\Vert A \Vert \leq c$.

For the other direction assume $f=e_0$, the eigenfunction which corresponds to the greatest eigenvalue $\lambda_0$.

$$\Vert Af \Vert^2=\Vert \lambda_0 f\Vert^2 = c^2$$

It follows that $\Vert A \Vert= c$. Where $c=\max\limits_{k\in\mathbb{Z}}\Big(\vert\frac{1}{(2k+1)\pi}\vert \Big)=\frac{1}{\pi}$.

  • $\begingroup$ The operator ist compact and self-afjoint. As I remarked in my question I have already shown this. $\endgroup$ Jun 27, 2019 at 5:23

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