Calculate the operator norm of $A: L^2[0,1] \to L^2[0,1]$ defined by $(Af)(x):=i\int_0^x f(t)\,dt-\frac{i}{2} \int_0^1 f(t) \,dt$

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 $\|A_k\|=\frac{1}{\pi}$.

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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 $\frac{1}{\pi}=0.3183099...$.