# The Volterra Operator

Here’s a proof that the Volterra operator is not normal which I thought was clever: Let ${T:L^2([0,1])\rightarrow L^2([0,1])}$ be defined by

$\displaystyle Tf(x) = \int_0^x f(t)\ \mathsf dt.$

If ${\|f\|_2=1}$ then

\displaystyle \begin{aligned} \|Tf\|_2^2 &= \int_0^1 \left|\int_0^x f(t)\ \mathsf dt\right|^2 \mathsf dx\\ &\leqslant \int_0^1 \int_0^x |f(t)|^2\ \mathsf dt\ \mathsf dx\\ &\leqslant \int_0^1 \int_0^1 |f(t)|^2\ \mathsf dt\ \mathsf dx\\ &= \int_0^1 \|f\|_2 \ \mathsf dx\\ &= \|f\|_2, \end{aligned}

so ${T}$ is a bounded operator. It follows that there exists a unique adjoint operator ${T^*:L^2([0,1])\rightarrow L^2([0,1])}$ satisfying ${\langle Tf, g\rangle = \langle f, T^*g\rangle}$ for all ${f,g\in L^2([0,1])}$. Since ${[0,1]}$ has finite Lebesgue measure, ${L^2([0,1])\subset L^1([0,1])}$, and thus

$\displaystyle \int\limits_{[0,1]^2}|f(t)g(x)|\ \mathsf d(x\times t)\leqslant \int\limits_{[0,1]^2}\|f\|_1\|g\|_1\ \mathsf d(x\times t)=\|f\|_1\|g\|_1<\infty.$

It follows that ${fg\in L^2([0,1]^2)}$ and hence by Fubini’s theorem,

\displaystyle \begin{aligned} \langle Tf, g\rangle &= \int_0^1 \int_0^x f(t) \mathsf dt\ g(x) \mathsf dx\\ &=\int\limits_{[0,1]\times[0,x]}f(t)g(x)\ \mathsf d(x\times t)\\ &=\int\limits_{[t,1]\times[0,1]}f(t)g(x)\ \mathsf d(x\times t)\\ &= \int_0^1\int_t^1 f(t)g(x)\ \mathsf dx \mathsf dt\\ &= \int_0^1 f(t)\left(\int_t^1 g(x) \mathsf dx\right) \mathsf dt\\ &= \left\langle f, \int_t^1 g(x) \mathsf dx\right\rangle, \end{aligned}

from which we see that ${T^*f(x) = \int_x^1 f(t)\ \mathsf dt}$.

If ${f\in L^2([0,1])}$ with ${\|f\|_2=1}$, then for any ${x,y\in[0,1]}$, Hölder’s inequality yields

\displaystyle \begin{aligned} |Tf(x)-Tf(y)| &= \left|\int_y^x f(t)\ \mathsf dt \right|\\ &\leqslant \int_y^x 1\cdot|f(t)|\ \mathsf dt\\ &\leqslant \left(\int_y^x 1\ \mathsf dt\right)^{\frac12}\left(\int_y^x |f(t)|^2\ \mathsf dt\right)^{\frac12}\\ &\leqslant |x-y|^{\frac12}\left(\int_0^1 |f(t)|^2\ \mathsf dt\right)^{\frac12}\\ &= \|f\|_2|x-y|^{\frac12}, \end{aligned}

so that ${Tf}$ is Hölder continuous. Moreover, the kernel of ${T}$ is ${K(x,t)=\chi_{(0,x)}(t)}$ and

$\displaystyle \int\limits_{[0,1]^2} |K(x,t)|^2\ \mathsf d(x\times t) = \int_0^1\int_0^x \ \mathsf dt\ \mathsf dx=\frac12<\infty,$

so ${K\in L^2([0,1]^2)}$. It follows that ${T}$ is a Hilbert-Schmidt operator and therefore ${T}$ is compact. It follows then from the Fredholm alternative (see \texttt{http://bit.ly/1Uh8BgP}) that if ${\lambda\in\sigma(T)}$, ${\lambda\ne 0}$ then ${\lambda}$ is an eigenvalue of ${T}$. That is, ${Tf(x)=\lambda f(x)}$ for all ${x\in[0,1]}$. If ${Tf=\lambda f}$, then as ${Tf\in C^0([0,1])}$ we must have ${f\in C^0([0,1])}$. By the fundamental theorem of calculus, ${Tf\in C^1([0,1])}$, and so ${f\in C^1([0,1])}$. Differentiating both sides of the eigenvalue equation yields ${f(x)=\lambda f'(x)}$, and hence ${f(x) = Ce^{\frac1\lambda x}}$ for some ${C\in\mathbb R}$. Since ${Tf(0)=0}$, we have ${f(0)=0}$, and therefore ${C=0}$. It follows that the spectral radius of ${T}$, ${r(T)}$, is zero.

If ${A}$ is a bounded operator on a Hilbert space and ${\|f\|=1}$ then

$\displaystyle \|A^*f\|^2 = \langle A^*f, A^*f\rangle = \langle AA^*f, f\rangle\leqslant \|AA^*f\|$

so

$\displaystyle \|A^*\|^2 \leqslant \|AA^*\|\leqslant \|A\|\|A^*\|$

and hence ${\|A^*\|\leqslant\|A\|}$. By symmetry, ${\|A\|\leqslant\|A^*\|}$, whence ${\|A^*\|=\|A\|}$. It follows that ${\|AA^*\|=\|A\|^2=\|A^*A\|}$. Recall Gelfand’s formula for the spectral radius of a bounded operator ${A}$:

$\displaystyle r(A) = \lim_{n\rightarrow\infty} \|A^n\|^{\frac1n},$

If ${A}$ is a self-adjoint operator and ${\|f\|=1}$ then

$\displaystyle \|Af\|^2 = \langle Af, Af\rangle = \langle A^2f, f\rangle\leqslant\|A^2f\|\|f\|=\|A^2f\|$

which implies ${\|A^2\|=\|A\|^2}$. By induction it follows that ${\|A^{2^n}\|=\|A\|^{2^n}}$ for all ${n}$ and hence

$\displaystyle r(A) = \lim_{n\rightarrow\infty} \|A^{2^n}\|^{\frac1{2^n}} = \lim_{n\rightarrow\infty}\|A\|=\|A\|.$

If ${A}$ is normal, then by induction ${\|(A^*A)^n\|=\|A^n\|^2}$, and as ${A^*A}$ is self-adjoint, ${r(A^*A)=r(A)^2=\|A\|^2}$, from which we see that the spectral radius of a normal operator is equal to its operator norm.

Since the spectral radius of the Volterra operator ${T}$ is zero and its operator norm is positive, we conclude that ${T}$ is not a normal operator.