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.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s