# Continuity of sesquilinear form on space of complex polynomials

I need help with the following:

Prove that sesquilinear form $$\Phi(p,q) = \int_0^1 p(x)\overline{q(x)}dx$$ defined on space of complex polynomials on $$[0,1]$$, with norm $$\|p\| = \int_0^1 |p(x)|dx$$, is not continuous, but is continuous for each variable individually.

I want to prove that, when $$\lim_{n \to \infty} p_n = p$$ and $$\lim_{n \to \infty} q_n = q$$, then $$\lim_{n \to \infty} \Phi(p_n,q) = \Phi(p,q)$$, $$\lim_{n \to \infty} \Phi(p,q_n) = \Phi(p,q)$$, but $$\lim_{n \to \infty} \Phi(p_n,q_n)$$ need not be $$\Phi(p,q)$$. Since $$|\Phi(p_n,q)-\Phi(p,q)| \leq \int_0^1 |p_n(x)-p(x)|q(x)dx \leq \|p_n-p\| \int_0^1 \overline{q(x)}dx ,$$ $$|\Phi(p,q_n)-\Phi(p,q)| \leq \int_0^1 p(x)|\overline{q_n(x)}-\overline{q(x)}|dx \leq \|\overline{q_n}-\overline{q}\|\int_0^1 p(x)dx,$$ $$\lim_{n \to \infty} \Phi(p_n,q) = \Phi(p,q)$$ and $$\lim_{n \to \infty} \Phi(p,q_n) = \Phi(p,q)$$. I don't know how to prove the third part, that $$\lim_{n \to \infty} \Phi(p_n,q_n)$$ is not $$\Phi(p,q)$$.

• It can happen that the third limit is actually correct, e.g. when the sequence is stationary. Therefore you have to find a counterexample. Hint: try $p_n = q_n$ Apr 8 at 15:11

You made a small mistake in your computation that $$\Phi$$ is continuous in either of the arguments --- but you're on the right track! One starts as you did by computing $$|\Phi(p_n,q)-\Phi(p,q)|=\Big|\int_0^1(p_n(x)-p(x))q(x)\,dx\Big|\leq\int_0^1|p_n(x)-p(x)|\,|q(x)|\,dx\,.\tag{1}$$ What we know is that $$p_n\to p$$ as $$n\to\infty$$, that is, $$\lim_{n\to\infty}\int_0^1|p_n(x)-p(x)|\,dx=0$$. This is almost the expression in (1) if it were not for $$q$$. Fortunately, as a polynomial $$q$$ is continuous meaning it is bounded on any bounded interval. This yields \begin{align*} |\Phi(p_n,q)-\Phi(p,q)|&\leq\int_0^1|p_n(x)-p(x)|\,|q(x)|\,dx\\ &\leq\int_0^1|p_n(x)-p(x)|\,\big(\max_{x'\in[0,1]}|q(x')|\big)\,dx\\ &=\underbrace{\big(\max_{x'\in[0,1]}|q(x')|\big)}_{<\infty}\;\underbrace{\int_0^1|p_n(x)-p(x)|\,dx}_{=\|p_n-p\|\to 0}\to 0 \end{align*} as $$n\to\infty$$. One argues similarly for the second component. Side note: the key estimate here is a special case of Hölder's inequality which would also deal with the domain of $$\Phi$$ being equipped with any other $$p$$-norm.
For discontinuity of $$\Phi$$ the hint @blamethelag gave in their comment is a good starting point. Just plugging this in suggests that if we can find polynomials $$(p_n)_{n\in\mathbb N}$$, $$p$$ such that $$\|p_n-p\|\to 0$$ but $$\Phi(p_n,p_n)=\int_0^1|p_n(x)|^2\,dx\not\to\int_0^1|p(x)|^2=\Phi(p,p)$$ then we are done. Also from the proof given above we know that any such counterexample has to satisfy $$\max_{x'\in[0,1]}|p_n(x')-p(x')|\not\to 0$$ as $$n\to\infty$$ (because else Hölder's inequality would show that $$\Phi(p_n,p_n)$$ converges to $$\Phi(p,p)$$). One of the standard counterexamples in this context is the sequence of polynomials $$p_n(x):=x^n$$ with limit $$p(x):=0$$. However, a quick computation shows this does not quite work yet, but maybe it does after tweaking things a bit.
Let us define $$p_n(x):=a_nx^n$$ where, for now, $$(a_n)_{n\in\mathbb N}$$ is an arbitrary sequence of numbers. By plugging this in we want to see what the conditions on $$a_n$$ are to make for a valid counterexample. In other words we need \begin{align*} 0=\lim_{n\to\infty}\|p_n-p\|&=\lim_{n\to\infty}\int_0^1|a_n x^n-0|\,dx\\ &=\lim_{n\to\infty}|a_n|\int_0^1 x^n\,dx=\lim_{n\to\infty}|a_n|\Big[\frac1{x^{n+1}}\Big]_0^1=\lim_{n\to\infty} \frac{|a_n|}{n+1} \end{align*} but \begin{align*} 0\neq\lim_{n\to\infty}|\Phi(p_n,p_n),\Phi(p,p)|&=\lim_{n\to\infty}\Big|\int_0^1|a_nx^n|^2\,dx\Big|\\ &=\lim_{n\to\infty}|a_n|^2\int_0^1x^{2n}\,dx=\lim_{n\to\infty}\frac{|a_n|^2}{2n+1}\,. \end{align*} Thus if we can find numbers which
grow slower than $$n+1$$ but not faster than $$2n+1$$ we are done. The simplest choice here is probably $$a_n=\sqrt{n}$$, that is, $$p_n(x):=\sqrt nx^n$$ converges to the zero function in $$\|\cdot\|$$, but $$\Phi(p_n,p_n)$$ does not converge to zero $$=\Phi(p,p)$$.