The Baire Principles of Functional Analysis — Examples

The second alternative of Baire dichotomy for a pointwise-defined family of bounded linear operators applies directly.

Therefore every nonempty open ball contains a singular vector, which is the condensation assertion.

Every $f_n$ is bounded and $\mathrm{\parallel }{f}_n\mathrm{\parallel }=n$ by testing the $n$th unit vector.

For fixed finitely supported $x$, $f_n(x)=0$ for all sufficiently large $n$, so ${\sup }_n\mathrm{\mid }{f}_n(x)\mathrm{\mid }<\mathrm{\infty }$.

The partial sums of $(2^{-k}{)}_{k\ge 1}$ lie in $c_{00}$ and are Cauchy in the sup norm but converge in its completion to a non-finitely-supported sequence. Thus the domain is incomplete and steps 1.1--1.2 refute the claim.

Since $\mathrm{\parallel }x{\mathrm{\parallel }}_{\mathrm{\infty }}\le \mathrm{\parallel }x{\mathrm{\parallel }}_1$, $I$ is bounded and bijective.

Its inverse is unbounded: for $x^{(m)}=(1\mathrm{/}m,\dots ,1\mathrm{/}m,0,\dots )$, $\mathrm{\parallel }{x}^{(m)}{\mathrm{\parallel }}_{\mathrm{\infty }}=1\mathrm{/}m$ but $\mathrm{\parallel }{x}^{(m)}{\mathrm{\parallel }}_1=1$.

The target is incomplete by Uniform boundedness fails on the incomplete space c_00. The partial sums of $(2^{-k}{)}_{k\ge 1}$ are also Cauchy in the ${\mathrm{\ell }}^1$ norm but have no limit in $c_{00}$, so the domain is incomplete as well.

If $I$ were open, its inverse would be continuous at $0$, hence bounded by linearity, contradicting step 1.2.

The equal-coordinate vectors from A bounded bijection of incomplete normed spaces need not be open show that $I$ is unbounded.

If $x_k\to x$ in sup norm and $x_k\to y$ in ${\mathrm{\ell }}^1$ norm, then every coordinate converges to both $x_j$ and $y_j$, hence $x=y$.

Thus the graph is closed in the product norm (The graph of a linear operator with a linear domain), while step 1.1 shows the operator is unbounded.

Newton--Leibniz Newton–Leibniz needs only continuity on $[a,b]$, differentiability on $(a,b)$, and a Riemann-integrable extension of the interior derivative gives $f_n(t)-f_n(0)={\int }_0^t{f}_n^{\mathrm{\prime }}(s)ds$. Passing to uniform limits yields $f(t)-f(0)={\int }_0^{t}g(s)ds$.

Since $g$ is continuous, the integral function is differentiable with derivative $g$; hence $f\in C^1$ and $Df=g$. The graph (The graph of a linear operator with a linear domain) is closed.

For integers $n\ge 1$, put $f_n(t)=\sin (nt)\mathrm{/}n$. Then $\mathrm{\parallel }{f}_n{\mathrm{\parallel }}_{\mathrm{\infty }}\le 1\mathrm{/}n$ whereas $\mathrm{\parallel }D{f}_n{\mathrm{\parallel }}_{\mathrm{\infty }}=1$, so $D$ is unbounded.