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The Baire category theorem states that a nonempty complete metric space cannot be a countable union of closed nowhere dense sets. The Uniform boundedness principle says that if $V$ is a Banach space, $W$ a normed vector space, and $\mathcal{T}\subseteq B(V,W)$ is pointwise bounded, meaning ${\sup }_{T\in \mathcal{T}}\Vert Tx\Vert <\infty$ for every $x\in V$, then ${\sup }_{T\in \mathcal{T}}\Vert T\Vert <\infty$.

To prove it, let $E_n=\{x:{\sup }_{T\in \mathcal{T}}\Vert Tx\Vert \le n\}$. Each set is closed by continuity of the individual operators, and their union is $V$. The Baire category theorem gives a ball $B(x_0,r)$ inside some $E_N$. For $\Vert h\Vert <r$, both $x_0+h$ and $x_0$ are in $E_N$, so $\Vert Th\Vert \le \Vert T(x_0+h)\Vert +\Vert T{x}_0\Vert \le 2N$ for every $T$. Scaling gives $\Vert T\Vert \le 2N/r$ uniformly. The zero-space and empty-family cases are immediate.

For the first family on a Hilbert space, fix $x$. The functionals $y\mapsto ⟨y,Tx⟩$ have operator norm $\Vert Tx\Vert$ by the Cauchy-Schwarz inequality and equality on the direction of $Tx$. Their values are pointwise bounded by the hypothesis, after taking complex conjugates if necessary. Apply the Uniform boundedness principle to obtain ${\sup }_{T\in \mathcal{E}}\Vert Tx\Vert <\infty$. Apply it a second time to the operator family itself to conclude ${\sup }_{T\in \mathcal{E}}\Vert T\Vert <\infty$.

For the diagonal hypothesis, the answer depends on the scalar field. Over $\mathbb{C}$, let $q_T(x)=⟨Tx,x⟩$, using the convention that the inner product is linear in its first variable. The complex polarization identity givesThe assumed bounds at these four fixed vectors make every matrix coefficient uniformly bounded, so the preceding argument applies. Thus the complex-space answer is yes, even without assuming self-adjointness.

Over $\mathbb{R}$, take $X={\mathbb{R}}^2$ andEvery diagonal value $⟨nJx,x⟩$ is zero, whereas $\Vert nJ\Vert =n$. Thus the real-space answer is no in general. Real diagonal values do not detect the skew-symmetric part of an operator. This is diagonal boundedness of operator families over real and complex Hilbert spaces.

With $\hat{f}(k)=(2\pi {)}^{-1}{\int }_{-\pi }^{\pi }f(s)e^{-iks}ds$, the classical Fourier partial sum is $s_N^f(t)={\sum }_{|k|\le N}\hat{f}(k)e^{ikt}$. Its Cesàro mean isFor the uniform Fejér theorem, extend the continuous endpoint-matching $f$ periodically. The given Fejér kernel is nonnegative, even and has total integral $\pi$. ThusGiven $\epsilon >0$, uniform continuity supplies $\delta >0$ so the bracket has modulus less than $\epsilon$ for $|u|<\delta$, uniformly in $t$. That part contributes at most $\epsilon$. On the remainder, $F_N(u)\le [2(N+1){\sin }^2(\delta /2){]}^{-1}$, and the bracket is at most $2\Vert f{\Vert }_{\infty }$. HenceLetting $\epsilon$ decrease to zero proves uniform convergence of the Fejér sums.

The later assertions attached to the printed factorial-frequency example are inconsistent. The following calculations resolve them while keeping the Fourier definition just given. Put $a_j=e^{i\sqrt{j+1}}-e^{i\sqrt{j}}$ and consider its formal trigonometric series. For $t=\pi p/q$, there is $J_0$ such that $\cos (j!t)=1$ for every $j\ge J_0$, because $2q$ divides $j!$. At $N_J=(J+1)!-1$, its ordinary partial sum therefore telescopes toThe exponential has subsequences tending to $1$ and $-1$: take $J+1$ to be integers nearest $(2\pi m{)}^2$ and $((2m+1)\pi {)}^2$. Thus the ordinary sums diverge at every $t\in \pi \mathbb{Q}$.

But their genuine Fourier Cesàro means also diverge there. Indeed,Since $|a_j|\le 2$ and ${\sum }_{j=0}^{J}j!\le 2J!$ for $J\ge 3$, its absolute value is at most $4/(J+1)\to 0$, uniformly in $t$. Thereforewhich has the same incompatible limit points. This is an instance of Cesàro sums at factorial cutoffs track ordinary sums. The printed Cesàro-convergence claim is false. Averaging the sparse series by its index $j$ would be a different operation from Fourier Cesàro summation by frequency.

There is a further issue: no $L^1$ function has exactly this Fourier series. Here is a direct proof using strongly lacunary polynomial L1-L2 bound. For a trigonometric polynomial $p$ supported on $\{\pm j!:j\ge 3\}$, consecutive positive frequencies have ratio at least four. Every zero sum of four signed frequencies pairs into cancelling equal frequencies: a largest unpaired frequency would exceed the sum of the other three. Expanding the fourth moment with normalized measure $dt/(2\pi )$ and bounding the three possible pairings givesThe Holder inequality gives $\Vert p{\Vert }_2\le \Vert p{\Vert }_1^{1/3}\Vert p{\Vert }_4^{2/3}$, hence $\Vert p{\Vert }_2\le \sqrt{3}\Vert p{\Vert }_1$.

If the asserted $f\in L^1$ existed, positivity and unit mass of the summation kernel would bound $\Vert {\sigma }_N^f{\Vert }_1\le \Vert f{\Vert }_1$. Subtract the finitely many modes from $j<3$; their norms stay uniformly bounded, and the remainder is a polynomial of the type above. Its $L^2$ norms would therefore be uniformly bounded. Orthogonality then forces ${\sum }_{j\ge 3}|a_j{|}^2<\infty$, by fixing any finite coefficient block and letting $N\to \infty$. However,so that sum diverges. This contradiction disproves the stated existence premise.

Even considered only formally, these coefficients rule out an $L^2$ function by Bessel's inequality, and hence rule out a bounded function on the finite interval. Thus a bounded continuous representative cannot satisfy the asserted series either. The asserted $L^1$ function does not exist.