Failure for a sequence to tend to the delta function does not imply the existence of a continuous function whose functional value tends to infinity.

Question

In the book Functional Analysis by Peter D. Lax the following definition is given.

A sequence $\{k_n\}$ of continuous functions on $[-1,1]$ tends to the $\delta$ function if $$lim_{n\rightarrow \infty}\int_{-1}^{1}f(t)k_n(t)dt = f(0)$$ for all continuous functions on $[-1,1]$.

Next a theorem is given which says:

A sequence $\{k_n\}$ over $C([-1,1])$ tends to the $\delta$ function if and only if it satisfies the following

1. $\lim_{n\rightarrow \infty}\int_{-1}^{1}k_n(t)dt = 1$.
2. For every $g\in C^{\infty}$ whose support does not include an interval about $0$ we have $\lim_{n\rightarrow \infty}\int_{-1}^{1}g(t)k_n(t)dt = 0$
3. There is a constant $c$ for which $\int_{-1}^{1}|k_n(t)|dt\leq c$

I understand the proof which is given in the book but my question concerns a corollary which states that if 3) is violated then there exists a continuous function $f$ for which

$$\lim_{n\rightarrow \infty}\int_{-1}^{1}f(t)k_n(t)dt = \infty.$$

I think this is false. Since $\sup_{\|f\|=1}\left|\int_{-1}^{1}f(t)k_n(t)dt\right|=\int_{-1}^{1}|k_n(t)|dt$ (this I know how to show) boundedness of the set

$$\left\{\int_{-1}^{1}f(t)k_n(t)dt\mid f\in C([-1,1])\right\}$$

for each fix $n$ would imply, by the principle of uniform boundedness, that $\int_{-1}^{1}|k_n(t)|dt$ would be bounded for every $n$. However even if this set is not bounded this does not imply that there exists some continuous function $f$ such that $\lim_{n\rightarrow \infty}\int_{-1}^{1}f(t)k_n(t)dt = \infty$ in fact I think I have a counterexample.

Say that $\{h_n\}$ is some sequence which tends to the $\delta$ function and therefore satisfies 1), 2) and 3) above (I guess that the Fejèr kernels would work). Next let $\{q_n\}$ be a sequence which satisfies 1) and 2) but not three, that is $\{q_n\}$ satisfies that $\forall M>0$ we can find $n\in \mathbf{N}$ such that $$\int_{-1}^{1}|q_n(t)|dt>M.$$

(I guess the Dirichlet kernels would work here).

Now let $\{k_n\}$ be the sequence obtained by taking every other element to be $h_n$ and every other element to be $q_n$ that is

$$k_n=\begin{cases} h_{n/2}&\text{if $n$ is even} \\ q_{(n+1)/2}&\text{otherwise} \end{cases}.$$

It is direct that $\{k_n\}$ satisfies 1) and 2) but not 3). Clearly if $f\in C([-1,1])$ then

$$\lim_{n\rightarrow \infty}\int_{-1}^{1}f(t)k_{2n}(t)dt = \lim_{n\rightarrow \infty}\int_{-1}^{1}f(t)h_{n}(t)dt = f(0)\neq \infty$$

and therefore the sequence can not tend to infinity. This means that even if a sequence does not satisfy 3) but satisfies 1) and 2) there does not have to exist some function $f$ such that $\lim \int fk_n dt = \infty$.

Does the author mean that there exists some continuous function such that the sequence in question diverges for this is obviously true?