2.6

Part a.) Define:

$g_n(x) = \Bigg\lbrace \begin{array}{cc} (n^2 2^n x + n) & x \in [-\frac{1}{n2^n},0) & \\ (-n^2 2^n x + n) & x \in [0,\frac{1}{n2^n}] \\ \end{array}$

Certainly, $\int_{\mathbb{R}} g_n = \frac{1}{2^n}$, for every $n$. Now, just define: $$f_m(x) = \sum_{n=1}^m g_n(x - n)$$ Now we can define $f(x) = \lim_{m \to \infty} f_m(x)$. Since each $f_m$ is positive, and $f_m \nearrow f$, monotonicly, we have --by the Monotone Convergence Theorem that $f$ must be measurable, and: $$ \lim_{m \to \infty} \int_{\mathbb{R}} f_m(x) = \int_{\mathbb{R}} f $$ By construction, we can directly calculate $\int_{\mathbb{R}} f$ = $\sum_{n=1}^\infty \frac{1}{2^n} = 1$. Thus, $f$ is $L^1$. Lastly, observe that the $\limsup_{x \to \infty} f(x) = \infty$, since $f(n) = n$ for all $n \in \mathbb{N}$.

Part b.) Assume to the contrary that there exists a function $f$ such that $f$ is $L^1$, uniformly continuous on $\mathbb{R}$, and $f(x) \nrightarrow 0$ as $x \to \infty$.

($\dagger$) Certainly, $\limsup_{x \to \infty} |f(x)| = \alpha > 0$.

Since $f$ is uniformly continuous, we know $f$ must be finite on any compact set. Define the sequence $\lbrace x_n \rbrace_{n=1}^\infty$ such that $x_n \in [n, n+1]$ and $|f(x_n)| = \max_{x \in [n, n+1]}{(|f(x)|)}$.

By $\dagger$, we know $\exists N \in \mathbb{N}$ such that $\forall n \geq N$ $\hspace{0.25cm} |f(x_n)| > \frac{\alpha}{2}$. Fix a particular $\epsilon$ such that $0 < \epsilon < \frac{\alpha}{4}$. Since $f$ is uniformly continuous, we know that for this particular $\epsilon$, $\exists \delta > 0$ such that $|x - y| < \delta \hspace{0.25cm} \Rightarrow |f(x) - f(y)| < \epsilon$.

Now, consider again our sequence $\lbrace x_n \rbrace_{n=1}^\infty$. For our particular epsilon, redefine the $\delta$ determined by $\epsilon$ such that $\delta = \min\lbrace\delta(\epsilon), \frac{1}{2} \rbrace$.

We have, by the reverse triangle inequality, that for every $n \geq N$: $$\big||f(x_n)| - |f(x_n + \delta)|\big| \leq |f(x_n) - f(x_n + \delta)| \leq \epsilon < \frac{\alpha}{4}$$ Define the sets: $$E_n = \lbrace x \hspace{0.25cm} | \hspace{0.25cm} x_{2n} \leq x \leq x_{2n} + \delta \rbrace$$ I skipped every other $x_n$ to ensure $\lbrace E_n \rbrace$ will be pair-wise disjoint.

...and now we can finally observe that: $$\int_\mathbb{R} |f| \geq \int_{(\bigcup_{n=N}^\infty E_n)} \frac{\alpha}{4} = \sum_{n = N}^\infty \frac{\delta\alpha}{4} = \infty $$ ...contradicting the assumption that $f \in L^1$.