## Does the function need to be Lipschitz or is Lebesgue integrable enough?

Let $$f:(0,1) to mathbb{R}$$, $$mathbb{Z}_n ={0,1,dots,n-1}$$ and $$delta >0$$.

Which is the least I can assume about $$f$$ so that

$$begin{equation*} begin{split} left| frac{1}{n} sum_{ x in mathbb{Z}_n} f(tfrac{x}{n}) – int_{(0,1)}f(u) du right| leq delta end{split} end{equation*}$$
for sufficiently large $$n$$?

I know for a fact that if $$f$$ is Lipschitz continuous with constant $$K$$, then the inequality holds as

$$begin{equation*} begin{split} left| frac{1}{n} sum_{ x in mathbb{Z}_n} f(tfrac{x}{n}) – int_{(0,1)}f(u) du right| & = left| frac{1}{n} sum_{ x in mathbb{Z}_n} f(tfrac{x}{n}) – sum_{x in mathbb{Z}_n}int_{tfrac{x}{n}}^{tfrac{x+1}{n}}f(u) du right| \ & = left| frac{1}{n} sum_{ x in mathbb{Z}_n} int_{tfrac{x}{n}}^{tfrac{x+1}{n}} (f(tfrac{x}{n}) – f(u) ) du right|\ & leq sum_{ x in mathbb{Z}_n} int_{tfrac{x}{n}}^{tfrac{x+1}{n}} left|f(tfrac{x}{n}) – f(u)right| du \ & leq sum_{ x in mathbb{Z}_n} int_{tfrac{x}{n}}^{tfrac{x+1}{n}} K left| tfrac{x}{n}- u right| du leq K sum_{ x in mathbb{Z}_n} int_{tfrac{x}{n}}^{tfrac{x+1}{n}} tfrac{1}{n} du\ & = K sum_{ x in mathbb{Z}_n} tfrac{1}{n^2}= frac{K}{n}. end{split} end{equation*}$$

So I guess my question is: Can I assume some (relevant) weaker condition about $$f$$ so that the inequality holds? In particular, is the inequality true if $$f$$ is just integrable?

## analysis – How do I check if the function is integrable (Lebesgue) on R?

Is f(x) = $$sum_{n=0}^{infty}((-1)^{n}/n) X_{(n,n+1)}$$ Lebesgue integrable on R? X is the characteristic function.

I’m confused about how to handle the sum and the alternating series part. The sum is clearly conditionally convergent. But it might not be non-negative. Is it still integrable over R? (Lebesgue).

New here, sorry for the crap formatting.

## partial differential equations – If \$S(t)\$ is a \$C_0\$-semigroup, is \$S(t-s)f(s)\$ Bochner integrable?

Let $$X$$ be a Banach space and let $$S(t)$$, $$t geq 0$$, be a $$C_0$$-semigroup on $$X$$.

Assume that $$f : (0,+infty) rightarrow X$$ is Bochner integrable.

Is $$S(t-s)f(s)$$ Bochner integrable on $$(0,t)$$ and does $$t mapsto int_0^t S(t-s)f(s)ds in C^0((0,+infty),X)$$ ?

The function $$t mapsto int_0^t S(t-s)f(s)ds$$ arises when we define the notion of weak solution to an inhomogeneous evolution PDE $$partial_t u(t) = Au(t) + f(t), quad u(0) = u_0$$

where $$A$$ is the infinitesimal generator of $$S(t)$$.

If $$f$$ is continuous, I know that the result is true, but I’m interested in the non-continuous case. I would expect this to be true.

Any proof or reference is welcomed.

## real analysis – Interval of \$p\$ that makes the \$p\$ power of a function integrable

For a measurable real-valued function $$f$$ on $$(0,infty)$$ let $$P(f):={pin(0,infty):|f|^pinmathcal{L}^1(mathbb{R},mathcal{B},lambda)}$$, where $$mathcal{B}$$ is the Borel $$sigma$$-algebra. Show that for every subinterval $$J$$ of $$(0,infty)$$, which may be open or closed on the left and on the right, there is some $$f$$ such that $$P(f)=J$$. Hint: Consider functions such as $$x^a|log x|^b$$ on $$(0,1)$$ and on $$(1,infty)$$, where $$b=0$$ unless $$a=-1$$.

I don’t understand the hint. $$f(x)=x^anotinmathcal{L}^1(mathbb{R},mathcal{B},lambda)$$ for any $$a$$. The log part is also not integrable:

$$begin{equation} begin{split} & int_0^infty x^{-1}|log x|^bdx \ = & int_0^infty |log x|^bdlog x \ = & int_0^1 (-log x)^bdlog x+int_1^infty (log x)^bdlog x \ = & int_{-infty}^0 (-x)^bdx+int_0^infty x^bdx \ = & -int_{-infty}^0 (-x)^bd(-x)+int_0^infty x^bdx \ = & -int_infty^0 x^bdx+int_0^infty x^bdx \ = & int_0^infty x^bdx+int_0^infty x^bdx \ end{split} end{equation}$$

But for $$J=(0,infty)$$, I can find such a $$f(x)$$:
$$begin{equation} f(x) = begin{cases} x-1 &text{1

## real analysis – A necessary and sufficient condition of Lebesgue integrable.

Let $$lgeqq 2$$ be natural number.
And let $$f$$ be a Lebesgue-measurable function and suppose $$f$$ is non negative.

Prove that $$f$$ is Lebesgue-integrable if and only if $$displaystylesum_{n=-infty}^{infty} l^n cdot m(A_n)$$ converges where $$m$$ is Lebesgue measure and $$A_n={xin mathbb{R^n} | l^n leqq f(x)

I’m not sure my proof is correct.

My proof

First, suppose $$displaystylesum_{n=-infty}^{infty} l^n cdot m(A_n)$$ converges.

Let $$A=displaystylebigcup_{n=-infty}^{infty} A_n.$$

Note that $$displaystyleint_{bigcup_{n=-infty}^{infty} A_n} f(x) dx=sum_{n=-infty}^{infty} int_{A_n} f(x) dx$$ since $$A_jcap A_k=emptyset$$ for $$jneq k$$ and $$A^c=emptyset$$ since $$A={ xin mathbb{R^n} | f(x)geqq 0}.$$

Then,
begin{align} displaystyleint |f(x)| dx&=displaystyleint f(x) dx\ &=int_{A} f(x) dx +int_{A^c} f(x) dx\ &=int_{bigcup_{n=-infty}^{infty} A_n} f(x) dx +int_{emptyset} f(x) dx\ &=sum_{n=-infty}^{infty} int_{A_n} f(x) dx +0\ &
Thus $$f$$ is Lebesgue-integrable.

Conversely, suppose $$f$$ is Lebesgue-integrable.

Then,
begin{align} sum_{n=-infty}^{infty} l^{n} cdot m(A_n) &=sum_{n=-infty}^{infty} l^{n} displaystyleint_{A_n} dx \ &=sum_{n=-infty}^{infty} displaystyleint_{A_n} l^{n} dx \ &leqq sum_{n=-infty}^{infty} displaystyleint_{A_n} f(x) dx \ &=displaystyleint_{bigcup_{n=-infty}^{infty} A_n} f(x) dx \ &=displaystyleint_{A} f(x) dx \ &=displaystyleint_{A} f(x) dx +displaystyleint_{emptyset} f(x) dx \ &=displaystyleint_{A} f(x) dx +displaystyleint_{A^c} f(x) dx \ &=displaystyleint f(x) dx \ &=displaystyleint |f(x)| dx \ &

Is this proof correct?

## real analysis – Prove that if \$ f \$ is Riemann Integrable then it is bounded

I get the idea of how to do this proof, essentially I would choose a pathological partition $$P$$ such that it includes a point $$c in P$$ where $$f(c) Delta_c – l geq epsilon$$. However I’m stuck on how to make this work in a $$delta,epsilon$$ proof.

Suppose the contrary that $$f$$ is Riemann Integrable but $$f$$ is not bounded. It suffices I show

$$exists epsilon > 0, forall delta > 0 text{s.t} exists P, m(P) < delta mathrm{but} |R(f,P) – l| geq epsilon$$

• $$m(P)$$ is the mesh of the partition $$P$$

Since $$f$$ is not bounded, choose $$c$$ such that $$f(c) > frac{l + epsilon}{delta}$$ and let $$c in P$$

Then I initially I had this:

$$|R(f,P) – l| = left|sum_{{s_j} in P} f(s_j) Delta_j – lright|$$
$$geq left|sum_{{s_j} in P} f(s_j) delta – lright|$$
$$geq left|f(c) delta – lright|$$
$$geq left|frac{l + epsilon}{delta} delta – lright|$$
$$= epsilon$$

However, clearly the inequality here is nowhere near correct, since the difference is an absolute value. How would I go about this proof?

(I’m following the Stephen Krantz, Real Analysis and Foundations Textbook)

## measure theory – Bochner integrability of composition of continuous and Bochner integrable function

Let $$(varOmega, varSigma ,mu)$$ be a finite measure space and $$X$$ be a separable Banach space. For $$1 leq p leq infty$$, let $$L_{p}(mu,X)$$ denote the class of all $$mu$$-Bochner integrable functions $$f$$ such that $$|f|_{p} = (int_{varOmega} |f|^{p})^{1/p} < infty$$.

If $$Phi in L_{p}(mu,X)$$ and $$h: X rightarrow X$$ is a continuous function, then is $$h circ Phi in L_{p}(mu,X)$$?

## calculus – Determine if the function \$f(x)\$ is integrable in \$[0,1]\$

Determine if the function $$f(x)$$ is integrable in $$(0,1)$$

$$displaystyle f( x) =begin{cases} frac{1}{x^{2}} & x >0\ 0 & x=0 end{cases}$$

Attempt:

Let’s check the limits of each side $$0^+,0^-$$ to show that $$frac{1}{x^2}$$ is not bounded.

$$displaystyle lim _{xrightarrow 0^{+}}frac{1}{x^{2}} =infty$$

$$displaystyle lim _{xrightarrow 0^{-}}frac{1}{x^{2}} =infty$$

Therefore, the function is not bounded when $$lim_{xrightarrow 0^{+/-}}$$
, and is not integrable in $$(0,1)$$

In addition, there is a well-known theorem:

If $$f$$ is a continuous function except for the finite number of unconsciously points $$Rightarrow$$ $$f$$ is integrable.

## real analysis – Proving f Riemann integrable implies |f| Riemann integrable by contradiction.

I know that there exists a theorem telling us that if a real valued function $$f$$ defined over an interval $$(a,b)$$ is Riemann integrable, then |$$f$$| is too (moreover I believe the converse to be false). I’ve tried to look at the proofs, but they seem a bit beyond my ability at the moment. My first thought was to suppose there existed a Riemann integrable function, let us say $$g$$, such that |$$g$$| was not Riemann integrable, and show that this leads to some sort of contradiction.

I thought about this for a while and have not been able to think of such a function/proof. Would anybody know of such an example?

## real analysis – prove \$log x\$ is integrable near the origin

Prove that $$log(x) in L^1((0,1))$$.which has been shown in this post.

The first idea is using the integral $$int_epsilon^1 log(x) dx$$ to approximate it.which may be the definition in classical analysis.I was bit confused for why this limit: $$lim_{epsilon to 0} I(epsilon) = int_0^1 log x dx$$ holds in Lebesgue integral. Maybe we need to take an absolute value first $$int_epsilon^1 |log(x)|dx$$ first then using the Monotone convergence theorem for non-negative function, rest of the proof are the similar ,is my interpretation correct?

The second idea also shows in this post:using the fact that $$x^{-alpha} in L^1((0,1))$$ for $$alpha <1$$ then near the origin ,we have $$x^{alpha}log x to 0$$ for all $$alpha >0$$ which means exist a small neiborhood near origin says $$(0,delta)$$ such that if $$xin (0,delta)$$ we have $$|x^alog x|le 1/2$$ i.e. $$|log x|le frac{1}{2}x^{-a}$$ since the RHS is integrable hence log is also integrable near origin,is my interpretation correct?